The Real And Complex Number Systems

∑ m,n
e −m2 n 2
converges since ∑ m,n
e −m2 n 2
∑ m
e −m2 ∑ n
e −n2 .
Remark: ∑ m,n1
e −m2 n 2
∑
m1
e −m2 ∑ n1
e −n2 e
e 2 −1
8.30 Asume that the double series ∑ m,n
anx mn converges absolutely for |x| 1. Call
its sum Sx. Show that each of the following series also converges absolutely for |x| 1
and has sum Sx :
∑
n1
Proof: ByTheorem 8.42,
So, ∑
n1
∑
n1
that
∑
m,n
an x n
1 − x n , ∑
anx mn ∑
n1
n1
an ∑
m1
Anx n ,whereAn ∑ ad.
d|n
x mn ∑ an
n1
2
.
x n
1 − xn if |x| 1.
an xn converges absolutely for
1−x
|x| 1 and has sum Sx.
n
Since every term in ∑ m,n
anx mn , the term appears once and only once in
Anx n . The converse also true. So, by Theorem 8.42 and Theorem 8.13, we know
∑
n1
Anx n ∑ anx mn Sx.
m,n
8.31 If is real, show that the double series ∑ m,n
m in − converges absolutely if,
p q
and only if, 2. Hint. Let sp, q ∑ m1
∑ n1
|m in| − . The set
m in : m 1,2,...p, n 1, 2, . . . , p
consists of p 2 complex numbers of which one has absolute value 2 , three satisfy
|1 2i| ≤ |m in| ≤ 2 2, five satisfy |1 3i| ≤ |m in| ≤ 3 2, etc. Verify this
geometricall and deduce the inequlity
p
p
2 −/2 ∑
2n − 1
n ≤ sp, p ≤ ∑
2n − 1
n1
n1 n 2 1 . /2
Proof: Since the hint is trivial, we omit the proof of hint. From the hint, we have
p
∑
n1
2n − 1
n 2 ≤ sp, p ∑
m1
p
p
∑
n1
p
|m in| − ≤ ∑
n1
2n − 1
1 n 2 /2 .
Thus, it is clear that the double series ∑ m,n
m in − converges absolutely if, and only if,
2.
8.32 (a) Show that the Cauchy product of ∑
n0
−1 n1 / n 1 with itself is a
divergent series.
Proof: Since