The Real And Complex Number Systems

converges by ∑ a n converges.
Conversly, since a n is monotonic, it must be decreasing since ∑ a n converges. So,
a n ≥ a n1 for all n. Hence,
a n a n1 1/2 ≥ a n1 for all n.
So, ∑ a n converges since ∑a n a n1 1/2 converges.
8.25 Given that ∑ a n converges absolutely. Show that each of the following series
also converges absolutely:
(a) ∑ a n
2
Proof: Since∑ a n converges, then a n → 0asn → . So, given 1, there exists a
positive integer N such that as n ≥ N, wehave
|a n| 1
which implies that
a n2 |a n| for n ≥ N.
So, ∑ a n2 converges if ∑|a n| converges. Of course, ∑ a n2 converges absolutely.
(b) ∑ an
1a n
(if no a n −1)
Proof: Since∑|a n| converges, we have lim n→ a n 0. So, there exists a positive
integer N such that as n ≥ N, wehave
1/2 |1 a n|.
Hence, as n ≥ N,
a n
2|a
1 a n|
n
which implies that ∑
(c) ∑ a n 2
1a n
2
Proof: It is clear that
an
1a n
By (a), we have proved that ∑ a n 2
converges. So, ∑ an
1a n
1a n
2
a2
n
≤ a
1 a2 n2 .
n
converges absolutely.
converges absolutely.
8.26 Determine all real values of x for which the following series converges.
∑
n1
Proof: Consider its partial sum
n
∑
k1
1 1 2 ... 1 n
1 1 2
... 1 k
k
sin nx
n .
sin kx
as follows.
As x 2m, the series converges to zero. So it remains to consider x ≠ 2m as
follows. Define
and
a k 1 1 2
... 1 k
k