The Real And Complex Number Systems

(b) Prove that ∑ n1
−1 n−1 /n s 1 − 2 1−s s if s 1.
Proof: Let
n
S n ∑ j1
−1 j−1
j s
, and thus consider its subsequence S 2n as follows:
2n
S 2n ∑
j1
2n
∑
j1
1
j s
1
j s
n
− 2 ∑
j1
n
− 2 1−s ∑
j1
1
2j s
which implies that
lim n→
S 2n 1 − 2 1−s s.
Since S n converges, we know that S 2n also converges and has the same value. Hence,
∑
n1
1
j s
−1 n−1 /n s 1 − 2 1−s s.
8.22 Given a convergent series ∑ a n , where each a n ≥ 0. Prove that ∑ a n n −p
converges if p 1/2. Give a counterexample for p 1/2.
Proof: Since
a n n −2p
2
≥ a n n −2p a n n −p ,
we have ∑ a n n −p converges if p 1/2 since
∑ a n converges and ∑ n −2p converges if p 1/2.
and
For p 1/2, we consider a n
1
nlogn 2 , then
∑ a n converges by Cauchy Condensation Theorem
∑ a n n −1/2 ∑ 1
n log n
diverges by Cauchy Condensation Theorem.
8.23 Given that ∑ a n diverges. Prove that ∑ na n also diverges.
n
Proof: Assume ∑ na n converges, then its partial sum ∑ k1
ka k is bounded. Then by
Dirichlet Test, we would obtain
∑ka k 1
k
∑ a k converges
which contradicts to ∑ a n diverges. Hence, ∑ na n diverges.
8.24 Given that ∑ a n converges, where each a n 0. Prove that
∑a n a n1 1/2
also converges. Show that the converse is also true if a n is monotonic.
Proof: Since
a n a n1
≥ a
2
n a n1 1/2 ,
we know that
∑a n a n1 1/2