MATHEMATICS 320, FALL 2016, PROBLEM SET 6
Solutions
1. Find all complex
z such that the following power series converge:
P∞ numbers
1
n+ n
−2 n
n z .
(a) (3 marks) n=1 3
1
2
n+ n
−2
If cn = 3
n , then lim(cn )1/n = 3 · 31/n (n1/n )−2 → 3 as n → ∞ because n1/n → 1
and 31/n → 1 as n → ∞ (Elementary Limits). Note also that the last limit implies the
2
subsequence 31/n → 1 as well. Therefore the power series has R = 1/3 and so will converge
1
if |z| < 1/3, and diverge
If |z| = 1/3, then |3n+ n n−2 z n | ≤ 31/n n−2 ≤ n32 . We
P∞ if3 |z| >P1/3.
−2
know (p test) that n=1 n2 = 3 ∞
converges. So by the comparison test, the power
n=1 n
series converges for |z| = 1/3 (even absolutely). So to conclude we have shown that the
power series converges
|z| ≤ 1/3.
P∞ iff−n
n
(b) (3 marks) n=0 2 z 2 .
In this case the power series has coefficients equal to 0 for powers which are not of the
n
form 2n . As a result it is simpler to just apply the root test with an = 2−n z 2 (rather than
n
the formula for radius of convergence). Then |an |1/n = 2−1 |z|2 /n . Since lim 2n /n = +∞
(elementary limit), we see that


if |z| < 1
0
1/n
lim sup |an | = 1/2 if |z| = 1


∞
if |z| > ∞.
It follows by the root test that the series converges (and in fact converges absolutely) iff
|z| ≤ 1.
2. (4 marks) Let an ≥ 0 for all n ∈ N. Prove that
(
)
∞
X
X
an = sup
an : F ⊂ N finite .
n=1
n∈F
Is this still true without the assumption that an ≥ 0? Prove your answer.
Pk
Assume
P∞ that an ≥ 0 for all n, and let sk = n=1 an . Then the sequence {sk } is increasing
and n=1 an = limk→∞ sk = sup{sk : k ∈ N} ≥ sk for any k ∈ N.
P
P
• We have sk = n∈FP
an with Fk = {1, . . . , k}, so that sup
≥
n∈F an : F ⊂ N finite
k
∞
sup{sk : k ∈ N} = n=1 an .
• Now let
P F be an arbitrary
P∞ finite subset of N, then F ⊂ {1, . . . , k} for some k ∈ N.
Then n∈F an ≤ sk ≤ n=1 an . Taking supremum over F , we get
(
)
∞
X
X
sup
an : F ⊂ N finite ≤
an .
n=1
n∈F
1
If an are not necessarily nonnegative, the result is not true. For example, consider the
1
series 1 − 1 + 12 − 12 + 13 − 31 + . . . . We have sP
2k = 0 and s2k−1 = k , so the series converges to
0. However, if Fk = {1, 3, . . . , 2k − 1}, then n∈Fk an = 1 + 12 + · · · + k1 → ∞ as k → ∞.
P
P∞
3. (5 marks) Let ∞
n=1 an and
n=1 bn be two series such that an > 0, bn > 0, and
bn+1
an+1
≤
an
bn
P∞
P
for all n ∈ N. Prove that if n=1 bn converges, then also ∞
n=1 an converges.
We claim that
an
bn
≤
(1)
a1
b1
for all n ∈ N. We will prove this by induction. For n = 1, we have aa11 = 1 = bb11 . Suppose
that (1) is true for n = k, then
an+1
an an+1
bn bn+1
bn+1
=
≤
=
.
a1
a1 an
b1 bn
b1
P∞ a1
P
Using (1), we have an ≤ ab11 bn . Since the series ∞
n=1 b1 bn , and
n=1 bn converges, so does
P∞
therefore so does n=1 an by the comparison test.
4. Decide whether the following series are convergent or divergent. Prove your answers.
∞
X
n2
(a) (2 marks)
: applying the root test, we get
n
2
n=1
2 1/n
1
n
(n1/n )2
→
,
=
2n
2
2
where at the last step we used Rudin, 3.20(c). Since 21 < 1, the series converges.
n2
∞ X
n
(b) (3 marks)
: Applying the root test again, we get
n+1
n=1
1/n n n 1 −n
1
n n2
=
= 1+
→ ,
n+1
n+1
n
e
where at the last step we used Rudin, 3.31. Since
1
e
< 1, the series converges.
P∞
5.(a) (3 marks) Let
a
≥
a
≥
a
≥
·
·
·
≥
0.
Prove
that
the
series
1
2
3
n=1 an converges if and
P
2
only if the series ∞
ka
converges.
k
k=1
We have
∞
X
an = (a1 ) + (a2 + a3 + a4 ) + (a5 + · · · + a9 ) + . . .
n=1
∞
∞
X
X
2
2
≥
(k − (k − 1) )ak2 =
(2k − 1)ak2
≥
k=1
∞
X
k=1
kak2 .
k=1
2
P
P∞
Therefore, if ∞
k=1 kak2 = ∞, then
n=1 an = ∞.
On the other hand, we also have
∞
X
an = (a1 + a2 + a3 ) + (a4 + · · · + a8 ) + (a9 + · · · + a15 ) + . . .
n=1
≤
∞
X
(((k + 1)2 − 1) − (k 2 − 1))ak2 =
k=1
≤
∞
X
∞
X
(2k + 1)ak2
k=1
3kak2 .
k=1
Therefore, if
P∞
k=1
kak2 < ∞, then
P∞
n=1
an < ∞.
∞
X
1
√ converges or diverges.
(b) (2 marks) Apply (a) to determine if the series
2 n
n=1
P∞ k
P
With an = 2√1 n , we have ∞
k=1 2k . We have
k=1 kak2 =
k
2k
1/k
where we again used Rudin, 3.20(c). Since
1
k 1/k
→ ,
=
2
2
1
2
< 1, the series converges.
3