The Real And Complex Number Systems

n
c n ∑ a k b n−k
k0
n
∑
k0
−1 k
k 1
n
−1 n ∑
k0
−1 n−k
n − k 1
1
k 1 n − k 1
and let fk n − k 1k 1 −k − n 2 2 n2
2 2 ≤ n2 for k 0, 1, . . . , n.
2
Hence,
n
|c n| ∑
k0
1
k 1 n − k 1
2n 1
≥ → 2asn → .
n 2
That is, the Cauchy product of ∑
n0
−1 n1 / n 1 with itself is a divergent series.
(b) Show that the Cauchy product of ∑
n0
−1 n1 /n 1 with itself is the series
2 ∑ −1
n1
1
n 1
1 2 ... 1 n .
n1
Proof: Since
n
c n ∑ a k b n−k
we have
∑
n0
k0
n
∑
k0
−1 n ∑
k0
2−1n
n 2
−1 n
n − k 1k 1
c n ∑
n
n0
2 ∑
n0
2 ∑
n1
∑ k0
1
n 2
n
2−1
n
n 2
1
k 1 ,
∑ n
k0
1
k 1 1
n − k 1
1
k 1
−1
n
n 2 1 1 2 ... 1
n 1
−1
n1
n 1
(c) Does this converge Why
Proof: Yes by the same argument in Exercise 8.26.
1 1 2 ... 1 n .
8.33 Given two absolutely convergent power series, say ∑
n0
a n x n and ∑ n0
b n x n ,
having sums Ax and Bx, respectively, show that ∑ n0
c n x n AxBx where