Series: Guide to Investigating
Convergence
Understanding the Convergence of
a Series
A series converges to λ if the limit
of the sequence of the partial sums
of the series is equal to λ
Example (1)
The Geom etric Series

 ar
n 1
n 1
Where a and r are fixed real num bers
Consider the geom etric series

s
n 0

n
  ar n 1
n 0
The sequenceof the n  th partial sum of the series is  Tn  ,
where :
Tn  a  ar  ar 2    ar n  2  ar n 1
(1)
A form ula for Tn is arrived at by m ultiplying equation(1)
by r and subtracting it from the resulting equation
rTn  ar  ar 2  ar 3    ar n 1  ar n
(2)
Subtracting (1) from (2) , we get
rTn  Tn  ar n  a
 Tn (r  1)  a (r n  1)
a (r n  1) ar n
a
 Tn 


r 1
r 1 r 1
Thus; The sequenceof the n  th partial sum of the series is :
ar n
a
 Tn  


r 1 r 1
The lim Tn of this sequencedependson r :
n 
1. If r  1, we get :
ar n
a
a
a
lim Tn  lim(

)  0(
)
n 
n  r  1
r 1
r 1 1 r
a
and so the series convergesto
1 r

a
n 1
We write  ar 
; r 1
1 r
n 0
1. If r  1, we get :
lim Tn does not exist
n 
and so the series diverges
Warning
Please, distinguish between:
1. The sequence Tn  of the n  th partial sum s of the series
ar n
a
 Tn   


r 1 r 1
&
2. The sequence sn  of the n  th term of the series
 sn    ar n 1 
Notice that :
a
1. For r  1 : lim Tn 
, While lim sn  lim ar n 1  0
n 
n 
n 
r 1
2. For r  1 and a positive: lim Tn   , While lim sn  lim a  a
n 
n 
n 
The Sum of the series
By, definition, the sum of the series, when exists, is equalto :
The l i m i t of the sequence Tn  of the n  th partial sum s of the series
Thus,

t
nk
n
 lim Tn
n 
Questions
Check, whether the given series is convergent, and if convergent find
its sum

n2

3 n 1

2 n2
5
1.  2 n 1
n 0 7
2
2.  n  2
n 0 7
2
3.  3 n 1
n 1 3
Example (2)
Telescoping Series
Exam ples(1)

Consider the series  t n ,
nk
Assum ethat there is a sequence sn , such that
t n  sn  sn 1 ; n  k
Let  Tn  be the sequenceof the n  th partial sum of the series
n
t
i k
n
So,
Tn  t k  t k 1  t k  2  t k 3    t n  2  t n 1  t n
 ( sk  sk 1 )  ( sk 1  sk  2 )  ( sk  2  sk 3 )    ( sn 1  sn )  ( sn  sn 1 )
 sk   sn 1
Thus,
 Tn   sk  sn 1 
and so if  sn  convergesto a real num ber , then  Tn  convergesto sk  
Warning
Please, distinguish between:
1. The sequence Tn  of the n  th partial sum s of the series
 Tn    s k  s n 1 
&
2. The sequence t n  of the n  th term of the series
 tn 
Notice that :
If  s n  converges, then :
lim Tn  s k  lim s n ,
n 
n 
While
lim t n  0
n 
Examples of this type of telescoping series
A Convergent Telescoping Series
1.


7
2
n4
n  4 n  5n  6
7
7
We have, 2

n  5n  6 (n  2)(n  3)
7
7


( show that!)
n2 n3
7
Let  sn 

n2
t n  sn  sn 1
Consider the series  t n  

7
Thus  2
 s4  lim sn
n 
n  4 n  5n  6
7
7

0 
42
6
Solutions

2. Let

2
n 1 n ( n  1)
 tn  
n 1
we have:
2
2
2
2
tn 

 
n(n  1) n  1 n n(n  1)
Let
2
< s n > < >
n
T hen t n  s n  s n 1
But < s n > convergest o 0
T hus,

2
 s1  lim s n  2  0  2

n 
n 1 n ( n  1)
Exam ples(2)

Consider the series  t n ,
nk
Assum ethat there is a sequence sn , such that
t n  sn 1  sn
;nk
Let  Tn  be the sequenceof the n  th partial sum of the series
n
t
i k
n
So,
Tn  t k  t k 1  t k  2  t k 3    t n  2  t n 1  t n
 ( sk 1  sk )  ( sk  2  sk 1 )  ( sk 3  sk  2 )    ( sn  sn 1 )  ( sn 1  sn )
  sk  sn 1
Thus,
 Tn   sn 1  sk 
and so if  sn  convergesto a real num ber , then  Tn  convergesto   sk
Examples of this type of telescoping series
A Divergent Telescoping Series


1
Consider the series  t n   ln(1  )
n
n 1
n 1
1
n 1
We have, ln(1  )  ln(
)  ln(n  1)  ln n
n
n
Let  s n  ln n 
Then,
t n  s n 1  s n
Thus the sequence Tn  of the n  th partial sum s of the series is :
 Tn   s1  s n 1  0  ln(n  1)  ln(n  1) 
 lim Tn  ln(n  1)  
n 
Questions I
Check, whether the given series is
convergent, and if convergent find its sum

2
1. 
n 1 n( n  1)

2.
n
n2

2
1
n
3. 
(n  1)!

4
 [arctann  arctan(n  1)]
n 1
Questions II
Show that the following series is a telescoping series, and then determine
whether it is convergent

1.
(
n  n  1)
n 1

2.
 sin n
n2
H int
2 sin n sin 12 cos(n  12 )  cos(n  12 )
sin n 

1
2 sin 2
2 sin 12

cos(
2n  1
2n  1
2n  1
2n  1
)  cos(
) cos(
) cos(
)
2
2
2
2


1
1
2 sin 2
2 sin 2
2 sin 12
Let :
sn 
2n  1
)
2
2 sin 12
cos(
then, s n 1 
Thus, sin n  s n 1  s n
cos(
2(n  1)  1
2n  1
) cos(
)
2
2

2 sin 12
2 sin 12

1.
(
n  n  1)
n 1

 sin n
2.
n2
H int
sin n 

cos(
2 sin n sin 12 cos(n  12 )  cos(n  12 )

2 sin 12
2 sin 12
2n  1
2n  1
2n  1
2n  1
)  cos(
) cos(
) cos(
)
2
2
2
2


2 sin 12
2 sin 12
2 sin 12
Let :
sn 
2n  1
)
2
2 sin 12
cos(
then, sn 1 
Thus, sin n  sn 1  sn
cos(
2(n  1)  1
2n  1
) cos(
)
2
2

2 sin 12
2 sin 12


5
2. Let  t n  
n 1
n 1 ( n  2)(n  3)
we have:
5
5
5
5
tn 



(n  2)(n  3) n  2 n  3 (n  2)(n  3)
Let
5
< s n > <
>
n2
T hen t n  s n  s n 1
But < s n > convergesto 0
T hus,

5
5
5
 s1  0 
0 

12
12
n 1 ( n  2)(n  3)
The Integral Test
Let a function f be positive decreasing and continuouson [ K , ).
& let  sn  be a sequencesatisfying : sn  f (n) ; n  K
; for som e
naturalnum berK


n K
K
Then either the series  sn and the im properintegral  f ( x)dx
both convergeor both diverge
Example (3)
The Hyperharm onic Series
(The p  series)

1

p
n 1 n
Where p  0
Consider the p  series


1
p
n 0 n
 sn  
n 1
Write :
1
; n N
np
Consider the function;
1
f ( x)  p ; x  [1, )
x
We have f is positive, continuousand decreasin g
f ( n)  s n 
By the int egral test :


1
1
converges
iff
the
im
proper
int
egral
dx converges

p
p

n 0 n
1 x

But, we know that

Thus
1

p
n 0 n
1
1 x p dx
converges if p  1 and diverges if p  1
converges if p  1 and diverges if p  1
Warning
Please, distinguish between:


1
p
n 1 n
1. The convergence of the series  sn  
n 1
&
2. The convergence of the sequence sn   
1
 of the n  th term of the series
p
n
Notice that :


1
diverges
p
n 1 n
1. For p  , the series  sn  
1
3
n 1
While lim sn  lim
n 
1
n 
n
1
3
 lim 3
n 
1
0
n
Questions
Check, whether the given series is convergent.

1.

n 1
1
3
 5
3.

n 1

n
5
n
n
2.

n 1
1
5
 5
2
4.

n 1
n
3
n

n
2
Algebra of Series Convergence

s
1. If both
n 1

n
and  t n are convergent
n 1

Then so is  (c1sn  c1t n )
n 1

2. If
s
n 1

n
is convergentand  t n are divergent
n 1

Then  (c1sn  c1t n ) is divergent
n 1

1. If both
s
n 1

n
and  sn are divergent
n 1

Then it does not follows that  (c1sn  c1t n ) is divergent
n 1
(nor it follows that it is convergent)
Questions
1. Investigate the convergence of each of the following series :
5
4 2 n1
a.  [ 3n2   ]
n
n 1 3

2
7 2 n1
b.  [ 3n2  2 ]
n
n 1 3
2. Give an exam pleof two divergent series


s
n 1
n


n 1
n 1
and  t n , such that the series  ( sn  t n ) is :
a. a convergentseries
b. a divergent series
Divergence Test
If lim sn  0
n

Then  sn diverges
n 1
Notice :

If lim sn  0 , it does not follow that  sn converges
n
n 1
Questions
Check, whether the given series is convergent.

1
1 n
1. [ 3  (1  n ) ]
n 1 n
5n  2n  3
2.  8
5
n 1 3n  6n  1


8
1
3.  n
n
n 1
3
Convergence Tests
Convergence Tests for Series of Positive
Terms
1. Comparison Test
2. Limit Comparison Test
3. Ratio Test
4. Root Test
The Comparison test

Let
s
n 1

n
and  t n be series of positive term s,
n 1
and s n  t n ; n  N .
Then :

1. If
t
n 1

n

1. If
s
n 1
converges, then  s n converges
n 1

n
diverges, then  t n diverges
n 1
Examples
Example (1)
Investigate whether each of the following series converges

1.
3
n 1
4
5

2.
3
n 1
n8  2
4
8
n5  2
Solution
4
4
4 1
1.

  8
3 5 n8  2 3 5 n8 3 n 5

The series of positive term s 
n 1

4
3 5 n8
4  1
( because
  8 and
8
5
3 n 1 5
n 1 3 n
n
and so, by the com parisontest :
4

The series of positive term s 
n 1
converges


n 1
1
n
4
3 n 2
5
8
8
5
converges
converges.
Why ?)
4
4
4 1
2.

  5
5
5
8
8
3
3 n 2 3 n
n8

The series of positive term s 
n 1

4
8
3 n
5
4  1
( because
  5 and
5
8
3 n 1 8
n 1 3 n
n
and so, by the com parisontest :
4

The series of positive term s 
n 1
diverges

1

n 1
n
4
38 n5  2
5
8
diverges
diverges.
Why ?)
Definition
Order of Magnitude of a Series

Let
s
n 1

n
and  t n be series of positive term s
n 1
Then :
sn
is a positive num ber, then we say that the series
n  t
n
1. If lim

s
n 1

n
and  t n have the sam eorder of m agnitude
n 1
sn
 0 , then we say that the series
n  t
n
2. If lim

s
n 1

n
has a lesser order of m agnitudethan  t n
n 1
sn
  , then we say that the series
n  t
n
3. If lim

s
n 1

n
has a greater order of m agnitudethan  t n
n 1
Question
Are there cases, where
sn
lim
is a negative num ber
n  t
n
Or
sn
lim  
n t
n
Why not ?
The Limit Comparison test

Let
s
n 1

n
and  t n be series of positive term s
n 1
Then :

1. If
s
n 1

n
and  t n have the sam e order of m agnitude, then
n 1
either they both convergeor both diverge.

2. If
s
n 1

n
has a lesser order of m agnitudethan  t n then
n 1

the convergence of
t
n 1

3. If
s
n 1

n
leads to the convergence of
s
n 1

n
n
.
has a greater order of m agnitudethan  t n then
n 1

the divergenceof
s
n 1

n
leads to the divergenceof
t
n 1
n
.
Examples
5n  2n  3
1.  12
5
n 1 3n  6n  1

7
3
5n  2n  3
2.  12
5
n 1 3n  6n  1

11
3
Solution
5n 7  2 n 3  3
1. Let  s n   12
 6n 5  1
n 1
n 1 3n
Consider the series of positive term s :




1
5
n 1
n 1 n
We have
 tn  
5n 7  2n 3  3
12
5
sn
5n12  2n 8  3n 5
3
n

6
n

1
lim  lim
 lim
n  t
n 
n   3n 12  6 n 5  1
1
n
n5
5
  [0, )
3

5n 7  2 n 3  3
Thus  12
has the sam e order of m agnitudeas
5
3
n

6
n

1
n 1

1
the series  5 which converges ( Why ? )
n 1 n
Therefore;
5n 7  2 n 3  3
The series of positive term s 12
converges
5
 6n  1
n 1 3n

5n11  2n 3  3
2.
Let  s n   12
 6n 5  1
n 1
n 1 3n
Consider the series of positive term s :




1
n 1
n 1 n
We have
 tn  
5n11  2n 3  3
12
5
sn
5n12  2n 4  3n
3
n

6
n

1
lim  lim
 lim
n  t
n 
n   3n 12  6 n 5  1
1
n
n
5
  [0, )
3

5n11  2n 3  3
Thus  12
has the sam e order of m agnitudeas
5
3
n

6
n

1
n 1

1
the series  which diverges ( Why ? )
n 1 n
Therefore;
5n11  2n 3  3
The series of positive term s  12
diverges
5
3
n

6
n

1
n 1

The Ratio test

Let
s
n 1
n
be a series of positive term s
Then :
s n 1
1. If lim
 1 then
n  s
n

s
n 1
n
converges
s n 1
2. If lim
 1 or   , then
n  s
n

s
n 1
n
diverges
Examples

n
n
1. 
n 1 n!

(3n)
2.  n
n 1 5

n
3.  n
n 1 5
n
The Root test

Let
s
n 1
n
be a series of positiveterm s
Then :
1. If lim n s n  1 then
n 

s
n 1
2. If lim n s n  1 or  
n 
n
converges

then
s
n 1
n
diverges
Examples
 2n  3 
1.  

n 1  5n  2 


n
3


2.  

n 1  ln 2 n  5 
n
Definition
Alternating Series
An alternating series is a series of of either of the
following form s

n
(

1
)
 sn
n1
Or

n1
(

1
)
sn

n1
where  sn  is a
sequenceof positive term s
Alternating Series Convergence
Test


Let  t n   (1) sn or
n
n 1
n 1
where  sn  is a


 t   (1)
n 1
n
n 1
n 1
sn
sequenceof positive term s
If  sn  is decreasin g and convergentto 0

Then  t n converges
n 1
Example
Consider the series


 s   (1)
n 1
n
n 1
n
1
n
We have:
1
 sn   
 is a decreasin g sequenceof positive term s
n
1
and lim sn  lim
n
n
n
Therefore;

n
(

1
)

n 1
1
convergesby the alternating series convergence test
n
Definition
Absolute and Conditional Convergence

1. We say that the series  s n converges absolutely
n 1

if
s
n 1
n
converges.

2. We say that the series  s n converges conditionally
n 1

if it converges, but
s
n 1
n
diverges
Example (1)
Consider the series


 s   (1)
n 1
n
n 1
n
1
n

We have shown that
 (1)
n
n 1
1
converges
n
On the other hand the series


1
1
n
(1)

diverges
why ?

n n 1
n
n 1
and so the series convergesconditionally
Example (2)
Consider the series


sin 2n
s n   (1)

n3
n 1
n 1
We have:
n
sin 2n
1
s n  (1)
 3
3
n
n
n

sin 2n
and so the series  (1)
converges(Why ?)
3
n
n 1
n

Thus the series  (1) n
n 1
sin 2n
convergesabsolutely
3
n
The Ratio test for Absolute
Convergence
s n 1
1. If lim
 1 then
n  s
n

s
n 1
n
convergesabsolutely
s n 1
2. If lim
 1 or   , then
n  s
n

s
n 1
n
diverges
Examples:
Investigate the absolute convergence of
the following series
(2n  1)!
2.  (1)
n!
n 1


n
5
1.  (1) n
n!
n 1

3.
n
(

1
)

n
5 cos(3n)
n9

n 1 ( 2n  1)!
5.  (1)
(n  1)!
n 1
n 1
5

4.
n 1
(

1
)

n 1
5n
73 n 4  6
More Examples on the Integral Test
Example (1)


n 1
n 1
Investigate the convergence of the series sn   e  n
Solution:
Let f ( x)  e  x
We have:
f is positive (Why ?), positive (Why ?) and continuouson [1,)
and sn  f (n) ; n  1
&

The im properintegral I 

f ( x) dx   e  x dx converges ( Showthat!)
1


n 1
n 1

 the series  sn   e  n converges
1
Example (2)


1
2
n 1 1  n
Investigate the convergence of the series sn  
n 1
Solution:
Let f ( x) 
1
1 x2
We have:
f is positive (Why ?), decreasing (Why ?) and continuouson [1,)
and sn  f (n) ; n  1
&

The im properintegral I 

1


f ( x) dx 
1
1 1  x 2 dx converges (Showthat!)
1
converges
2
n 1 1  n
 the series  sn  
n 1

Example (3)


1
3
n  2 n ln n
Investigate the convergence of the series s n  
n2
Solution:
Let f ( x) 
1
x ln 2 x
We have:
f is positive (Why ?), decreasing (Why ?) and continuouson [2,)
and s n  f (n) ; n  2

We will show that
1
convergesby showing that

3
n
ln
n
n2

the im properintegral
1
2 x ln 3 x dx converges
we have:

1
2 x ln 3 x dx  lim
t  
 lim [
t  
t
t
1
dx
3
dx

ln
x

2 x ln 3 x lim

x
t   2
1 
1
1
1
]



2
t lim
ln 2 x
ln 2 t ln 2 2 ln 2 2
