The Real And Complex Number Systems

then
b k sin kx,
a k1 − a k 1 1 2 ... 1 k 1
k1
k 1
− 1 1 2
... 1 k
k
k1 1 ... 1 1 − k 11 1 2 k k1 2
kk 1
k
k1
− 1 1 ... 1
2 k
0
kk 1
and
n
∑ b k ≤ 1
k1
sin x .
2
So, by Dirichlet Test, we know that
∑
k1
a k b k ∑
k1
1 1 ... 1
2 k
sin kx
k
... 1 k
converges.
From above results, we have shown that the series converges for all x ∈ R.
8.27. Prove that following statements:
(a) ∑ a n b n converges if ∑ a n converges and if ∑b n − b n1 converges absolutely.
Proof: Consider summation by parts, i.e., Theorem 8.27, then
n
∑
k1
a k b k A n b n1 − ∑
k1
n
A k b k1 − b k .
Since ∑ a n converges, then |A n| ≤ M for all n. In addition, by Theorem 8.10, lim n→ b n
exists. So, we obtain that
(1). lim n→
A n b n1 exists
and
n
(2). ∑
k1
n
|A k b k1 − b k | ≤ M ∑
k1
|b k1 − b k | ≤ M ∑
k1
|b k1 − b k |.
(2) implies that
n
(3). ∑ A k b k1 − b k converges.
k1
n
By (1) and (3), we have shown that ∑ k1
a k b k converges.
Remark: In 1871, Paul du Bois Reymond (1831-1889) gave the result.
(b) ∑ a n b n converges if ∑ a n has bounded partial sums and if ∑b n − b n1 converges
absolutely, provided that b n → 0asn → .
Proof: Bysummation by parts, wehave
n
∑
k1
a k b k A n b n1 − ∑ A k b k1 − b k .
k1
Since b n → 0asn → and ∑ a n has bounded partial sums, say |A n| ≤ M for all n. Then
n