# Solutions to Assignment 6, Math 220 1

```Solutions to Assignment 6, Math 220
1 For each of the following series, indicate whether they converge or
P
1
a: ∞
k 2 +k
Pk=1
−k
b: P ∞
k=1 k
∞
1
c: k=2 k log
k
P∞
d: P k=1 sin(π/k)
1
e: ∞
k=3 k 2 −3k+2
P∞
f: k=1 kk!k
Solution
P
1
a: Converge by comparison to ∞
k=1 k 2
b: Converge by the root test.
c: Diverge by the integral test. P
1
d: Diverge by comparison with ∞
P∞k=1 2k
e: Converge by comparison to k=3 k2
f: Converge by, say, the ratio test
2 For each of the following series, indicate whether they converge or
P
(−1)k
√
a: ∞
k=1
k
P∞
k
b: k=1(−1)k k+1
P∞
c: k=1 (−1)k k2k+1
3 −4
P∞
k 2k k 2
d: k=1(−1) k!
Solution
a: Converge by the alternating series test, conditionally by the
integral test.
k
b: Diverge since (−1)k k+1
fails to approach 0
P
3
c: Converge absolutely by comparison to ∞
k=3 k 2
d: Converge absolutely by the ratio test
P
P 2
3 If
an with an &gt; 0 is convergent, then is
an always convergent?
Either prove it or give a counterexample.
Pn
Pn
2
Solution This is true. If we let sn =
k=1 ak and Sn =
k=1 ak ,
then the triangle inequality tells us that Sn ≤ s2n . Since lim sn exists
it follows that sn is bounded above and hence, so is SnP
. Since Sn is
monotone and bounded, it is convergent and thus, so is
a2n .
1
2
P
P√
4 If
an with an &gt; 0 is convergent, then is
an always convergent? Either prove it or give a counterexample.
Solution A counterexample follows by taking an =
1
.
n2
P
5 Can youPgive an example
of
a
convergent
series
xn and a diverP
gent series
yn such that (xn + yn ) is convergent.
so does their difference
yn .
P
P
(xn + yn ) and
xn converge, then
P
6 Find an explicit expression for the nth partial sum of ∞
n=2 log(1 −
2
1/n ) to show that this series converges to − log 2. Is this convergence
absolute?
P
n+1
2
. We prove
Solution We claim that N
log(1
−
1/n
)
=
log
n=2
2n
this by induction. Since N = 2 implies that the sum is just log(3/4),
we may suppose that the claim is true for N = k ≥ 2. Then
2
k+1
X
k+1
k + 2k
2
log(1 − 1/n ) = log
+ log
2k
k 2 + 2k + 1
n=2
and the result follows by simple algebra.
P∞
2
The above formula tells us that
n=2 log(1 − 1/n ) converges to
log(1/2)
= − log 2. Since all the terms in the series are negative,
P∞
2
n=2 | log(1 − 1/n )| thus converges also, to log 2.
7 Show that the series
1 1 1 1 1 1 1 1
1+ − + + − + + − +&middot;&middot;&middot;
2 3 4 5 6 7 8 9
is divergent.
Solution If we let sn denote the partial sums of the above series, we
have that
1
1
1 1
1 1 1
1
s3n = 1 + −
+ −
+
−
+
+ &middot; &middot; &middot;+
.
2 3
4 5 6
3n − 2 3n − 1 3n
It follows that
1
1 1
s3n &gt; 1 + + + &middot; &middot; &middot; +
4 7
3n − 2
and so
1
1
1 1
s3n &gt;
1+ + +&middot;&middot;&middot;+
.
3
2 3
n
P
1
Since ∞
n=1 n diverges, it follows that s3n (and thus sn ) does likewise.
```