Sequence and Series
Engineering Mathematics - I
2
1.1 Sequence
A function f:N → S, where S is any nonempty set is called a Sequence
i.e., for each n ∈ N, ∃ a unique element f(n) ∈ S. The sequence is written as f(1), f(2),
f(3), ......f(n)...., and is denoted by {f(n)}, or <f(n)>, or (f(n)). If f(n) = an , the sequence is
written as a1 , a2 .....an and denoted by , {an } or < an > or ( an ) . Here f(n) or an are the
nth terms of the Sequence.
Ex. 1.
1 , 4 , 9 , 16 ,......... n 2 ,.....(or) < n 2 >
N
S
2
3
1
4
9
n
n2
.
.
.
.
Ex. 2.
1 1 1
1
⎛ 1 ⎞
, 3 , 3 ,..... 3 ....(or ) ⎜ 3 ⎟
3
1 2 3
n
⎝n ⎠
Ex. 3.
1, 1, 1......1..... or <1>
.
.
.
.
Sequences and Series
Ex 4:
3
( −1)
1 , –1, 1, –1, ......... or
n −1
Note : 1. If S ⊆ R then the sequence is called a real sequence.
2. The range of a sequence is almost a countable set.
1.1.1 Kinds of Sequences
1. Finite Sequence: A sequence < an > in which an = 0 ∀n > m ∈ N is said to
be a finite Sequence. i.e., A finite Sequence has a finite number of terms.
2. Infinite Sequence: A sequence, which is not finite, is an infinite sequence.
1.1.2 Bounds of a Sequence and Bounded Sequence
1. If ∃ a number ‘M’ ∋ an ≤ M, ∀n ∈ N, the Sequence < an > is said to be
bounded above or bounded on the right.
1 1
Ex. 1, , , ,....... here an ≤ 1 ∀n ∈ N
2 3
2. If ∃ a number ‘m’ ∋ an ≥ m, ∀n ∈ N, the sequence < an > is said to be
bounded below or bounded on the left.
Ex. 1 , 2 , 3 ,.....here an ≥ 1 ∀n ∈ N
3. A sequence which is bounded above and below is said to be bounded.
1⎞
n⎛
Ex. Let an = ( −1) ⎜ 1 + ⎟
⎝ n⎠
n
1
2
3
4
......
an
-2
3/2
-4/3
5/4
......
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From the above figure (see also table) it can be seen that m = –2 and M =
∴ The sequence is bounded.
3
.
2
1.1.3 Limits of a Sequence
A Sequence < an > is said to tend to limit ‘l’ when, given any + ve number ' ∈ ',
however small, we can always find an integer ‘m’ such that an − l <∈, ∀n ≥ m , and we
write Lt an = l or an → l
n →∞
Ex.
If an =
1
n2 + 1
then < an >→ .
2
2
2n + 3
1.1.4 Convergent, Divergent and Oscillatory Sequences
1. Convergent Sequence: A sequence which tends to a finite limit, say ‘l’ is
called a Convergent Sequence. We say that the sequence converges to ‘l’
2. Divergent Sequence: A sequence which tends to ±∞ is said to be Divergent
(or is said to diverge).
3. Oscillatory Sequence: A sequence which neither converges nor diverges ,is
called an Oscillatory Sequence.
3 4 5
1
Ex. 1. Consider the sequence 2 , , , ,..... here an = 1 +
n
2 3 4
The sequence < an > is convergent and has the limit 1
1
1
1
1
an − 1 = 1 + − 1 = and <∈ whenever n >
n
n
n
∈
1
Suppose we choose ∈= .001 , we have < .001 when n > 1000.
n
1
n
< an > converges to 3.
Ex. 2. If an = 3 + ( −1)
'n'
Sequences and Series
5
Ex. 3.
If an = n 2 + ( −1) .n, < an > diverges.
Ex. 4.
If an =
n
1
n
+ 2 ( −1) , < an > oscillates between -2 and 2.
n
1.2 Infinite Series
If < un > is a sequence, then the expression u1 + u2 + u3 + ........ + un + ..... is called an
∞
infinite series. It is denoted by Σ un or simply Σun
n =1
The sum of the first n terms of the series is denoted by sn
i.e.,
sn = u1 + u2 + u3 + ...... + un ; s1 , s2 , s3 ,....sn are called partial sums.
1.2.1 Convergent, Divergent and Oscillatory Series
Let Σun be an infinite series. As n → ∞, there are three possibilities.
(a) Convergent series: As n → ∞, sn → a finite limit, say ‘s’ in which case the
series is said to be convergent and ‘s’ is called its sum to infinity.
Thus Lt sn = s (or) simply Ltsn = s
n →∞
∞
This is also written as u1 + u2 + u3 + ..... + un + ...to∞ = s. (or) Σ un = s (or)
n =1
simply Σun = s.
(b) Divergent series: If sn → ∞ or −∞ , the series said to be divergent.
(c) Oscillatory Series: If sn does not tend to a unique limit either finite or infinite it
is said to be an Oscillatory Series.
Note: Divergent or Oscillatory series are sometimes called non convergent series.
1.2.2 Geometric Series
The series, 1 + x + x 2 + .....x n −1 + ... is
(i) Convergent when x < 1 , and its sum is
(ii) Divergent when x ≥ 1 .
1
1− x
(iii) Oscillates finitely when x = -1 and oscillates infinitely when x < -1.
Proof: The given series is a geometric series with common ratio ‘x’
1 − xn
∴ sn =
when x ≠ 1
[By actual division – verify]
1− x
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(i) When x < 1:
⎛ xn ⎞
1
⎛ 1 ⎞
⎡⎣since x n → 0 as n → ∞ ⎤⎦
−
=
Lt sn = Lt ⎜
Lt
⎜
⎟
⎟
n →∞
n →∞ 1 − x
⎝
⎠ n →∞ ⎝ 1 − x ⎠ 1 − x
1
∴ The series converges to
1− x
n
x −1
and sn → ∞ as n → ∞
(ii) When x ≥ 1: sn =
x −1
∴ The series is divergent.
(iii) When x = –1: when n is even, sn → 0 and when n is odd, sn → 1
∴ The series oscillates finitely.
(iv) When x < −1, sn → ∞ or −∞ according as n is odd or even.
∴ The series oscillates infinitely.
1.2.3 Some Elementary Properties of Infinite Series
1. The convergence or divergence of an infinites series is unaltered by an addition or
deletion of a finite number of terms from it.
2. If some or all the terms of a convergent series of positive terms change their signs,
the series will still be convergent.
3. Let Σun converge to ‘s’
Let ‘k’ be a non – zero fixed number. Then Σkun converges to ks.
Also, if Σun diverges or oscillates, so does Σkun
4. Let Σun converge to ‘l’ and Σvn converge to ‘m’. Then
(i) Σ(un + vn ) converges to ( l + m ) and (ii) Σ(un + vn ) converges to ( l – m )
1.2.4 Series of Positive Terms
Consider the series in which all terms beginning from a particular term are +ve.
Let the first term from which all terms are +ve be u1 .
Let Σun be such a convergent series of +ve terms. Then, we observe that the
convergence is unaltered by any rearrangement of the terms of the series.
1.2.5 Theorem
If Σun is convergent, then Lt un = 0 .
n →∞
Proof :
sn = u1 + u2 + ...... + un
sn −1 = u1 + u2 + ...... + un −1, , so that, un = sn − sn −1
Sequences and Series
7
Suppose Σun = l then Lt sn = l and Lt sn −1 = l
n →∞
∴ Lt un = Lt ( sn − sn −1 ) ;
n →∞
n →∞
n →∞
Lt sn − Lt sn −1 = l − l = 0
n →∞
n →∞
Note: The converse of the above theorem need not be always true. This can be
observed from the following examples.
1 1
1
1
(i) Consider the series, 1 + + + ....... + + .... ; un = , Lt un = 0
n n →∞
2 3
n
But from p-series test (1.3.1) it is clear that Σ
(ii)
Consider the series,
1
is divergent.
n
1 1 1
1
+ 2 + 2 + ..... + 2 + ......
2
1 2 3
n
1
1
, Lt u = 0, by p series test, clearly Σ 2 converges,
2 n →∞ n
n
n
Note : If Lt un ≠ 0 the series is divergent;
un =
n →∞
Ex.
un =
2n − 1
, here Lt un = 1 ∴
n →∞
2n
Σun is divergent.
1.3 Tests for the Convergence of an Infinite Series
In order to study the nature of any given infinite series of +ve terms regarding
convergence or otherwise, a few tests are given below.
1.3.1 P-Series Test
∞
1
1
1
1
= p + p + p + ......, is
p
n
1
2
3
(i) Convergent when p > 1, and (ii) Divergent when p ≤ 1 .
The infinite series, Σ
n =1
Proof :
Case (i) Let
p > 1,3 p > 2 p ; ⇒
p > 1;
∴
Similarly,
1
1
< p
p
3
2
1
1
1
1
2
+ p < p+ p = p
p
2
3
2
2
2
1
1
1
1
1
1
1
1
4
+ p + p+ p < p+ p + p+ p = p
p
4
5
6
7
4
4
4
4
4
1
1
1
8
+ p + .... + p < p , and so on.
p
8
9
16
8
Adding we get
(JNTU 2002, 2003)
Engineering Mathematics - I
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Σ
1
2
4
8
< 1 + p + p + p + ....
p
n
2
4
8
1
1
1
1
< 1 + ( p −1) + 2( p −1) + 3( p −1) + ......
p
n
2
2
2
The RHS of the above inequality is an infinite geometric series with common
1
ratio p −1 < 1( since p > 1) The sum of this geometric series is finite.
2
Σ
i.e.,
∞
1
is also finite.
n =1 n p
Hence Σ
∴ The given series is convergent.
1
1 1 1
= 1 + + + + ......
np
2 3 4
1 1 1 1 1
+ > + =
We have,
3 4 4 4 2
1 1 1 1 1 1 1 1 1
+ + + > + + + =
5 6 7 8 8 8 8 8 2
1 1
1
1 1
1 1
+ + ....... > + + ..... = and so on
9 10
16 16 16
16 2
1
⎛1 1⎞ ⎛1 1 1 1⎞
Σ p = 1 + ⎜ + ⎟ + ⎜ + + + ⎟ + .....
∴
n
⎝ 2 3⎠ ⎝ 4 5 6 7 ⎠
1 1 1
≥ 1 + + + + .....
2 2 2
The sum of RHS series is ∞
n −1 n +1
⎛
⎞
and Lt sn = ∞ ⎟
=
⎜ since sn = 1 +
n →∞
2
2
⎝
⎠
Case (ii) Let p =1;
Σ
∞
1
( p = 1 ) diverges.
n =1 n p
∴ The sum of the given series is also ∞ ; ∴ Σ
1
1
1
= 1 + p + p + .....
p
n
2
3
1
1 1 1
p < 1, p > , p > , ...... and so on
Since
2
2 3
3
1
1 1 1
∴
Σ p > 1 + + + + .......
n
2 3 4
From the Case (ii), it follows that the series on the RHS of above inequality is
divergent.
Case (iii) Let p<1,
Σ
Sequences and Series
∴
9
Σ
1
is divergent , when P < 1
np
Note: This theorem is often helpful in discussing the nature of a given infinite series.
1.3.2 Comparison Tests
1. Let Σun and Σvn be two series of +ve terms and let Σvn be convergent.
Then Σun converges,
(a) If un ≤ vn , ∀ n ∈ N
u
(b) or n ≤ k ∀n ∈ N where k is > 0 and finite.
vn
u
(c) or n → a finite limit > 0
vn
Proof : (a) Let Σvn = l (finite)
Then, u1 + u2 + ..... + un + ...... ≤ v1 + v2 + .....vn + ..... ≤ l > 0
Since l is finite it follows that Σun is convergent
un
≤ k ⇒ un ≤ kvn , ∀n ∈ N , since Σvn is convergent and k (>0) is finite,
vn
Σkvn is convergent ∴ Σun is convergent.
u
u
(d) Since Lt n is finite, we can find a +ve constant k ,∋ n < k ∀n ∈ N
n →∞ v
vn
n
∴ from (2) , it follows that Σun is convergent
2. Let Σun and Σvn be two series of +ve terms and let Σvn be divergent. Then
Σun diverges,
* 1. If un ≥ vn , ∀n ∈ N
u
or * 2. If n ≥ k , ∀n ∈ N where k is finite and ≠ 0
vn
u
or * 3. If Lt n is finite and non-zero.
n →∞ v
n
Proof:
1. Let M be a +ve integer however large it may be. Science Σvn is divergent, a
number m can be found such that
v1 + v2 + ..... + vn > M , ∀n > m
(c)
∴
∴
u1 + u2 + ..... + un > M , ∀n > m ( un ≥ vn )
Σun is divergent
Engineering Mathematics - I
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2. u1 ≥ kvn ∀n
Σvn is divergent ⇒ Σkvn is divergent
∴
Σun is divergent
u
u
3. Since Lt n is finite, a + ve constant k can be found such that n > k , ∀n
n →∞ v
vn
n
(probably except for a finite number of terms )
∴ From (2), it follows that Σun is divergent.
Note :
(a)
In (1) and (2), it is sufficient that the conditions with * hold ∀n > m ∈ N
Alternate form of comparison tests : The above two types of comparison tests
2.8.(1) and 2.8.(2) can be clubbed together and stated as follows :
un
= k, where k is
n →∞ v
n
If Σun and Σvn are two series of + ve terms such that Lt
non- zero and finite, then Σun and Σvn both converge or both diverge.
(b)
1. The above form of comparison tests is mostly used in solving problems.
2. In order to apply the test in problems, we require a certain series Σvn whose
nature is already known i.e., we must know whether Σvn is convergent are
divergent. For this reason, we call Σvn as an ‘auxiliary series’.
3. In problems, the geometric series (1.2.2.) and the p-series (1.3.1) can be
conveniently used as ‘auxiliary series’.
Solved Examples
EXAMPLE 1
Test the convergence of the following series:
3 4 5
6
4 5 6 7
+
+ ..... (b)
+ + + + .....
(a) + +
1 8 27 64
1 4 9 16
∞
(c) Σ ⎡( n 4 + 1)
n =1 ⎢
⎣
14
− n⎤
⎥⎦
SOLUTION
(a)
Step 1: To find "un " the nth term of the given series. The numerators 3, 4, 5,
6......of the terms, are in AP.
nth term tn = 3 + ( n − 1) .1 = n + 2
Denominators are 13 , 23 ,33 , 43.....nth term = n3 ; ∴ un =
n+2
n3
Step 2: To choose the auxiliary series Σvn . In un , the highest degree of n in the
numerator is 1 and that of denominator is 3.
Sequences and Series
∴ we take, vn =
Step 3: Lt
n →∞
11
1
1
= 2
3−1
n
n
un
n+2 2
n+2
⎛ 2⎞
= Lt
× n = Lt
= Lt ⎜1 + ⎟ = 1, which is non- zero and
3
n
n
n
→∞
→∞
→∞
vn
n
n
⎝ n⎠
finite.
un
=1
n →∞ v
n
Step 4: Conclusion: Lt
∴ Σun and Σvn both converge or diverge (by comparison test). But Σvn = Σ
convergent by p-series test (p = 2 > 1); ∴ Σun is convergent.
(b)
1
is
n2
4 5 6 7
+ + + + .....
1 4 9 16
Step 1: 4 , 5, 6, 7, .....in AP , tn = 4 + ( n − 1)1 = n + 3
∴ un =
n+3
n2
1
be the auxiliary series
n
u
⎛ n +3⎞
⎛ 3⎞
Step 3: Lt n = Lt ⎜ 2 ⎟ × n = Lt ⎜1 + ⎟ = 1 , which is non-zero and finite.
→∞
n →∞ v
n →∞
n
⎝ n ⎠
⎝ n⎠
n
Step 4: ∴ By comparison test, both Σun and Σvn converge are diverge together.
1
But Σvn = Σ is divergent, by p-series test (p = 1); ∴ Σun is divergent.
n
Step 2: Let Σvn =
∞
(c)
Σ ⎡( n + 1)
⎣⎢
n =1
4
14
1
1
⎡
⎤
4
4
⎧
⎫
1
1
⎛
⎞
⎛
⎞
4
⎢
⎤
− n = ⎨n ⎜1 + 4 ⎟⎬ – n = n ⎜1 + 4 ⎟ – 1⎥
⎢⎝ n ⎠
⎥
⎦⎥ ⎩ ⎝ n ⎠⎭
⎣⎢
⎦⎥
⎡
⎤
1⎛1 ⎞
− 1⎟
⎜
⎢
⎥
1
3
4 4 ⎠ 1
⎡ 1
⎤
. 8 + ..... − 1⎥ = n ⎢ 4 −
= n ⎢1 + 4 + ⎝
+ .....⎥
8
4
2!
4
32
n
n
n
n
⎢
⎥
⎣
⎦
⎢⎣
⎥⎦
1
3
1 ⎡1
3
⎤
= 3−
+ .... = 3 ⎢ −
+ .....⎥
n ⎣ 4 32n 4
4n 32n 7
⎦
1
Here it will be convenient if we take vn = 3
n
Engineering Mathematics - I
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un
1
⎛1
⎞ 1
= Lt ⎜ −
+ ..... ⎟ = , which is non-zero and finite
4
n →∞ v
n →∞ 4
32n
⎝
⎠ 4
n
Lt
∴ By comparison test, Σun and Σvn both converge or both diverge. But by pseries test Σvn =
1
is convergent. (p = 3 > 1); ∴ Σun is convergent.
n3
EXAMPLE 2
If u n =
3
4
3n 2 + 1
2n3 + 3n + 5
show that Σun is divergent
SOLUTION
As n increases, un approximates to
3
4
3n 2
2n3
=
3
2
1
3
1
×
4
∴ If we take vn =
n
2
n
3
3
=
4
[(or) Hint: Take vn =
1
12
1
n
2
1
3
1
.
4
1
n
1
12
1
1
n
3
l1 − l2
u
33
, Lt n = 1 which is finite.
n →∞ v
n
2 4
, where l1 and l2 are indices of ‘n’ of the largest terms
1
in denominator and nominator respectively of un . Here vn =
n
3 2
−
4 3
=
1
n
1
12
]
By comparison test, Σvn and Σun converge or diverge together. But Σvn = Σ
divergent by p – series test ( since p =
∴ Σun is divergent.
1
n
1
<1)
12
EXAMPLE 3
1
2
3
4
+
+
+
+ ......
2
3
4
5
Test for the convergence of the series.
SOLUTION
Here, un =
n
;
n +1
1
Take vn =
n
1 1
−
2 2
=
u
1
1
= 1 (finite)
= 1 , Lt n = Lt
0
→∞
→∞
n
n
1
vn
n
1+
n
1
12
is
Sequences and Series
13
Σvn is divergent by p – series test. (p = 0 < 1)
∴ By comparison test, Σun is divergent, (Students are advised to follow the procedure
given in ex. 1.2.9(a) and (b) to find “ un ” of the given series.)
EXAMPLE 4
Show that 1 +
1 1
1
+ + ....... + + ..... is convergent.
1 2
n
SOLUTION
1
(neglecting 1st term )
n
1
1
1
=
<
= n −1
1.2.3......n 1.2.2.2.....n − 1times (2 )
1 1
1
Σun < 1 + + 2 + 3 + ......
∴
2 2
2
1
which is an infinite geometric series with common ratio < 1
2
1
∴
Σ n −1 is convergent. (1.2.3(a)). Hence Σun is convergent.
2
un =
EXAMPLE 5
Test for the convergence of the series,
1
1
1
+
+
+ .......
1.2.3 2.3.4 3.4.5
SOLUTION
un
= Lt
n →∞ v
n →∞
n
n3
= 1 (finite)
1 ⎞⎛ 2 ⎞
3⎛
n ⎜ 1 + ⎟⎜ 1 + ⎟
⎝ n ⎠⎝ n ⎠
∴ By comparison test, Σun , and Σvn converge or diverge together. But by p-series test,
1
Σvn = Σ 3 is convergent ( p = 3 > 1 ); ∴ Σun is convergent .
n
un =
1
;
n ( n + 1)( n + 2 )
Take vn =
1
n3
Lt
EXAMPLE 6
If un = n 4 + 1 − n 4 − 1, show that Σun is convergent.
SOLUTION
1
1
1 ⎞2
1 ⎞2
⎛
⎛
un = n 2 ⎜ 1 + 4 ⎟ − n 2 ⎜ 1 − 4 ⎟
⎝ n ⎠
⎝ n ⎠
[JNTU, 2005]
Engineering Mathematics - I
14
⎡⎛
1
1
1
1
1
1
⎞ ⎛
⎞⎤
= n 2 ⎢⎜ 1 + 4 − 8 +
− .... ⎟ − ⎜1 − 4 − 8 −
− ..... ⎟ ⎥
12
12
⎠ ⎝ 2n 8n 16n
⎠⎦
⎣⎝ 2n 8n 16n
1
1
⎡1
⎤ 1 ⎡
⎤
= n 2 ⎢ 4 + 12 + ....⎥ = 2 ⎢1 + 10 + ....⎥
⎣ n 8n
⎦ n ⎣ 8n
⎦
u
1
vn = 2 , hence Lt n = 1
n
→∞
n
vn
Take
∴ By comparison test, Σun and Σvn converge or diverge together. But Σvn =
convergent by p –series test (p = 2 > 1) ∴ Σun is convergent.
EXAMPLE 7
Test the series
1
1
1
+
+
+ ..... for convergence.
1+ x 2 + x 3 + x
SOLUTION
un =
1
;
n+ x
take vn =
1
,
n
then
un
1
n
=
=
vn n + x 1 + x
n
⎛
⎞
⎜ 1 ⎟
1
Lt ⎜
= 1; Σvn = Σ is divergent by p-series test (p =1 )
⎟
n →∞
n
⎜1+ x ⎟
n⎠
⎝
∴ By comparison test, Σun is divergent.
EXAMPLE 8
∞
⎛1⎞
Show that Σ sin ⎜ ⎟ is divergent.
n =1
⎝n⎠
SOLUTION
1
⎛1⎞
take vn =
un = sin ⎜ ⎟ ;
n
⎝n⎠
⎛1⎞
sin ⎜ ⎟
u
⎝ n ⎠ = Lt sin t (where t = 1 ) = 1
Lt n = Lt
n
n →∞ v
n →∞ ⎛ 1 ⎞
t →0 t
n
⎜ ⎟
⎝n⎠
∴ Σu , Σvn both converge or diverge . But Σvn = Σ
n
( p -series test, p = 1 ); ∴ Σun is divergent.
1
is divergent
n
1
is
n2
Sequences and Series
15
EXAMPLE 9
⎛1⎞
Test the series Σ sin −1 ⎜ ⎟ for convergence.
⎝n⎠
SOLUTION
1
1
un = sin −1 ;
Take
vn =
n
n
⎛1⎞
−1⎜ ⎟
1
u
⎛ θ ⎞ ⎛
⎞
sin ⎝ n ⎠ ; =
Lt ⎜
= 1⎜ Taking sin −1 = θ ⎟
Lt n = Lt
⎟
θ
→
0
n →∞ v
n →∞ ⎛ 1 ⎞
n
⎝ sin θ ⎠ ⎝
⎠
n
⎜ ⎟
n
⎝ ⎠
But Σvn is divergent. Hence Σun is divergent.
EXAMPLE 10
Show that the series 1 +
1 22 33
+ + + ..... is divergent.
22 33 43
SOLUTION
1 2 2 33
+
+
+ ..... . Therefore
2 2 33 4 4
nn
nn
1
;
=
=
n=
n
n
+
+
n
n
1
1
( )( )
⎛ 1⎞ n⎛ 1⎞
⎛ 1 ⎞⎛ 1 ⎞
n ⎜ 1 + ⎟⎜1 + ⎟
n ⎜1 + ⎟ .n ⎜1 + ⎟
⎝ n⎠ ⎝ n⎠
⎝ n ⎠⎝ n ⎠
Neglecting the first term, the series is
un =
nn
( n + 1)
Take vn =
n +1
1
n
un
1
1
1
= Lt
= Lt
=
n
n
n
n
→∞
→∞
vn
e
⎛ 1 ⎞⎛ 1 ⎞
⎛ 1⎞
⎜ 1 + ⎟⎜1 + ⎟
⎜1 + ⎟ .1
⎝ n ⎠⎝ n ⎠
⎝ n⎠
1
which is finite and Σvn = Σ
is divergent by p –series test ( p = 1)
n
∴
Σun is divergent.
∴
Lt
n →∞
EXAMPLE 11
Show that the series
1
3
5
+
+
+ .......∞ is convergent.
1.2.3 2.3.4 3.4.5
SOLUTION
1
3
5
+
+
+ .......∞
1.2.3 2.3.4 3.4.5
(JNTU 2000)
Engineering Mathematics - I
16
1⎞
⎛
2− ⎟
⎜
2n − 1
1
n⎠
⎝
= 2.
nth term = un =
n ( n + 1)( n + 2 ) n ⎛ 1 ⎞ ⎛ 2 ⎞
⎜1 + ⎟ ⎜1 + ⎟
⎝ n ⎠⎝ n ⎠
1
vn = 2
Take
n
1⎞
⎛
2− ⎟
⎜
u
1
n⎠
⎛ 1 ⎞
⎝
Lt n = Lt 2
÷⎜ 2 ⎟
n →∞ v
n →∞ n
⎝n ⎠
1+ 1 1+ 2
n
n
n
u
2−0
Lt n =
= 2 which is finite and non-zero
n →∞ v
(1 + 0 )(1 + 0 )
n
(
But
∑v
n
1
∑n
=
)
∑ v converge or diverge together
is convergent. ∴ ∑ u is also convergent.
∴ By comparison test
∑u
)(
n
and
n
n
2
EXAMPLE 12
∞
Test whether the series
∑
n =1
1
n + n +1
is convergent
(JNTU 1997, 1999, 2003)
SOLUTION
∞
∑
The given series is
1
n + n +1
1
un =
n + n +1
n =1
=
(
n +1 − n
n + n +1
)(
n +1 − n
)
= n +1 − n
1
⎧⎛
1
1
⎪⎧⎛ 1 ⎞ 2 ⎪⎫
⎞ ⎫
un = n ⎨⎜1 + ⎟ − 1⎬ = n ⎨⎜1 +
− 2 + ..... ⎟ − 1⎬
n⎠
⎠ ⎭
⎩⎝ 2n 8n
⎩⎪⎝
⎭⎪
1
⎧1
⎫ 1 ⎧1 1
⎫
u n = n ⎨ – 2 + ...⎬ =
+ ....⎬
⎨ –
n ⎩ 2 8n
⎩ 2 n 8n
⎭
⎭
Sequences and Series
1
n
vn =
Take
17
un
1 ⎧1 2
⎫ ⎛ 1 ⎞ 1
= Lt
− + ......⎬ ÷ ⎜
⎨
⎟=
n →∞ v
n →∞
n ⎩ 2 8n
⎭ ⎝ n⎠ 2
n
Lt
which is finite and non-zero .
un and
Using comparison test
∑
1
∑v = ∑ n
But
n
∑u
∴
∑v
n
converge or diverge together.
(
is divergent since p = 1
2
)
is also divergent.
n
EXAMPLE 13
∞
∑ ⎡⎣
Test for convergence
n =1
n3 + 1 − n ⎤
⎦
[JNTU 1996, 2003, 2003]
(
)
1
⎡
⎤
1 1 −1
⎡⎛
1⎞3 ⎤
1
1
3
3
⎢
. 6 + ..... − 1⎥
un = n ⎢⎜ 1 + 3 ⎟ − 1⎥ = n 1 + 3 +
⎢
⎥
3
1.2
n
n
n
⎠
⎢⎣⎝
⎥⎦
⎣
⎦
th
n term
Lt
Then
un
n →∞
=
1
1
1 ⎛1 1
1
⎞
− 5 + ...... = 2 ⎜ − 3 + ...... ⎟ ; Let vn = 2
2
n
3n 9n
n ⎝ 3 9n
⎠
vn
= Lt 1 − 1 3 + .... = 1 ≠ 0
3
3
n →∞
9n
∑v
n
)
(
∴ By comparison test,
But
3
∑u
n
and
∑v
n
both converge or diverge.
is convergent by p -series test ( since p = 2 > 1) ∴
∑u
n
is convergent.
EXAMPLE 14
Show that the series,
p≤2
2 3
4
+ p + p + ....... is convergent for p > 2 and divergent for
p
1 2
3
SOLUTION
nth term of the given series = un =
Let us take vn =
1
n
p −1
; Lt
n →∞
un
vn
(
) (
1
1+ 1
n +1 n 1+ n
n
=
=
p
p
p −1
n
n
n
= 1 ≠ 0;
)
Engineering Mathematics - I
18
∑ u and ∑ v
But ∑ v = ∑ 1
n
∴
n
n
n
both converge or diverge by comparison test.
converges when p -1>1 ; i.e., p >2 and diverges when
p −1
p − 1 ≤ 1 i.e p ≤ 2 ; Hence the result.
EXAMPLE 15
⎛ 2n + 3 ⎞
Test for convergence ∑ ⎜ n
⎟
n =1 ⎝ 3 + 1 ⎠
∞
SOLUTION
(
(
⎡ 2n 1 + 3
⎢
2n
un = ⎢
n
1
⎢⎣ 3 1 + 3n
)
)
1
2
(JNTU 2003)
1
⎤ 2
⎥
⎥ ;
⎥⎦
2
vn = n ;
3
Take
un
= 1 ≠ 0 ; ∴ By comparison test,
n →∞ v
n
∑u
Lt
n
∞
⎛
un ⎜ 1 +
=
vn ⎜ 1 +
⎝
n
n
and
∑v
n
⎞
2 ⎟
1 n⎟
3 ⎠
3
1
2
n
behave the same way.
3
2 2 ⎛2⎞ 2
⎛2⎞ 2
+ + ⎜ ⎟ + ....., which is a geometric series with
But ∑ vn = ∑ ⎜ ⎟ =
3 3 ⎝3⎠
n =1 ⎝ 3 ⎠
common ratio 2 (<1) ∴ ∑ vn is convergent. Hence ∑ un is convergent.
3
EXAMPLE 16
Test for convergence of the series,
1
4
9
+
+
+ ......
4.7.10 7.10.13 10.13.16
(JNTU 2003)
SOLUTION
and
∴
4, 7, 10,..............is an A . P;
tn = 4 + ( n − 1) 3 = 3n + 1
7, 10, 13,............is an A . P;
tn = 7 + ( n − 1) 3 = 3n + 4
10 , 13 , 16 ,.............is an A. P;
tn = 10 + ( n − 1) 3 = 3n + 7
un =
=
n2
=
( 3n + 1)( 3n + 4 )( 3n + 7 ) 3n 1 + 1
(
(
27 n 1 + 1
3n
)(
1
1+ 4
3n
)(
1+ 7
3n
)
;
n2
3n
) .3n (1 + 4 3n ) .3n (1 + 7 3n )
Sequences and Series
Taking vn =
19
1
, we get
n
un
1
=
≠ 0 ; ∴ By comparison test, both
n →∞ v
27
n
Lt
manner. But by p –series test,
∑v
n
∑u
n
and
∑v
n
is divergent, since p = 1. ∴
behave in the same
∑u
n
is divergent .
EXAMPLE 17
Test for convergence
2 n 2 − 5n + 1
∑ 4n 3 − 7 n 2 + 2
(JNTU 2003)
SOLUTION
nth term of the given series = un =
Let
vn =
2 n 2 − 5n + 1
4n3 − 7 n 2 + 2
1
n2
⎡ n 2− 5 + 1
⎤
⎡ 2− 5 + 1
2
un
n2 ⎥
n
n
⎢
⎢
n2
n
= Lt . ⎢
× ⎥ = Lt ⎢
Lt
n →∞ v
n →∞
n →∞
3
1
7
7
2
2
n
⎢⎣ n 4 − n + n3
⎥⎦
⎢⎣ 4 − n + n3
∴ By comparison test, ∑ un and ∑ vn both converge or diverge.
)
(
But
∑v
n
)
(
is convergent. [p series test – p = 2 > 1] ∴
∑u
n
⎤
2
⎥
⎥= 4 ≠0
⎥⎦
is convergent .
EXAMPLE 18
Test the series
∑u
n
, whose nth term is
( 4n
1
2
− i)
SOLUTION
⎡
⎤
un
n2
⎢
⎥ 1
un =
;
Lt
= Lt ⎢
2
⎥= 4≠0
n →∞ v
n →∞
2
i
( 4n − i )
n
⎢⎣ n 4 − n 2 ⎥⎦
∴ ∑ un and ∑ vn both converge or diverge by comparison test. But ∑ vn is
1
1
Let vn = 2 ,
n
convergent by p –series test ( p = 2 > 1) ; ∴
∞
Note: Test the series
∑ 4n
n =1
1
2
–1
(
∑u
n
is convergent.
)
Engineering Mathematics - I
20
EXAMPLE 19
⎛1⎞
⎝n⎠
⎛1⎞
⎝n⎠
If un = ⎜ ⎟ .sin ⎜ ⎟ , show that
∑u
n
is convergent .
SOLUTION
1
, so that
n2
Let vn =
⎛u
Lt ⎜ n
n →∞ v
⎝ n
∑v
is convergent by p –series test .
n
( )
( )
sin 1
⎞
n = Lt ⎛ sin t ⎞
=
Lt
⎟ n →∞
⎜
⎟
t →0
1
⎝ t ⎠
⎠
n
⎛ un
⎝ vn
where t = 1/n, Thus Lt ⎜
n →∞
∑u
∴ By comparison test,
⎞
⎟ =1≠ 0
⎠
n
is convergent.
EXAMPLE 20
Test for convergence
1
tan( 1 )
n
n
∑
SOLUTION
Take vn = 1
n
3
u
; Lt ⎡ n
n →∞
2
⎢⎣
Hence by comparison test,
⎤ = 1 ≠ 0 ( as in above example)
vn ⎥⎦
∑u
n
converges as
∑v
n
converges.
EXAMPLE 21
∞
Show that
∑ sin
n =1
2
⎛1⎞
⎜ ⎟ is convergent.
⎝n⎠
SOLUTION
⎛1⎞
un = sin ⎜ ⎟ ;
⎝n⎠
2
Let
1
Take vn = 2 ,
n
( ) ⎤⎥
⎡ sin 1
⎛ un ⎞
n
Lt ⎜ ⎟ = Lt ⎢
n →∞ v
n →∞ ⎢
1
⎝ n⎠
n
⎣
2
⎛ sin t ⎞
= Lt ⎜
⎟
→
0
t
⎥
⎝ t ⎠
⎦
u
t = 1 ; Lt ⎛⎜ n ⎞⎟ = 12 = 1 ≠ 0
n n→∞ ⎝ vn ⎠
∴ By comparison test, ∑ un and ∑ vn behave the same way.
where
But
∑v
n
is convergent by p- series test, since p = 2 > 1; ∴
∑u
n
is convergent.
2
Sequences and Series
21
EXAMPLE 22
∞
∑
Show that
1
n=2
log ( n n )
is divergent.
SOLUTION
un = 1
; log 2 < 1 ⇒ 2 log 2 < 2 ⇒ 1
>1 ;
n log n
2 log 2
2
1
> 1 ..... 1
> 1 ,n∈ N
Similarly
3log 3
3,
n log n
n
∴
∑ 1
> ∑ 1 ; But ∑ 1 is divergent by p-series test.
n log n
n
n
By comparison test, given series is divergent. [If ∑ vn is divergent and un ≥ vn ∀n then
∑u
n
is divergent.]
(Note : This problem can also be done using Cauchy’s integral Test.
EXAMPLE 23
∞
∑ (c + n) ( d + n)
Test the convergence of the series
−r
−s
, where c, d, r, s are all +ve.
n =1
SOLUTION
The nth term of the series = un =
vn =
Let
Lt
n →∞
1
n
r +s
un
= 1 ≠ 0 , ∴ ∑ un and
vn
∑v
But by p-series test,
∴
un
=
vn
Then
∑u
n
n
1
(c + n) ( d + n)
r
s
.
nr +s
r
⎛ c⎞
⎛ d⎞
n ⎜ 1 + ⎟ .n s ⎜1 + ⎟
⎝ n⎠
⎝ n⎠
r
∑v
n
s
=
1
r
⎛ c⎞ ⎛ d⎞
⎜1 + ⎟ ⎜1 + ⎟
⎝ n⎠ ⎝ n⎠
both converge are diverge, by comparison test.
converges if (r + s) > 1 and diverges if (r + s) ≤ 1
converges if ( r + s ) > 1 and diverges if ( r + s ) ≤ 1.
EXAMPLE 24
∞
Show that
∑n
(
− 1+ 1
n
) is divergent.
1
SOLUTION
un = n
(
− 1+ 1
n
)= 1
1
n.n
s
Take
n
vn =
u
1
1
; Lt n = Lt 1 = 1 ≠ 0
n
n
→∞
→∞
n
vn
n n
Engineering Mathematics - I
22
Lt
For let
1
1
= y say; log y = Lt − .log n = − Lt n = 0
n →∞ n
n →∞ 1
1
n →∞
n
1
n
⎛∞⎞
⎟ using L Hospitals rule)
⎝∞⎠
converge or diverge. But p-series test,
By comparison test both ∑ un and ∑ vn
∴
y = e0 = 1
∑v
n
(⎜
diverges (since p =1); Hence
∑u
n
diverges.
EXAMPLE 25
(n + a)
∑
p
q
n =1 ( n + b ) ( n + c )
r
∞
Test for convergence the series
SOLUTION
un =
Take vn =
(n + a )
(n + b ) p (n + c )q
r
1
n
p+ q −r
=
(
np
n
q
q
un
=1≠ 0 ;
n →∞ v
n
1
=
n
p+ q −r
.
(
(
1+ b
)(
p
n
n
)
r
1+ c
n
)
q
;
; Lt
But by p-series test,
∑u
p
1+ a
r
nr 1 + a
Applying comparison tests both
Hence
)
n
(1 + b n ) n (1 + c n )
, a, b, c , p, q, r, being +ve.
∑v
n
∑u
n
and
∑v
n
converge or diverge.
converges if ( p+ q –r ) > 1and diverges if ( p + q –r) ≤ 1.
converges if ( p + q – r ) > 1 and diverges if ( p + q – r) ≤ 1.
EXAMPLE 26
Test the convergence of the following series whose nth terms are:
( 3n + 4 )
( 2n + 1)( 2n + 3)( 2n + 5)
(a)
(d)
1
;
( 3 + 5n )
n
;
1
;
n
(b)
tan
(e)
1
n.3n
(c)
⎛ 1 ⎞ ⎛ n +1 ⎞
⎜ 2 ⎟⎜
⎟
⎝ n ⎠⎝ n + 3 ⎠
SOLUTION
(a)
Hint : Take
vn =
⎛u ⎞ 3
1
; v is convergent; Lt ⎜ n ⎟ = ≠ 0 (Verify)
2 ∑ n
n →∞ v
n
⎝ n⎠ 8
Apply comparison test:
un is convergent [the student is advised to work out this problem fully ]
∑
n
Sequences and Series
(b)
23
Proceed as in Example 8;
∑u
n
is convergent.
⎛u ⎞
(1 + 1n ) = e = 1 ≠ 0
1
Hint : Take vn = 2 ; Lt ⎜ n ⎟ = Lt
n
n →∞ v
n
e3 e 2
⎝ n ⎠ n→∞ 1 + 3
n
1
vn = 2 is convergent (work out completely for yourself )
n
n
(c)
(d)
(
un =
∑u
n
⎛u ⎞
1
1
1
1
; Take vn = n ; Lt ⎜ n ⎟ = 1 ≠ 0
= n.
n
n
n →∞ v
5
3 +5
5 ⎡ ⎛3⎞ ⎤
⎝ n⎠
⎢1 + ⎜ ⎟ ⎥
⎣⎢ ⎝ 5 ⎠ ⎦⎥
n
and
∑v
n
behave the same way. But
geometric series with common ratio
∴
(e)
)
∑u
n
∑v
n
is convergent since it is a
1
<1
5
is convergent by comparison test .
1
1
≤ n , ∀n ∈ N , since
n
n.3
3
1
1
∴
∑ n.3n ≤ ∑ 3n
n.3n ≥ 3n ;
…..(1)
The series on the R.H .S of (1) is convergent since it is geometric series with
r=
1
<1.
3
∴ By comparison test
1
∑ n.3
n
is convergent.
EXAMPLE 27
Test the convergence of the following series.
1+ 2
1+ 2 + 3
1+ 2 + 3 + 4
+ ...............
+ 2
+ 2
2
2
2
2
1 + 2 1 + 2 + 3 1 + 22 + 32 + 42
(a)
1+
(b)
12 + 22 12 + 22 + 32 12 + 22 + 32 + 42
1 + 3 3 + 3 3 3 + 3 3 3 3 + ..............
1 +2 1 +2 +3 1 +2 +3 +4
Engineering Mathematics - I
24
SOLUTION
n
1 + 2 + 3 + .... + n
un = 2
=
1 + 22 + 32 + .....n 2
( n + 1)
3
2
=
( 2n + 1) ( 2n + 1)
n ( n + 1)
6
un
1
n
3
⎛
⎞ 3
Take vn =
; Lt
= Lt ⎜
= ≠0
⎟
n →∞ v
n →∞ 2n + 1
n
⎝
⎠ 2
n
(a)
∑u
(b)
and
n
∑v
n
behave alike by comparison test.
But
∑v
un =
1 + 2 + .... + n
=
13 + 23 + .....n3
n
2
is diverges by p-series test. Hence
2
2
n ( n + 1)
n2
( 2n + 1)
6
( n + 1)
2
=
∑u
n
is divergent.
2 ( 2n + 1)
3n ( n + 1)
4
1
Hint : Take vn = and proceed as in (a) and show that
n
∑u
n
is divergent.
Exercise 1.1
1. Test for convergence the infinite series whose nth term is:
(a)
1
n− n
[Ans : divergent]
(b)
n +1 − n
n
[Ans : convergent]
(c)
n2 + 1 − n
[Ans : divergent]
(d)
(e)
(f)
(g)
(h)
n
n −1
2
n3 + 1 − n3
1
n ( n + 1)
n
n +1
2n 3 + 5
4n 5 + 1
2
[Ans : convergent]
[Ans : divergent]
[Ans : divergent]
[Ans : convergent]
[Ans : convergent]
Sequences and Series
25
2. Determine whether the following series are convergent or divergent.
1
2
3
+
+
+ ............
−1
−2
1+ 3
1+ 3
1 + 3−3
12 22 32
2 + 10n
+ 3 + 3 + ........ +
+ ..... .......
3
1
2
3
n3
1
1
1
+
+
+ ...... ........
1+ 2
2+ 3
3+ 4
(a)
(b)
(c)
2 3 4
+ + + ...... .........
32 42 52
1 1 1
+ + + ..........
12 23 34
(d)
(e)
∞
∑
(f)
n =1
4
4n 2 + 2 n + 3
∑ (8
∞
(g)
n2 + 1
3
1
n
[Ans : convergent]
[Ans : divergent]
[Ans : divergent]
[Ans : convergent]
...............
[Ans : divergent]
)
− 1 ....................
1
[Ans : divergent]
[Ans : divergent]
(h)
3n3 + 8
∑1 5n5 + 9 ........................
[Ans : convergent]
(i)
1
2
3
+
+
+ ...........
1.3 3.5 5.7
[Ans : divergent]
∞
1.3.3 D’ Alembert’s Ratio Test
Let (i)
∑u
n
be a series of +ve terms and (ii) Lt
Then the series
n →∞
∑u
n
un +1
= k ( ≥ 0)
un
is (i) convergent if k < 1 and (ii) divergent if k > 1.
Proof :
un +1
= k ( < 1)
n →∞ u
n
Case (i) Lt
From the definition of a limit, it follows that
∃m > 0 and l ( 0 < l < 1) ∋
un +1
< l∀n ≥ m
un
Engineering Mathematics - I
26
um +1
u
< l , m + 2 < l ,..........
um
um +1
i.e.,
∴
⎡ u
⎤
u
um + um +1 + um + 2 + ...... ........ = um ⎢1 + m +1 + m + 2 + .....⎥
um
um
⎣
⎦
um 1
um 1
um
um
um
< um (1 + l + l 2 + ...) = um .
But um .
2
1
.
um 1
um
.....
1
( l < 1)
1− l
1
is a finite quantity ∴
1− l
∞
∑u
n=m
n
is convergent
By adding a finite number of terms u1 + u2 + ...... + um −1 , the convergence of the
∞
series is unaltered.
∑u
n=m
n
is convergent.
un +1
= k >1
n →∞ u
n
Case (ii) Lt
There may be some finite number of terms in the beginning which do not satisfy
the condition
∋
un +1
≥ 1 . In such a case we can find a number ‘m’
un
un +1
≥ 1, ∀n ≥ m
un
Omitting the first ‘m’ terms, if we write the series as u1 + u2 + u3 + ........., we
have
u
u2
u
≥ 1, 3 ≥ 1, 4 ≥ 1 .......... and so on
u1
u2
u3
∴
⎛ u u u
⎞
u1 + u2 + ...... + un = u1 ⎜1 + 2 + 3 . 2 + ..... ⎟ (to n terms)
⎝ u1 u2 u1
⎠
≥ u1 (1 + 1 + 1.1 +.......to n terms)
= nu1
Lt
n →∞
n
∑u
n =1
n
≥ Lt n.u1 which → ∞ ; ∴
n →∞
∑u
n
is divergent .
Sequences and Series
27
∞
Note: 1 The ratio test fails when k = 1. As an example, consider the series,
1
∑n
n =1
p
p
p
⎛ 1 ⎞
un +1
⎛ n ⎞
⎜
⎟ =1
= Lt ⎜
=
Lt
Lt
⎟
n →∞ u
n →∞ n + 1
n →∞ ⎜
1+ 1 ⎟
⎝
⎠
n
n⎠
⎝
Here
i.e.,
k = 1 for all values of p,
But the series is convergent if p > 1 and divergent if p ≤ 1 , which shows that
when k = 1, the series may converge or diverge and hence the test fails .
Ratio test can also be stated as follows:
Note: 2
If
∑u
n
un
= k , then
n →∞ u
n +1
is series of +ve terms and if Lt
∑u
n
is convergent
If k > 1 and divergent if k < 1 (the test fails when k = 1).
Solved Examples
Test for convergence of Series
EXAMPLE 28
x
x2
x3
+
+
+ .................
1.2 2.3 3.4
(a)
SOLUTION
un =
xn
x n +1
; un +1 =
;
n ( n + 1)
( n + 1)( n + 2 )
un +1
=x
n →∞ u
n
∴ By ratio test
∑u
When x = 1, un =
n
n
is convergent When |x| < 1 and divergent when | x | > 1;
u
1
1
; Take vn = 2 ; Lt n = 1
n (1 + 1 n )
n n→∞ vn
2
∴ By comparison test
∑u
n ( n + 1)
un +1
x n +1
1
=
=
x.
.
n
un
⎛ 2⎞
( n + 1)( n + 2 ) x
⎜1 + ⎟
⎝ n⎠
Lt
Therefore
Hence
(JNTU 2003)
∑u
n
is convergent.
is convergent when x ≤ 1 and divergent when x > 1 .
Engineering Mathematics - I
28
1 + 3x + 5 x 2 + 7 x 3 + .......
(b)
SOLUTION
un = ( 2n − 1) x n −1 ;
∴ By ratio test
∑u
n
∑u
Lt
is convergent when x < 1 and divergent when x > 1
When x = 1: un = 2n − 1; Lt un = ∞ ; ∴
Hence
un +1
⎛ 2n + 1 ⎞
= Lt ⎜
⎟x = x
n →∞ u
n →∞ 2n − 1
⎝
⎠
n
un +1 = ( 2n + 1) x n ;
n →∞
∑u
n
is divergent.
is convergent when x < 1 and divergent when x ≥ 1
n
∞
(c)
xn
...........
∑
2
n =1 n + 1
SOLUTION
un =
xn
;
n2 + 1
un +1 =
x n +1
( n + 1)
2
+1
.
(
)
⎡ 2
⎤
1 2 ⎥
n
1
+
⎢
un +1
un +1 ⎛ n + 1 ⎞
n
⎥ ( x) = x
Hence
= Lt ⎢
=⎜ 2
⎟ x , nLt
→∞ u
n →∞
2
un ⎝ n + 2n + 2 ⎠
⎛
⎢ n2 1 + + 2 ⎞ ⎥
n
⎢⎣ ⎜⎝ n n 2 ⎟⎠ ⎥⎦
∴ By ratio test, ∑ un is convergent when x < 1 and divergent when x > 1 When
2
1
1
; Take vn = 2
n +1
n
∴ By comparison test, ∑ un is convergent when x ≤ 1 and divergent when x > 1
x = 1: un =
2
EXAMPLE 29
⎛ n2 − 1 ⎞ n
∑ ⎜ 2 ⎟x , x > 0 for convergence.
n →∞ ⎝ n + 1 ⎠
∞
Test the series
SOLUTION
⎡ ( n + 1)2 − 1 ⎤ n +1
⎛ n2 − 1 ⎞ n
un = ⎜ 2 ⎟ x ; un +1 = ⎢
⎥x
2
⎢⎣ ( n + 1) + 1 ⎥⎦
⎝ n +1 ⎠
Sequences and Series
29
⎡⎛ n 2 + 2n ⎞ ( n 2 + 1) ⎤
un +1
⎥ .x
= Lt ⎢⎜ 2
Lt
⎟ 2
n →∞ u
n →∞
⎢⎣⎝ n + 2n + 2 ⎠ ( n − 1) ⎥⎦
n
(
(
∴ By ratio test,
∑
)
⎡
⎤
n 4 (1 + 2 n ) 1 + 1 n 2
⎥=x
= Lt ⎢ 4
2
n →∞ ⎢ n 1 + 2 n + 2 n 2
n
1
–
1
⎥⎦
⎣
un is convergent when x < 1 and divergent when x > 1 when x = 1,
)(
)
n2 − 1
1
Take vn = 0
2
n +1
n
un =
Applying p-series and comparison test, it can be seen that
∴
∑u
n
∑u
n
is divergent when x = 1.
is convergent when x < 1 and divergent x ≥ 1
EXAMPLE 30
Show that the series 1 +
2 p 3p 4 p
+ +
+ ..... , is convergent for all values of p.
2
3
4
SOLUTION
( n + 1)
np
un =
; un +1 =
n
n +1
p
⎡ ( n + 1) p n ⎤
⎧⎪ 1 ⎛ n + 1 ⎞ p ⎫⎪
un +1
Lt
= Lt ⎢
× p ⎥ = Lt ⎨
⎜
⎟ ⎬
n →∞ u
n →∞
⎢⎣ n + 1 n ⎥⎦ n→∞ ⎩⎪ ( n + 1) ⎝ n ⎠ ⎭⎪
n
p
∑u
1
⎛ 1⎞
× Lt ⎜1 + ⎟ = 0 < 1 ;
= Lt
n →∞ ( n + 1)
n →∞
⎝ n⎠
n
is convergent for all ‘ p ‘ .
EXAMPLE 31
Test the convergence of the following series
1 1
1
1
+ p + p + p + ............
p
1 3 5
7
SOLUTION
un =
1
( 2n − 1)
p
;
un +1 =
1
( 2n + 1)
p
Engineering Mathematics - I
30
2 p.n p (1 − 1 2n )
un +1 ( 2n − 1)
=
=
;
p
p
un
( 2n + 1) 2 p n p (1 + 1 2n )
p
p
∴ Ratio test fails.
1 u
np
Take vn = p ; n =
=
n vn ( 2n − 1) p
which is non – zero and finite
∴ By comparison test,
un and
∑
But by p – series test, ∑ v
n
when p ≤ 1
∴
∑u
n
=
1
1 ⎞
⎛
2 p ⎜1 − ⎟
⎝ 2n ⎠
∑v
n
1
∑n
p
p
un +1
=1
n →∞ u
n
Lt
un
1
= p,
n →∞ v
2
n
; Lt
both converge or both diverge.
converges when p > 1 and diverges
is convergent if p > 1 and divergent if p ≤ 1 .
EXAMPLE 32
∞
Test the convergence of the series
∑
n =1
SOLUTION
un =
(n + 1)x n ; x > 0
n3
( n + 1) x n ; u ( n + 2 ) x n+1
n +1
3
n3
( n + 1)
un +1
n + 2 n +1
n3
⎛ n + 2 ⎞⎛ n ⎞
=
.
.
=⎜
.x
x
3
n
un
( n + 1) x ⎝ n + 1 ⎟⎠ ⎜⎝ n + 1 ⎟⎠
( n + 1)
3
2⎞
⎛
⎜ 1+ n ⎟
un +1
1
= Lt ⎜
.x = x
Lt
⎟
3
n →∞ u
n →∞
1
1
⎛
⎞
n
⎜ 1+ ⎟ 1+
⎝ n ⎠ ⎜⎝ n ⎟⎠
∴ By ratio test,
∑u
When x = 1, un =
n
converges when x < 1 and diverges when x > 1 .
n +1
n3
1
; By comparison test ∑ un is convergent ( give proof )
n2
∴ ∑ un is convergent if x ≤ 1 and divergent if x > 1.
Take vn =
Sequences and Series
31
EXAMPLE 33
Test the convergence of the series
∞
(i)
⎛n
2
⎝
n
∑⎜ 2
n −1
+
(JNTU 2002)
2.5.8 2.5.8.11
1 1.2 1.2.3
1 ⎞
(ii) 1 +
+
+
+
+ ... (iii) +
2 ⎟
1.5.9 1.5.9.13
3 3.5 3.5.7
n ⎠
SOLUTION
⎛ n2 1 ⎞
∑
⎜ n+ 2⎟ =
n ⎠
n −1 ⎝ 2
∞
(i)
un +1
( n + 1)
=
2
2n +1
∞
n2 ∞ 1
+∑ 2
∑
n
n =1 2
n =1 n
Let un =
( n + 1) . 2n
u
; n +1 =
un
2n +1 n 2
2
n2
1
; vn = 2
n
2
n
2
u
1 ⎛ 1⎞ 1
Lt n +1 = Lt . ⎜1 + ⎟ = < 1
n →∞ u
n →∞ 2
2
⎝ n⎠
n
∑ u is convergent. By p –series test, ∑ v
∴ The given series ( ∑ u + ∑ v ) is convergent.
∴ By ratio test
n
n
n
(ii)
is convergent.
n
Neglecting the first term, the series can be taken as,
2.5.8 2.5.8.11
+
+
1.5.9 1.5.9.13
Here, 1st term has 3 fractions ,2nd term has 4 fractions and so on .
∴ nth term contains ( n + 2 ) fractions
2. 5. 8.......are in A. P.
∴ ( n + 2)
th
term = 2 + ( n + 1 ) 3 = 3n + 5 ;
∴ 1. 5. 9,.......are in A. P.
th
∴ ( n + 2 ) term = 1 + ( n + 1 ) 4 = 4n + 5
∴
un =
2.5.8..... ( 3n + 5 )
1.5.9..... ( 4n + 5 )
un +1 =
un +1
un
2.5.8..... ( 3n + 5 )( 3n + 8 )
1.5.9..... ( 4n + 5 )( 4n + 9 )
( 3n + 8) ;
=
( 4n + 9 )
∴ By ratio test,
∑u
n
8⎞
⎛
n⎜3+ ⎟
u
n⎠ 3
= <1
Lt n +1 = Lt ⎝
n →∞ u
n →∞ ⎛
9⎞ 4
n
n⎜4 + ⎟
n⎠
⎝
is convergent.
Engineering Mathematics - I
32
(iii)
1, 2, 3, ........ are in A. P nth term = n ; 3. 5. 7..........are in A.P. nth term = 2n + 1
⎡ 1.2.3.....n ⎤
un = ⎢
⎥
⎣ 3.5.7..... ( 2n + 1) ⎦
⎡
⎤
1.2.3.....n ( n + 1)
un +1 = ⎢
⎥
⎣ 3.5.7..... ( 2n + 1)( 2n + 3) ⎦
∴
un +1 ⎛ n + 1 ⎞
=⎜
⎟
un ⎝ 2n + 3 ⎠
∴
⎛ 1⎞
n. ⎜1 + ⎟
u
n⎠ 1
= <1
Lt n +1 = Lt ⎝
n →∞ u
n →∞ ⎛
3⎞ 2
n
n⎜2 + ⎟
n⎠
⎝
By ratio test, ∑ un is convergent.
EXAMPLE 34
1.3.5.... ( 2n − 1) n −1
.x ( x > 0 )
2.4.6....2n
n =1
∞
Test for convergence
∑
SOLUTION
The given series of +ve terms has un =
and
un +1 =
1.3.5.... ( 2n + 1) n
x
2.4.6.... ( 2n + 2 )
(JNTU 2001)
1.3.5.... ( 2n − 1) n −1
.x
2.4.6....2n
(
(
)
)
2n 1 + 1
un +1
⎛ 2n + 1 ⎞
2n .x = x
x
= Lt ⎜
=
Lt
⎟
n →∞ u
n →∞ 2n + 2
n →∞
⎝
⎠
2n 1 + 2
n
2n
∴ By ratio test, ∑ un is converges when x < 1 and diverges when x > 1 when x = 1, the
Lt
test fails.
1.3.5.... ( 2n − 1)
< 1 and Lt un ≠ 0
n →∞
2.4.6.....2n
∴ ∑ un is divergent. Hence ∑ un is convergent when x < 1, and divergent when x ≥ 1
Then un =
Sequences and Series
33
EXAMPLE 35
⎛ 2n − 2 ⎞ n −1
2
6 2
Test for the convergence of 1 + x + x + ........ + ⎜ n
⎟ x + ..... ( x > 0 )
5
9
⎝ 2 +1 ⎠
(JNTU 2003)
SOLUTION
⎛ 2n − 2 ⎞ n −1
⎟ x , ( n ≥ 2 ) and ' un ' are all +ve.
n
⎝ 2 +1 ⎠
Omitting 1st term, un = ⎜
(2
=
(2
n +1
un +1
− 2)
⎛ un +1 ⎞
⎛ 2n +1 − 2 ⎞ ⎛ 2n + 1 ⎞
n
x
;
=
Lt
Lt
.
⎜
⎟
⎜ n +1
⎟×⎜ n
⎟ .x
n +1
n →∞
+ 1)
⎝ un ⎠ n→∞ ⎝ 2 + 1 ⎠ ⎝ 2 − 2 ⎠
) (
) (
(
(
)
)
⎡ 2n +1 1 − 1
2n 1 + 1 n ⎤
n
⎢
2
2 .x ⎥ = x ;
= Lt
.
⎥
n →∞ ⎢ n +1
n
1
2
⎢⎣ 2 1 + 2n +1 2 1 − 2n ⎥⎦
Hence, by ratio test, ∑ un converges if x < 1 and diverges if x > 1.
When x = 1, the test fails. Then un =
Hence
∑u
n
2n − 2
; Lt un = 1 ≠ 0 ; ∴
2n + 1 n→∞
∑u
n
diverges
is convergent when x < 1 and divergent x > 1
EXAMPLE 36
∞
Using ratio test show that the series
∑
( 3 − 4i )
n!
n=0
SOLUTION
un
( 3 − 4i )
=
n
n!
Hence, by ratio test,
;
∑u
un +1
n
( 3 − 4i )
=
n
converges
n +1
(n + 1)!
(JNTU 2000)
⎛ un +1 ⎞
⎛ 3 − 4i ⎞
⎟ = nLt
⎜
⎟ = 0 <1
⎝ un ⎠ →∞ ⎝ n + 1 ⎠
; Lt ⎜
n →∞
converges.
EXAMPLE 37
Discuss the nature of the series,
2
3 2 4 3
x+
x +
x + ........∞ ( x > 0 )
3.4
4.5
5.6
SOLUTION
Since x > 0 , the series is of +ve terms ;
(JNTU 2003)
Engineering Mathematics - I
34
un =
( n + 1) x n > u = ( n + 2 ) x n+1
n +1
( n + 2 )( n + 3)
( n + 3)( n + 4 )
⎡
2
2
2
un +1 ⎡ ( n + 2 ) .x ⎤
⎢ n (1 + 2 n ) .x
=⎢
Lt
⎥ = Lt ⎢
n →∞ u
+
+
1
4
n
n
(
)(
)
⎢
⎥⎦ n→∞ ⎢ n 2 1 + 5 + 4 2
n
⎣
n
n
⎣
Therefore by ratio test, ∑ un converges if x < 1 and diverges if x >1
(
When x = 1, the test fails; Then un =
Taking vn =
u
1
; Lt = n = 1 ≠ 0
→∞
n
n
vn
∴ By comparison test
∑u
n
and
)
⎤
⎥
⎥ = x;
⎥⎦
( n + 1) ;
( n + 2 )( n + 3)
∑v
n
behave same way. But
∑v
n
is divergent by p-
series test. (p = 1);
∴
un is diverges when x = 1
∑
∴ ∑u
n
is convergent when x < 1 and divergent when x ≥ 1
EXAMPLE 38
Discuss the nature of the series
3.6.9.....3n.5n
∑ 4.7.10.....( 3n + 1)( 3n + 2 )
SOLUTION
Here,
3.6.9.....3n
5n
;
un =
4.7.10..... ( 3n + 1) ( 3n + 2 )
un +1 =
3.6.9.....3n ( 3n + 3) 5n +1
;
4.7.10..... ( 3n + 1)( 3n + 4 )( 3n + 5 )
( 3n + 2 )( 3n + 3) .5
un +1
= Lt
n →∞ u
n →∞ ( 3n + 4 )( 3n + 5 )
n
Lt
(
)(
)(
⎡ 5.9n 2 1 + 2
1+ 3
n
3
3n
⎢
= Lt
2
n →∞ ⎢
9n 1 + 4
1+ 5
3n
3n
⎣
∴ By ratio test, ∑ un is divergent.
(
) ⎤⎥ = 5 > 1
) ⎥⎦
(JNTU 2003)
Sequences and Series
35
EXAMPLE 39
∞
Test for convergence the series
∑n
1− n
n =1
SOLUTION
un = n1− n ; un +1 = ( n + 1) ;
−n
u n +1 (n + 1)– n
nn
1⎛ n ⎞
=
=
= ⎜
⎟
1– n
n
un
n ⎝ n +1⎠
n
n(n + 1)
n
n
u
1 ⎛ 1 ⎞⎟
1
Lt n +1 = Lt . ⎜
= 0. = 0 < 1
n →∞ u
n →∞ n ⎜
e
1+ 1 ⎟
n
n⎠
⎝
∴ By ratio test ∑ un , is convergent
EXAMPLE 40
∞
2n3
Test the series ∑
, for convergence.
n =1 n
SOLUTION
2 ( n + 1)
2n3
; un +1 =
un =
n +1
n
3
(
3
2
1+ 1
un +1 2 ( n + 1)
n ( n + 1)
n
=
× 3=
=
3
un
n +1
n
n
2n
u
Lt n +1 = 0 < 1 ;
n →∞ u
n
∴ By ratio test,
∑u
n
is convergent.
EXAMPLE 41
Test convergence of the series
SOLUTION
un =
2n n !
∑ nn
2n +1 ( n + 1) !
2n n !
u
;
=
;
n +1
n +1
nn
( n + 1)
)
2
;
Engineering Mathematics - I
36
n +1
un +1 2 ( n + 1) ! n n
⎛ n ⎞
. n = 2⎜
=
⎟
n +1
un
2 n!
⎝ n +1 ⎠
( n + 1)
un +1
1
= 2 Lt
n →∞ u
n →∞
n
1+ 1
Lt
∴ By ratio test,
(
∑u
n
n2 + 1
3n + 1
∑u
(b)
1 + 2n
(d)
)
=
2
< 1 (since 2 < e < 3)
e
is convergent.
EXAMPLE 42
Test the convergence of the series
(a)
n
n
n
(e)
1 + 3n
n
where un is
x n −1
( 2n + 1)
a
, ( a > 0)
(c)
⎛ 1.2.3....n
⎞
⎜
⎟
⎝ 4.7.10.....3n + 3 ⎠
⎛ 3n3 + 7 n 2 ⎞ n
⎜
⎟x
9
n
5
11
+
⎝
⎠
SOLUTION
(a)
⎡ ( n + 1)2 + 1 3n + 1 ⎤
⎛ un +1 ⎞
Lt ⎜
× 2 ⎥
⎟ = Lt ⎢ n +1
n →∞
n + 1 ⎥⎦
⎝ un ⎠ n→∞ ⎢⎣ 3 + 1
⎡ n 2 ⎛1 + 2 + 2 ⎞ 3 n ⎛1 +
⎜
⎟
⎢ ⎜⎝
n
n2 ⎠
⎝
.
=
⎢
n +1 ⎛
n → ∞ ⎢ n 2 ⎛1 + 1 ⎞
3 ⎜1 +
⎜
⎟
n2 ⎠
⎝
⎝
⎣
Lt
=
∴ By ratio test,
(b)
⎞⎟ ⎤
3n ⎠ ⎥
⎥
⎞
1
n +1 ⎟ ⎥
3 ⎠⎦
1
1
<1
3
∑u
n
is convergent.
a
⎡ xn
2n + 1) ⎤
⎛ un +1 ⎞
(
Lt ⎜
×
⎥
⎟ = Lt ⎢
a
n →∞
x n −1 ⎥⎦
⎝ un ⎠ n →∞ ⎣⎢ ( 2n + 3)
⎡ 2a n a 1 + 1 a ⎤
⎢
2n . x ⎥ = x
= Lt
a
⎢
⎥
n →∞
a a
3
⎢⎣ 2 n 1 + 2n
⎥⎦
(
(
By ratio test,
∑u
n
)
)
convergence if x < 1 and diverges if x > 1.
2
Sequences and Series
37
When x = 1, the test fails; Then, un =
⎛u
Lt ⎜ n
n →∞ v
⎝ n
1
( 2n + 1)
a
a
⎞
1
⎛ n ⎞
⎟ = nLt
⎜
⎟ = nLt
→∞ 2n + 1
→∞
⎝
⎠
⎠
2+ 1
(
∴ By comparison test,
∑u
and
n
n
)
∑v
n
a
=
; Taking vn =
1
we have,
na
1
≠ 0 and finite ( since a > 0 ).
2a
have same property
But p –series test, we have
(i)
∑v
n
convergent when a > 1
and (ii) divergent when a ≤ 1
∴ To sum up, (i) x < 1,
∑u
n
is convergent ∀a .
is divergent ∀a .
(ii)
x > 1,
(iii)
x = 1, a > 1,
(iv)
(c)
∑u
n
∑u
x = 1, a ≤ 1 , ∑ u
n
is convergent, and
n
is divergent.
⎡
1.2.3....n ( n + 1)
4.7.10.... ( 3n + 3) ⎤
u
×
Lt n +1 = Lt ⎢
⎥
n →∞ u
n →∞ 4.7.10.... ( 3n + 3 )( 3n + 6 )
1.2.3....n
n
⎣
⎦
⎡ ( n + 1) ⎤
1
= Lt ⎢
= <1 ;
⎥
n →∞ 3 ( n + 2 )
9
⎣⎢
⎦⎥
2
∴ By ratio test,
(d)
∑u
n
is convergent
⎡ (1 + 2n +1 ) (1 + 3n ) ⎤
un +1
⎥
Lt
= Lt ⎢
×
n +1
n
n →∞ u
n →∞
⎢⎣ (1 + 3 ) (1 + 2 ) ⎥⎦
n
1
2
⎡ n +1 ⎛
1 ⎞ n⎛
1 ⎞⎤
⎢ 2 ⎜1 + 2n +1 ⎟ 3 ⎜1 + 3n ⎟ ⎥
⎝
⎠× ⎝
⎠⎥
= Lt ⎢
n →∞
1
1
⎛
⎞
⎛
⎞
⎢ 3n +1 1 +
2n ⎜1 + n ⎟ ⎥
⎜
⎟
n
+
1
⎢⎣
⎝ 3 ⎠
⎝ 2 ⎠ ⎥⎦
∴ By ratio test,
∑u
n
is convergent.
1
2
⎛2⎞
=⎜ ⎟
⎝3⎠
1
2
<1
2
Engineering Mathematics - I
38
(e)
⎡ 3 ( n + 1)3 + 7 ( n + 1)2 5n9 + 11 ⎤
un +1
= Lt ⎢
× 3
× x⎥
Lt
9
n →∞ u
n →∞
3
7
+
n
+
+
5
1
11
n
(
)
⎢
⎥⎦
n
⎣
(
)
(
3
⎡ 3
2
1
1
⎢ 3n 1 + n + 7 n 1 + n
= Lt ⎢
9
n →∞
+ 11
5n 9 1 + 1
⎢
n
⎣
(
(
)
)
(
)
2
⎤
)
(
⎥
n
5
×
× x⎥
3n (1 + 7
⎥
3n )
⎦
5n9 1 + 11
9
3
3
)
)
)
(
(
3
2
⎡ 3⎧
⎤
1 + 7 1 + 1 ⎫⎬ 5n9 1 + 11
3
1
n
+
⎨
⎢
⎥
9
n
n
3n
5n × x ⎥ = x
⎩
⎭×
= Lt ⎢
9
n →∞
11 ⎫
⎧
⎢
⎥
3n3 1 + 7 3
+ 9⎬
5n 9 ⎨ 1 + 1
n
3
n
⎢⎣
⎥⎦
n
5
⎩
⎭
∴ By ratio test, ∑ un converges when x < 1 and diverges when x > 1.
(
When x = 1, the test fails,
(
3n3 1 + 7
Then
Taking
)
)
(
)
7
3 1 + 3n
un =
= 6
5n 1 + 11
5n9 1 + 11 9
5n
5n9
u
1
3
vn = 6 , we observe that Lt n = ≠ 0
→∞
n
vn 5
n
(
3n
)
(
)
∴ By comparison test and p series test, we conclude that
∴
∑u
n
∑u
n
is convergent.
is convergent when x ≤ 1 and divergent when x > 1.
Exercise – 1.2
1. Test the convergency or divergency of the series whose general term is :
(a)
(b)
(c)
(d)
(e)
xn
...............................
n
nx n −1 ...........................
⎛ 2n − 2 ⎞ n −1
⎜ n
⎟ x ..............
⎝ 2 +1 ⎠
⎛ n2 + 1 ⎞ n
⎜ 2 ⎟ x ...................
⎝ n −1 ⎠
n
nn
........................
[Ans : x < 1cgt , x ≥ 1dgt ]
[Ans : x < 1cgt , x ≥ 1dgt ]
[Ans : x < 1cgt , x ≥ 1dgt ]
[Ans : x < 1cgt , x ≥ 1dgt ]
[Ans: cgt.]
Sequences and Series
4n. n
nn
(f)
..........................
( n + 1)
( 3 + 1)
[Ans: dgt.]
n
3
(g)
39
n
.......................
[Ans: cgt.]
2. Determine whether the following series are convergent or divergent :
x
x2
x3
+
+
+ ..............
1.2 3.4 5.6
x x 2 x3
1 + 2 + 2 + 2 + ..............
2 3 4
1
x
x2
+
+
+ ......
1.2.3 4.5.6 7.8.9
x x 2 x3
xn
1 + + + + ..... 2
+ ....
n +1
2 5 10
1.2 2.3 3.4
+ 2 + 3 + ...............
x
x
x
(a)
(b)
(c)
(d)
(e)
[Ans : x ≤ 1cgt , x > 1dgt ]
[Ans : x ≤ 1cgt , x > 1dgt ]
[Ans : x ≤ 1cgt , x > 1dgt ]
[Ans : x ≤ 1cgt , x > 1dgt ]
[Ans : x > 1cgt , x ≤ 1dgt ]
1.3.4 Raabe’s Test
Let
∑u
n
Then
(i) If k > 1,
Proof :
⎧⎪ ⎛ un
⎞ ⎫⎪
− 1⎟ ⎬ = k
⎪⎩ ⎝ un +1 ⎠ ⎪⎭
be series of +ve terms and let Lt ⎨n ⎜
n →∞
∑u
n
is convergent. (ii) If k < 1,
Consider the series
∑u
n
is divergent. (The test fails if k = 1)
1
∑v = ∑ n
n
p
⎡⎛ n + 1 ⎞ p ⎤
⎡⎛ 1 ⎞ p ⎤
⎡ vn
⎤
n⎢
− 1⎥ = n ⎢⎜
⎟ − 1⎥ = n ⎢⎜1 + ⎟ − 1⎥
⎢⎣⎝ n ⎠
⎥⎦
⎢⎣⎝ n ⎠
⎥⎦
⎣ vn +1 ⎦
⎡⎛
⎞ ⎤
p p ( p − 1) 1
. 2 + .... ⎟ − 1⎥
= n ⎢⎜ 1 + +
2
n
⎢⎣⎝ n
⎠ ⎥⎦
p ( p − 1) 1
⎧v
⎫
. + ..........
= p+
Lt n ⎨ n − 1⎬ = p
n →∞
2
n
⎩ vn +1 ⎭
Engineering Mathematics - I
40
⎧ un
⎫
− 1⎬ = k > 1
n →∞
⎩ un +1 ⎭
We choose a number ‘p’ ∋ k > p > 1 ; Comparing the series
Case (i) In this case, Lt n ⎨
∑v
n
which is convergent , we get that
∑u
n
∑u
n
with
will converge if after some
fixed number of terms
p
i.e. If,
i.e., If
un
v
⎛ n +1⎞
> n =⎜
⎟
un +1 vn +1 ⎝ n ⎠
p ( p − 1) 1
⎛ u
⎞
n ⎜ n − 1⎟ > p +
. + .........from (1)
∠
u
2
n
⎝ n +1 ⎠
⎛ u
⎞
Lt n ⎜ n − 1⎟ > p
n →∞
⎝ un +1 ⎠
i.e., If k > p, which is true . Hence
∑u
n
is convergent .The second case also
can be proved similarly.
Solved Examples
EXAMPLE 43
Test for convergence the series
1 x3 1.3 x5 1.3.5 x 7
. +
. + .....
x+ . +
2 3 2.4 5 2.4.6 7
(JNTU 2006, 2008)
SOLUTION
Neglecting the first tem ,the series can be taken as ,
1 x3 1.3 x5 1.3.5 x 7
. +
. +
. + .....
2 3 2.4 5 2.4.6 7
1.3.5....are in A.P. nth term = 1 + ( n − 1) 2 = 2n − 1
2.4.6...are in A.p. nth term = 2 + ( n − 1) 2 = 2n
3.5.7.....are in A.P nth term = 3 + ( n − 1) 2 = 2n + 1
∴
un ( nth term of the series) =
1.3.5.... ( 2n − 1) x 2 n +1
.
2.4.6.... ( 2n ) 2n + 1
Sequences and Series
41
1.3.5.... ( 2n − 1)( 2n + 1) x 2 n +3
un +1 =
.
2.4.6.... ( 2n )( 2n + 2 ) 2n + 3
un +1 1.3.5.... ( 2n + 1) x 2 n +3
2.4.6....2n ( 2n + 1)
.
.
.
=
2.4.6.... ( 2n + 2 ) ( 2n + 3) 1.3.5.... ( 2n − 1) x 2 n +1
un
( 2n + 1) x 2
( 2n + 2 )( 2n + 3)
2
=
2
1 ⎞
⎛
4n ⎜ 1 + ⎟
u
⎝ 2n ⎠
∴
Lt n +1 = Lt
x2 = x2
n →∞ u
n →∞
2
3
⎛
⎞
⎛
⎞
n
4n 2 ⎜ 1 + ⎟ ⎜ 1 + ⎟
⎝ 2n ⎠ ⎝ 2n ⎠
∴ By ratio test, ∑ un converges if x < 1 and diverges if x > 1
2
If x = 1 the test fails.
x2 = 1
Then
and
( 2n + 2 )( 2n + 3)
un
=
2
un +1
( 2n + 1)
( 2n + 2 )( 2n + 3) − 1 = 6n + 5
un
−1 =
2
2
un +1
( 2n + 1)
( 2n + 1)
⎧⎪ ⎛ un
⎞ ⎫⎪
⎛ 6n 2 + 5n ⎞
Lt ⎨n ⎜
− 1⎟ ⎬ = Lt ⎜ 2
⎟
n →∞
n →∞ 4n + 4n + 1
⎝
⎠
⎩⎪ ⎝ un +1 ⎠ ⎭⎪
5⎞
⎛
n2 ⎜ 6 + ⎟
3
n⎠
⎝
= Lt
= >1
n →∞
4 1 ⎞ 2
⎛
n2 ⎜ 4 + + 2 ⎟
n n ⎠
⎝
By Raabe’s test, ∑ un converges. Hence the given series is convergent when x ≤ 1 an
divergent when x > 1 .
EXAMPLE 44
Test for the convergence of the series
1+
3
3.6 2
3.6.9 3
x+
x +
x + .....; x > 0
7
7.10
7.10.13
(JNTU 2007)
Engineering Mathematics - I
42
SOLUTION
Neglecting the first term,
un =
3.6.9....3n
.x n
7.10.13....3n + 4
un +1 =
3.6.9....3n ( 3n + 3)
.x n +1
7.10.13.... ( 3n + 4 )( 3n + 7 )
un +1 3n + 3
u
.x ; Lt n +1 = x
=
n →∞ u
3n + 7
un
n
∑u
∴ By ratio test,
n
is convergent when x < 1 and divergent when x > 1.
When x = 1 The ratio test fails. Then
un
3n + 7 un
4
;
=
−1 =
3n + 3
un +1 3n + 3 un +1
⎧⎪ ⎛ u
⎞ ⎫⎪
⎛ 4n ⎞ 4
Lt ⎨n ⎜ n − 1⎟ ⎬ = Lt ⎜
⎟ = >1
n →∞
n →∞ 3n + 3
⎝
⎠ 3
⎩⎪ ⎝ un +1 ⎠ ⎭⎪
∴ By Raabe’s test,
∑u
n
is convergent .Hence the given series converges if x ≤ 1 and
diverges if x > 1.
EXAMPLE 45
Examine the convergence of the series
∞
12.52.92.... ( 4n − 3)
n =1
42.82.122.... ( 4n )
∑
2
2
SOLUTION
un =
12.52.92.... ( 4n − 3)
42.82.122.... ( 4n )
2
2
;
2
12.52.92.... ( 4n − 3) ( 4n + 1)
un +1 =
42.82.122.... ( 4n ) ( 4n + 4 )
2
2
2
( 4n + 1) = 1 (verify)
u
Lt n +1 = Lt
2
n →∞ u
n →∞
( 4n + 4 )
n
2
∴ The ratio test fails. Hence by Raabe’s test,
∑u
n
is convergent. (give proof)
Sequences and Series
43
EXAMPLE 46
Find the nature of the series
∑
( n)
2
xn , ( x > 0)
2n
(JNTU 2003)
SOLUTION
un
( n)
=
2
2n
n
.x ; un +1
( n + 1)
=
2
2n + 2
( n + 1)
un +1
x;
=
un
( 2n + 1)( 2n + 2 )
.x n +1
2
(
)
2
n2 1 + 1
un +1
x
n
Lt
.x =
= Lt
n →∞ u
n →∞
4
4n 2 1 + 1
1+ 2
n
2n
2n
(
∴ By ratio test,
∑u
n
)(
converges when
)
x
< 1 , i. e ; x < 4; and diverges when x >4;
4
When x = 4, the test fails.
( 2n + 1)( 2n + 2 )
un
=
2
un +1
4 ( n + 1)
un
−2n − 2
−1
;
−1 =
=
2
2 ( n + 1)
un +1
4 ( n + 1)
∴ By ratio test,
Hence
∑u
∑u
n
⎡ ⎛ u
⎞ ⎤ −1
Lt ⎢ n ⎜ n − 1⎟ ⎥ =
<1
n →∞
u
2
n
+
1
⎝
⎠
⎣
⎦
is divergent
is convergent when x < 4 and divergent when x > 4
n
EXAMPLE 47
Test for convergence of the series
SOLUTION
un =
∑
4.7.... ( 3n + 1) n
x
1.2.3....n
4.7.... ( 3n + 1) n
4.7.... ( 3n + 1)( 3n + 4 ) n +1
x ; un +1 =
x
1.2.3....n
1.2.3....n ( n + 1) .
⎡ ( 3n + 4 ) ⎤
un +1
= Lt ⎢
.x ⎥ = 3 x
n →∞ u
n →∞
+
n
1
(
)
n
⎣
⎦
Lt
(JNTU 1996)
Engineering Mathematics - I
44
∴ By ratio test
∑u
n
converges if 3 x < 1 i.e., x <
1
1
and diverges if x > ;
3
3
1
, the test fails
3
⎤
1 ⎡u
1
⎡ −1 ⎤
⎡ (n + 1)3 ⎤
x = , n ⎢ n – 1⎥ = n ⎢
– 1⎥ = n ⎢
When
=−
⎥
4⎞
3 ⎣ u n +1 ⎦ ⎣ 3n + 4 ⎦
⎛
⎣ 3n + 4 ⎦
⎜3+ ⎟
n⎠
⎝
⎡u
⎤
1
Lt n ⎢ n − 1⎥ = − < 1
n →∞
3
⎣ un +1 ⎦
If x =
∴ By Raabe’s test,
∴
∑u
n
∑u
n
is divergent.
is convergent when x <
1
1
and divergent when x ≥
3
3
EXAMPLE 48
Test for convergence 2 +
SOLUTION
The nth term
un =
3x 4 x 2 5 x3
+
+
+ ........... ( x > 0 )
2
3
4
( n + 1) x n−1 ;
n
un +1 =
(
(
( n + 2) xn ;
( n + 1)
)
)
(JNTU 2003)
un +1 n ( n + 2 )
=
.x
2
un
( n + 1)
n2 1 + 2
un +1
n .x = x
Lt
= Lt
2
n →∞ u
n →∞ 2
n
n 1+ 1
n
∴ By ratio test, ∑ un is convergent if x < 1 and divergent if x > 1
If x = 1, the test fails.
⎡ ( n + 1)2
⎤
⎡
⎤
⎡ un
⎤
1
Lt n ⎢
− 1⎥ = Lt n ⎢
− 1⎥ = Lt n ⎢
⎥ = 0 <1
n →∞
n →∞
⎣ un +1 ⎦ n→∞ ⎢⎣ n ( n + 2 ) ⎥⎦
⎣ n ( n + 2) ⎦
Then
∴ By Raabe’s test
∴
∑u
n
∑u
n
is divergent
is convergent when x < 1 and divergent when x ≥ 1
EXAMPLE 49
Find the nature of the series
3 3.6 3.6.9
+
+
+ ......∞
4 4.7 4.7.10
(JNTU 2003)
Sequences and Series
SOLUTION
un =
45
3.6.9.....3n ( 3n + 3)
3.6.9.....3n
; un +1 =
4.7.10..... ( 3n + 1)( 3n + 4 )
4.7.10..... ( 3n + 1)
(
(
)
)
3n 1 + 3
un +1 3n + 3
un +1
3n = 1
=
; Lt
= Lt
n
→∞
n
→∞
3n + 4
un
un
3n 1 + 4
3n
Ratio test fails.
⎡ ⎧u
⎫⎤
⎡ ⎛ 3n + 4 ⎞ ⎤
− 1⎟ ⎥
Lt ⎢ n ⎨ n − 1⎬⎥ = Lt ⎢ n ⎜
n →∞
n →∞
⎣ ⎝ 3n + 3 ⎠ ⎦
⎣ ⎩ un +1 ⎭⎦
n
n
1
= Lt
= Lt
= <1
n →∞ 3 ( n + 1)
n →∞
3
3n 1 + 1
n
∴ By Raabe’s test ∑ un is divergent.
∴
(
)
EXAMPLE 50
If p, q > 0 and the series
1+
1 p 1.3. p ( p + 1) 1.3.5 p ( p + 1)( p + 2 )
+
+
+ ....
2 q 2.4.q ( q + 1) 2.4.6 q ( q + 1)( q + 2 )
is convergent , find the relation to be satisfied by p and q.
SOLUTION
un =
1.3.5..... ( 2n − 1) p ( p + 1) ..... ( p + n − 1)
[neglecting 1st term]
2.4.6.....2n q ( q + 1) ..... ( q + n − 1)
un +1 =
1.3.5..... ( 2n − 1)( 2n + 1) p ( p + 1) ..... ( p + n − 1)( p + n )
2.4.6.....2n ( 2n + 2 ) q ( q + 1) ..... ( q + n − 1)( q + n )
un +1 ( 2n + 1) ( p + n )
=
;
un
( 2n + 2 ) ( q + n )
(
(
)
)
(
(
⎡
1
n 1+ p
⎢ 2 n 1 + 2n
n
un +1
Lt
.
= Lt ⎢
n →∞ u
n →∞
q
1
n
⎢ 2 n 1 + 2n n 1 + n
⎣
∴ ratio test fails.
Let us apply Raabe’s test
) ⎤⎥⎥ = 1
) ⎥⎦
Engineering Mathematics - I
46
⎡ ⎧⎪ ( q + n )( 2n + 2 ) ⎫⎪⎤
⎡ ⎛ u
⎞⎤
Lt ⎢ n ⎜ n − 1⎟ ⎥ = Lt ⎢ n ⎨
− 1⎬⎥
n →∞
n →∞
+
+
u
p
n
n
2
1
(
)(
)
⎪
⎣ ⎝ n +1 ⎠ ⎦
⎭⎪⎦⎥
⎣⎢ ⎩
⎡ ⎧
⎫⎤
⎢ ⎪ 2q ( n + 1) − p ( 2n + 1) + n ⎪⎥
Lt ⎢ n ⎨
⎬⎥
n →∞
⎪⎥
2+ 1
⎢ ⎪ n2 1 + p
n
n ⎭⎦
⎣ ⎩
(
(
)(
) (
)
)
⎡ 2q 1 + 1 − p 2 + 1 + 1 ⎤
n
n
⎥ = 2q − 2 p + 1
Lt ⎢
n →∞ ⎢
⎥
2
2
⎣
⎦
2q − 2 p + 1
Since ∑ un is convergent, by Raabe’s test,
>1
2
⇒ q − p > 1 , is the required relation.
2
Exercise 1.3
∞
1. Test whether the series
∑u
n
is convergent or divergent where
1
22.42.62..... ( 2n − 2 )
un =
.x 2 n
3.4.5..... ( 2n − 1)( 2n )
2
[Ans : x ≤ 1cgt , x > 1dgt ]
2. Test for the convergence the series
∞
∑
1
4.7.10..... ( 3n + 1) n
x
n
[Ans : x <
1
1
cgt , x ≥ dgt ]
3
3
3. Test for the convergence the series :
(i)
22.42 22.42.52.7 2 22.42.52.7 2.82.102
+
+
+ ...
32.32 32.32.62.62 32.32.62.62.92.92
(ii)
3.4
4.5 2 5.6 3
x+
x +
x + ..... ( x > 0 )
1.2
2.3
3.4
(iii)
1.3.5..... ( 2n − 1)
xn
.
∑ 2.4.6.....2n ( 2n + 2 ) ( x > 0 ) [Ans : cgt if x ≤ 1 dgt if , x > 1]
[Ans : divergent]
[Ans : cgt if x ≤ 1 dgt if , x > 1]
Sequences and Series
( 1)
1+
(iv)
2
2
47
( 2)
x+
2
x2
4
( 3)
+
2
x3
6
+ ...... ( x > 0 )
[Ans : cgt if x < 4 and dgt if , x ≥ 4 ]
1.3.5 Cauchy’s Root Test
Let
∑u
n
be a series of +ve terms and let Lt un
1
= l . Then
n
n →∞
∑u
n
is convergent when
l < 1 and divergent when l > 1
Proof : (i) Lt un
1
= l < 1 ⇒ ∃a +ve number ' λ ' ( l < λ < 1) ∋ un
n
n →∞
1
n
< λ , ∀n > m
un < λ n , ∀n > m
(or)
λ < 1, ∑ λ n is a geometric series with common ratio < 1 and
Since
therefore convergent.
Hence
un is convergent.
∑
(ii) Lt un
1
n
n →∞
= l >1
∴ By the definition of a limit we can find a number r ∋ un
i.e., un > ∀n > r
1
> 1∀n > r
n
i.e., after the 1st ‘r ‘ terms , each term is > 1.
Lt
un = ∞ ∴
un is divergent.
n →∞
When
Note :
Lt ( u
n →∞
∑
∑
1
n
n
) = 1, the root test can’t decide the nature of ∑ u
n
. The fact of
this statement can be observed by the following two examples.
1.
2.
Consider the series
Consider the series
∑
1
n3
: − Ltun
1
n
n →∞
∑ 1 n , in which Ltu
n →∞
In both the examples given above,
Ltu
n →∞
1
n
n
1
n
n
1
3
⎛ 1 ⎞
= Lt ⎜ 1 ⎟ = 1
n →∞ ⎝ n n ⎠
1
= Lt 1 = 1
n →∞ n n
⎛ 1⎞
= Lt ⎜ 3 ⎟
n →∞ ⎝ n ⎠
n
= 1 . But series (1) is convergent
(p-series test)
And series (2) is divergent. Hence when the limit=1, the test fails.
Engineering Mathematics - I
48
Solved Examples
EXAMPLE 51
Test for convergence the infinite series whose nth terms are:
(i)
1
n2n
(ii)
1
(log n) n
(iii)
1
⎡ 1⎤
⎢⎣1 + n ⎥⎦
(JNTU 1996, 1998, 2001)
n2
SOLUTION
(i)
(ii)
(iii)
1
1
1
1
1
, un n = 2 ; Lt un n = Lt 2 = 0 < 1;
2n
n →∞
n →∞ n
n
n
By root test ∑ un is convergent.
un =
1
1
1
1
1
; Lt un n = Lt
= 0 < 1;
; un n =
n
→∞
→∞
n
n
(log n)
log n
log n
∴ By root test , ∑ un is convergent.
un =
un =
1
⎛ 1⎞
⎜1 + ⎟
⎝ n⎠
n2
∴ By root test
; un
1
n
∑u
n
=
1
⎛ 1⎞
⎜1 + ⎟
⎝ n⎠
n
Ltu
n →∞
1
n
n
= Lt
n →∞
1
⎛ 1⎞
⎜1 + ⎟
⎝ n⎠
n
=
1
< 1;
e
is convergent.
EXAMPLE 52
Find whether the following series are convergent or divergent.
∞
1
(i) ∑ n
n =1 3 − 1
(iii)
SOLUTION
(i)
un
1
n
⎛ 1 ⎞
=⎜ n ⎟
⎝ 3 −1 ⎠
1
n
⎛
⎞
⎜
⎟
1
⎟
=⎜
⎜ 3n ⎛1 − 1 ⎞ ⎟
⎜ ⎜ 3n ⎟ ⎟
⎠⎠
⎝ ⎝
⎡⎣( n + 1) x ⎤⎦
∑
n n +1
n =1
∞
1 1 1
(ii) 1 + 2 + 3 + 4 + .....
2 3 4
1
n
n
Sequences and Series
Lt un
1
n
n →∞
49
⎛
⎜
1
= Lt ⎜
n →∞
1
⎜ n⎛
⎜ 3 ⎜1 − 3n
⎝ ⎝
⎞
⎟
⎟
⎞⎟
⎟⎟
⎠⎠
1
n
1
1
⎛ 1 ⎞
un = n ; Lt un n = Lt ⎜ n ⎟
→∞
→∞
n
n
n
⎝n ⎠
(ii)
(iii)
⎡( n + 1) x ⎤⎦
un = ⎣
n n +1
Lt un
1
n
n →∞
1
< 1 ; By root test,
3
∑u
= 0 < 1 ; By root test,
∑u
=
1
n
n
n
is convergent.
is convergent.
n
⎡ {( n + 1) x}n ⎤
⎥
= Lt ⎢
n →∞ ⎢
n n +1
⎥
⎣
⎦
1
n
1
⎡ ⎧ ( n + 1) x ⎫n 1 ⎤ n
⎛ n +1⎞ 1
Lt ⎢ ⎨
⎬ . ⎥ = nLt
⎜
⎟ x. 1
→∞
n →∞
⎝ n ⎠ n n
⎢⎣ ⎩ n ⎭ n ⎥⎦
⎛
⎞
1
1
⎛ 1⎞ 1
Lt ⎜ 1 + ⎟ x. 1 = Lt x. 1 = x
since Lt x. 1 = 1⎟
⎜
n →∞
n →∞
n →∞
⎝ n⎠ n n
n n
n n
⎝
⎠
∴ ∑ un is convergent if x < 1 and divergent if x > 1 and when x = 1 the
test fails.
Then un
( n + 1)
=
n
n
vn =
; Take
n +1
1
n
( n + 1) = ⎛1 + 1 ⎞ ;
un ( n + 1)
=
.n =
⎜
⎟
n +1
vn
n
nn
⎝ n⎠
∴ By comparison test, ∑ un is divergent.
n
∑v
Hence ∑ u
(
n
If un =
diverges by p –series test )
n
is convergent if x < 1 and divergent x ≥ 1
2
( n + 1)
n2
un
= e >1
n →∞ v
n
Lt
n
EXAMPLE 53
nn
n
, show that
∑u
n
is convergent.
Engineering Mathematics - I
50
Lt un
1
n
n →∞
⎡
n2
n
= Lt ⎢
n2
n →∞ ⎢
+
1
n
(
)
⎣
⎤
⎥
⎥
⎦
1
n
nn
; = Lt =
( n + 1)
n →∞
n
⎛ n ⎞
= Lt ⎜
⎟
n →∞ n + 1
⎝
⎠
n
n
⎛
⎞
⎜ 1 ⎟ 1
= Lt ⎜
⎟ = <1; ∴
n →∞
e
⎜ 1+ 1 ⎟
⎝ n⎠
∑u
n
converges by root test .
EXAMPLE 54
2
3
1 ⎛2⎞ ⎛3⎞
Establish the convergence of the series
+ ⎜ ⎟ + ⎜ ⎟ + ........
3 ⎝5⎠ ⎝7⎠
SOLUTION
n
⎛ n ⎞
un = ⎜
⎟ .........(verify);
⎝ 2n + 1 ⎠
By root test, ∑ un is convergent.
Lt un
n →∞
1
n
⎛ n ⎞ 1
= Lt ⎜
⎟ = <1
n →∞ 2 n + 1
⎝
⎠ 2
EXAMPLE 55
∞
Test for the convergence of
∑
n =1
n n
.x
n +1
SOLUTION
1
1
⎛
⎞2
⎛
⎞2
⎜
⎜ 1 ⎟ n
1
1 ⎟
.x ; Lt un n = Lt ⎜
un = ⎜
⎟ .x = x
⎟
n →∞
n →∞
1
1
⎜ 1+ ⎟
⎜ 1+ ⎟
⎝ n⎠
⎝ n⎠
∴ By root test, ∑ un is convergent if x < 1 and divergent if x > 1 .
When x = 1 : un =
1
n
, taking vn = 0 and applying comparison test , it can be
n
n +1
seen that is divergent
un is convergent if x < 1 and divergent if x ≥ 1 .
∑
EXAMPLE 56
∑(
∞
Show that
n =1
1
)
n
n n − 1 converges.
Sequences and Series
SOLUTION
(
un = n
Lt un
1
1
n
n →∞
∑u
∴
n
51
)
−1
n
(
)
1
(
= Lt n n − 1 = 1 − 1 = 0 < 1 since Lt n
n →∞
1
n
n →∞
)
=1 ;
is convergent by root test.
n
EXAMPLE 57
n
Examine the convergence of the series whose n
th
⎛n+2⎞ n
term is ⎜
⎟ .x
⎝ n+3⎠
SOLUTION
n
1
⎛n+2⎞ n
⎛n+2⎞
un = ⎜
un n = Lt ⎜
⎟x = x
⎟ .x ; nLt
n
→∞
→∞
⎝ n+3⎠
⎝ n+3⎠
∴ By root test, ∑ un converges when x < 1 and diverges when x > 1 .
n
⎛ 2⎞
n
⎜1 + ⎟
n⎠
⎛n+2⎞
un = Lt ⎝
When x = 1 : un = ⎜
⎟ ; nLt
n
→∞
→∞
n
⎝ n+3⎠
⎛ 3⎞
1
+
⎜
⎟
⎝ n⎠
e2 1
= ≠ 0 and the terms are all +ve .
=
e3 e
∴ ∑ un is divergent . Hence ∑ un is convergent if x < 1 and divergent if x ≥ 1 .
EXAMPLE 58
Show that the series,
−1
−2
−3
⎡ 22 2 ⎤
⎡ 33 3 ⎤
⎡ 44 4 ⎤
−
+
−
+
⎢ 12 1 ⎥
⎢ 23 2 ⎥
⎢ 34 − 3 ⎥ + ...... is convergent
⎣
⎦
⎣
⎦
⎣
⎦
−n
⎡ ( n + 1)n +1 n + 1 ⎤
−
un = ⎢
⎥ ;=
n +1
n ⎥⎦
⎢⎣ n
⎛ 1⎞
⎜1 + ⎟
⎝ n⎠
−n
−n
⎛ n +1⎞
⎜
⎟
⎝ n ⎠
−n
⎡⎛ n + 1 ⎞ n ⎤
⎢⎜
⎟ − 1⎥
⎢⎣⎝ n ⎠
⎥⎦
(JNTU 2002)
−n
−1
n
⎡⎛ 1 ⎞ n ⎤
⎤
1
⎛ 1 ⎞ ⎡⎛ 1 ⎞
n
+
−
=
+
+
;
u
1
1
1
1
⎢⎜
⎥
n
⎜
⎟ ⎢⎜
⎟ − 1⎥
⎟
⎝ n ⎠ ⎢⎣⎝ n ⎠
⎥⎦
⎢⎣⎝ n ⎠
⎥⎦
−1
Engineering Mathematics - I
52
=
1
1
⎛ 1 ⎞ ⎪⎧⎛ 1 ⎞ n ⎪⎫
⎜1 + ⎟ 1 +
− 1⎬
⎝ n ⎠ ⎨⎪⎜⎝ n ⎟⎠
⎩
⎭⎪
1
1 1
1
∴ Lt un n = .
=
<1
n →∞
1 e −1 e −1
∴ By root test, ∑ un is convergent.
EXAMPLE 59
∞
Test
∑u
m =1
m
for convergence when um =
e− m
(
1+ 2
m
)
− m2
SOLUTION
( )
Lt um
m →∞
1
m
(
⎡
2
⎢ 1+ m
= Lt ⎢
m →∞
em
⎢
⎣
Hence Cauchy’s root tells us that
)
m2
∑u
1
⎤
⎥
⎥
⎥
⎦
m
m
1⎛
2⎞
e2
1
+
=
= e >1
⎜
⎟
m →∞ e
e
⎝ m⎠
; Lt
is divergent.
m
EXAMPLE 60
Test the convergence of the series
n
∑e
n2
.
SOLUTION
1
Lt un
1
n →∞
n
n n
= Lt n = 0 < 1
n →∞ e
∴ By root test,
∑u
n
is convergent.
EXAMPLE 61
( n + 1) .x + ......, x > 0
2
32
Test the convergence of the series, 2 x + 3 x 2 + ...
1
2
n n +1
n
n
SOLUTION
Lt un
n →∞
1
n
⎡ ( n + 1)n .x n ⎤
= Lt ⎢
⎥
n +1
n →∞
⎢⎣ n
⎥⎦
1
n
⎡⎛ n + 1 ⎞ 1 ⎤
= Lt ⎢⎜
⎟ . 1 n .x ⎥
n →∞
n
⎝
⎠ n
⎣
⎦
Sequences and Series
53
1
⎡⎛ 1 ⎞ 1 ⎤
= Lt ⎢⎜1 + ⎟ . 1 .x ⎥ = 1.1.x = x ⎡since Lt n n = 1⎤
⎢
⎥⎦
n →∞
n →∞
⎣
⎣⎝ n ⎠ n n ⎦
∴ By root test, ∑ un converges if x < 1 and diverges when x > 1 .
When x = 1 , the test fails.
n
1
⎛ 1⎞ 1
un = ⎜ 1 + ⎟ . ; Take vn =
n
⎝ n⎠ n
Then
n
un
⎛ 1⎞
= Lt ⎜1 + ⎟ = e ≠ 0
n →∞ v
n →∞
⎝ n⎠
n
Lt
∴ By comparison test and p-series test,
Hence
∑u
∑u
n
is divergent.
is convergent when x < 1 and divergent when x ≥ 1 .
n
Exercise 1.4
1. Test for convergence the infinite series whose nth terms are:
(a)
(b)
1
2 −1
1
.............................
n
( log )
. ( n ≠ 1) ......................
2n
[Ans : convergent]
[Ans : convergent]
n
(c)
(d)
(e)
(f)
(g)
(h)
⎛ 3n + 1 ⎞
.x ⎟ .................................
⎜
⎝ 4n + 3 ⎠
xn
............................................
n
n
.............................................
nn
3n.∠n
........................................
n3
( 2n
(
2
− 1)
( 2n )
n
1
n
)
4
4
cgt , x ≥ dgt ]
3
3
[Ans : cgt for all x ≥ 0 ]
[Ans : convergent]
[Ans : convergent]
n
2n
−1
[Ans : x <
...................................
[Ans : convergent]
...................................
[Ans : convergent]
2n