Homework 4 Key
• My Problem 1, (4 points): Find the radius of convergence of each of
the power series
1. Apply a ratio or root test, R = 1/2.
2. Ratio test yields R = 0.
3. Ratio test yields R = ∞.
4. Note that the kth coefficient ck = 2k/3 if k is a multiple of 3, 0
else. Then lim sup |ck |1/k = 21/3 , so R = 2−1/3 .
• My Problem 2, (2 points) Note that
lim sup ak = lim sup(ak ; k ≥ n), lim sup bk = lim sup(bk ; k ≤ n).
n
n
It’s not hard to see that for any n
sup(ak + bk ; k ≥ n) ≤ sup(ak ; k ≥ n) + sup(bk ; k ≥ n).
Take the limit of both sides above (they exist, since the sequences are
nondecreasing) to find
lim sup(ak + bk ) ≤ lim sup ak + lim sup bk .
To show that strict inequality is possible, let ak be the sequence 1, 0, 1, 0, . . .
and bk the sequence 0, 1, 0, 1, . . ..
∑ √x
∑
• My Problem 3, (2 points) Prove that if k xk converges then k k k
converges, if xk ≥ 0.
A proof can be based on the Cauchy-Schwarz inequality |ab| ≤ a2 /2 +
√
b2 /2. We have xk /k ≤ xk /2 + 1/(2k 2 ), and so
n √
∑
xk
k=1
k
≤
n
∑
k=1
xk +
n
∑
1
.
2
2k
k=1
Of course as → ∞ both sums on the right converges (and so are
bounded), hence the sum on the left converges since its partial sums
are bounded.
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• My Problem 4, (5 points) Let xk be a nonincreasing (xk ≥ xk+1 ) sequence
of positive numbers that converges to zero and suppose that
∑∞
x
k=1 k converges. Prove that limk kxk = 0. Show this result is false
if the nonincreasing assumption is dropped.
I’ll give two proofs, one inspired by stuff done by Artem Yankov using
the Cauchy criterion.
∑
Here’s the Cauchy criterion proof. If
k ak converges then by the
Cauchy criterion we have that for any ϵ > 0 there exists an N so that
n
∑
ak < ϵ/2
k=m
for m, n ≥ N (no absolute values needed, since everything is nonnegative). Let us fix m = ⌈n/2⌉ (greatest integer function) and so
obtain
n
∑
ak < ϵ/2
k=⌈n/2⌉
for ⌈n/2⌉ ≥ N ; this last inequality certainly holds if n ≥ 2N + 1. Note
also that there are either (n + 2)/2 terms in the sum above (if n even)
or (n + 3)/2 terms (n odd). Either way, there’s more than n/2. We
can conclude that since ak ≥ ak+1 we have
n
∑
n
ak < ϵ/2
an <
2
k=⌈n/2⌉
if n ≥ 2N + 1. This yields
nan < ϵ.
for n ≥ N ′ = 2N + 1, which is precisely the definition of limn nan = 0.
Here’s another proof straight from the definition of convergence for
series and sequences. First, let’s do this by contradiction. Suppose
that limk kak is NOT zero (meaning it either doesn’t exist or equals
something other than zero). The precise meaning of limk kak = 0 is
that for each ϵ > 0 there is an N so that kak < ϵ for all k ≥ N (no
2
absolute values needed since kak > 0). The negation of this statement
is that for SOME ϵ > 0 there is NO N such that kak < ϵ for all k ≥ N .
Put even more plainly, for some ϵ > 0 we can obtain kak ≥ ϵ for
arbitrarily large values of k. Let’s fix this value of ϵ for the remainder
of the problem.
Take note of the fact that if for a specific N we have N aN ≥ ϵ then
aN ≥ ϵ/N and so for any k ≤ N we also have ak ≥ ϵ/N (since ak ≥ aN
by assumption). As a result we have, for any M < N that
N
∑
ak ≥
k=M +1
N
∑
N −M
ϵ
=
ϵ.
N
N
k=M +1
(1)
∑
I will show that the series k ak has partial sums that are unbounded.
Choose some M1 so that M1 aM1 ≥ ϵ (which we can do, by the above);
choose N1 ≥ 2M1 so that N1 aN1 ≥ ϵ. Then making use of equation (1)
with M = M1 and N = N1 shows that
N1
∑
ak ≥
k=M1 +1
1
N1 − M1
ϵ ≥ ϵ.
N1
2
Now choose M2 > N1 so that M2 aM2 ≥ ϵ and choose N2 ≥ 2M2 so that
N2 aN2 ≥ ϵ. We then obtain as above
N2
∑
k=M2
1
ak ≥ ϵ.
2
+1
Note that the ranges M1 ≤ k ≤ N1 and M2 ≤ k ≤ N2 do not overlap,
so that we may conclude (all ak are positive) that
N2
∑
ak ≥ ϵ.
k=1
Continue this process, choosing Mm and Nm ≥ 2Mm , with Mm+1 >
Nm . We obtain
Nj
∑
j
ak ≥ ϵ
2
k=1
3
∑
for all j. The sequence of partial sums is unbounded, so k ak cannot
converge.
∑
Thus if k ak converges it must be the case that lim kak = 0.
To show that strictly decreasing is needed, let ak be defined by
{
1/10n , k = 10n
ak =
1/2k , else
That is, a1 = 1, a10 = 1/10, a100 = 1/100 and so on, with all other
values filled in by a harmless geometric sequence. Then
∑
ak
k
converges (easy to check) but lim sup kak = 1 and lim inf kak = 0, so
kak doesn’t converge to zero.
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