MATH 1320 : Spring 2014
Lab Instructor : Kurt VanNess
Lab 4 Solutions
1. Suppose you know that a sequence {an } is an increasing sequence and all its terms lie between the
numbers 2 and 4. Explain why the sequence must have a limit? What can you say about the value of
the limit?
Solution
Since {an } is an increasing sequence, an < an+1 for all n ≥ 1. Because all of its terms lie between 2
and 4, {an } is a bounded sequence. By the Monotonic Sequence Theorem, {an } is convergent; that is,
{an } has a limit L. L must be great than 2 since {an } is increasing, so 2 < L ≤ 4.
2. Find the limit of the sequence
(
√
q
2,
√
2 2,
r q
)
√
2 2 2, . . . .
Solution
Let us define L as the limit of the sequence (i.e. lim an = L) and let an be an arbitrary term in the
n→∞
√
sequence. The next term of the sequence,
a
can
be written as an+1 = 2an . Taking the limit of
n+1
√
both sides as n → ∞, we have L = 2L, which is equivalent to L2 = 2L, or L(L − 2) = 0. Therefore,
we either have L = 0 or L = 2, but since the sequence is positive and increasing the limit must be
L= 2.
∞
X
3. Suppose that
an (an 6= 0) is known to be a convergent series. Prove that
n=1
∞
X
1
is a divergent
a
n=1 n
series.
Solution
Since we know that the series of an converges, then we know from Section 8.2, Theorem 6, that
lim an = 0. Let us define the terms of a sequence {bn } by bn = a1n . Since we know that lim an = 0,
n→∞
n→∞
1
1
then we must have lim bn = lim
=
6= 0. Therefore, by the Divergence Test, the
n→∞
n→∞ an
limn→∞ an
series of bn diverges, which means the series of 1/an diverges.
4. If
P
an and
P
bn are both divergent, is
P
(an + bn ) necessarily divergent?
Solution
We will show that it is not necessarily divergent by a counter-example. Let an = 1 and bn = −1. We
∞
∞
∞
∞
X
X
X
X
know that
an =
1 = ∞ and
bn =
−1 = −∞, and so they are both divergent. However,
∞
X
n=1
(an + bn ) =
n=1
∞
X
n=1
(1 + −1) =
n=1
∞
X
n=1
n=1
0 = 0, which is convergent.
n=1
Page 1 of 3
MATH 1320 : Spring 2014
Lab Instructor : Kurt VanNess
5. If the nth partial sum of a series
P∞
n=1
an is sn =
n−1
n+1 ,
then find an and
P∞
n=1
an .
Solution
For n = 1, a1 = 0 since the first partial sum s1 = 0. For n > 1,
an = sn − sn−1 =
Also,
∞
X
n − 1 (n − 1) − 1
n(n − 1) − (n − 2)(n + 1)
2
−
=
=
n + 1 (n − 1) + 1
n(n + 1)
n(n + 1)
an = lim sn = lim
n=1
n→∞
n→∞
6. Compute the sum of the series:
n−1
1 − 1/n
= lim
= 1.
n + 1 n→∞ 1 + 1/n
∞
X
2
. [Hint: Telescoping - Use partial fraction decomposition]
n(n + 2)
n=1
Solution
We recognize the series as a telescoping series, so we want to construct a formula for the N th partial
sum, SN , and find its limit as N → ∞. Begin by using partial fraction decomposition to obtain
A
B
1
1
2
= +
= −
n(n + 2)
n
n+2
n n+2
Then the series becomes
∞ X
1
n=1
N X
1
1
−
, and the formula for the N th partial sum SN is given by:
n n+2
1
n n+2
n=1
1 1
1 1
1 1
1 1
=
−
+
−
+
−
+
−
+
1 3
2 4
3 5
4 6
1
1
1
1
1
1
1
1
... +
−
−
−
−
+
+
+
N −3 N −1
N −2 N
N −1 N +1
N
N +2
1
1
1
=1 + −
−
2 N +1 N +2
∞
X
2
3
1
1
1
Taking the limit as N → ∞, we have
.
= lim 1 + −
−
=
N →∞
n(n
+
2)
2
N
+
1
N
+
2
2
n=1
SN =
−
7. Determine whether the following series converge or not by using either the Integral Test or the Comparison Test:
∞
∞
X
X
1
n
√
,
.
3
5
n(ln
n)
n +n+1
n=2
n=1
Solution
We use the Integral Test to determine whether or not the first series converges. Let f (x) = x(ln1x)3 .
The function f (x) is decreasing for x ≥ 2. We must now determine whether or not the following integal
Page 2 of 3
MATH 1320 : Spring 2014
Lab Instructor : Kurt VanNess
converges where we use a u-substitution of u = ln x:
Z ∞
Z R
1
1
dx
=
lim
dx
3
R→∞
x(ln
x)
x(ln
x)3
2
2
Z ln R
1
du
= lim
R→∞ ln 2 u3
ln R
−1
= lim
R→∞ 2u2
ln 2
1
1
= lim
−
R→∞ 2(ln 2)2
2(ln R)2
1
=
2(ln 2)2
Since the integral converges, then the series converges by the Integral Test.
We use the Comparison Test to determine whether the second series converges or not. We guess that the
n
series converges because of the highest powers in the numerator and denominator. Let an = √n5 +n+1
.
P
P
We must find a series
bn such that 0 ≤ an ≤ bn for n ≥ 1 and
bn converges. We notice that:
n
n
1
≤ √ = 3/2
5
n
+n+1
n
√
√
1
for all n ≥ 1 using the argument that n5 + n + 1 > n5 for n ≥ 1. So we choose bn = n3/2
. The
P∞
3
1
series n=1 n3/2 converges because it is a p-series with p = 2 > 1. Therefore, the series converges by
the Comparison Test.
0≤ √
n5
8. (Alternating Series)
(a) Show that the series
∞
X
(−1)n
converges.
n2n
n=1
k
X
(−1)n
. Then S(k) is the partial sum of the series. Compute k so
n2n
n=1
that S(k) is within 0.01 of the sum of the series.
(b) Let S(k) be defined by:
Solution
i) To show convergence, we use the Alternating Series Test. The function bn = n21n is positive and
1
decreasing for n ≥ 1. Furthermore, lim bn = lim
= 0. Therefore, the series converges.
n→∞
n→∞ n2n
ii) Now we compute k so S(k) is within 0.01 of the sum of the series. Let S be the exact sum of
the series. By the Alternating Series Estimation Theorem, the error |S − S(k)| is bounded as follows:
|S − S(k) ≤ bk+1 . We want the error to be smaller than 0.01, so we want to choose k to satisfy the
following inequality:
1
1
|S − S(k) ≤
<
.
(k + 1)2k+1
100
Using a trial and error process, we find that if k = 3 then
1
160 . So we choose k ≥ 4.
1
(3+1)23+1
=
1
64
and if k = 4 then
1
(4+1)24+1
=
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