The Real And Complex Number Systems

So, the series diverges.
(ii) As −1 < α < 0, say a n = α(α−1)···(α−n+1)
n!
. Then a n = (−1) n b n , where
with
b n =
b n+1
b n
−α (−α + 1) · · · (−α + n − 1)
n!
= n − α
n
> 0.
< 1 since − 1 < −α < 0
which implies that {b n } is decreasing with limit L. So, if we can show L = 0,
then ∑ a n converges by Theorem 8.16.
Rewrite
n∏
(
b n = 1 − α + 1 )
k
k=1
and since ∑ α+1
k
diverges, then by Theroem 8.55, we have proved L = 0.
In order to show the convergence is conditionally, it suffices to show the
divergence of ∑ b n . The fact follows from
b n
1/n
=
−α (−α + 1) · · · (−α + n − 1)
(n − 1)!
≥ −α > 0.
(iii) As α = 0, it is clearly that the series converges absolutely.
(iv) As α > 0, we consider ∑ |( α n)| as follows. Define a n = |( α n)| , then
a n+1
a n
= n − α
n + 1
< 1 if n ≥ [α] + 1.
It implies that na n − (n + 1) a n = αa n and (n + 1) a n+1 < na n . So, by
Theroem 8.10,
∑
an = 1 ∑
nan − (n + 1) a n
α
converges since lim n→∞ na n exists. So, we have proved that the series converges
absolutely.
9.35 Show that ∑ a n x n converges uniformly on [0, 1] if ∑ a n converges.
Use this fact to give another proof of Abel’s limit theorem.
Proof: Define f n (x) = a n on [0, 1] , then it is clear that ∑ f n (x) converges
uniformly on [0, 1] . In addition, let g n (x) = x n , then g n (x) is unifomrly
bouned with g n+1 (x) ≤ g n (x) . So, by Abel’s test for uniform convergence,
33