The Real And Complex Number Systems

Proof: If x = −1, we consider three cases: (i) α < 0, (ii) α = 0, and (iii)
α > 0.
(i) As α < 0, then
∞∑
( α n) (−1) n =
n=0
∞∑
n=0
(−1) n α (α − 1) · · · (α − n + 1)
,
n!
say a n = (−1) n α(α−1)···(α−n+1)
n!
, then a n ≥ 0 for all n, and
a n
1/n
=
−α (−α + 1) · · · (−α + n − 1)
(n − 1)!
≥ −α > 0 for all n.
Hence, ∑ ∞
n=0 (α n) (−1) n diverges.
(ii) As α = 0, then the series is clearly convergent.
(iii) As α > 0, define a n = n (−1) n ( α n) , then
a n+1
a n
= n − α
n
≥ 1 if n ≥ [α] + 1. (*)
It means that a n > 0 for all n ≥ [α] + 1 or a n < 0 for all n ≥ [α] + 1. Without
loss of generality, we consider a n > 0 for all n ≥ [α] + 1 as follows.
Note that (*) tells us that
and
So,
m∑
n=[α]+1
a n > a n+1 > 0 ⇒ lim
n→∞
a n exists.
a n − a n+1 = α (−1) n ( α n) .
(−1) n ( α n) = 1 α
m∑
n=[α]+1
(a n − a n+1 ) .
By Theorem 8.10, we have proved the convergence of the series ∑ ∞
n=0 (α n) (−1) n .
(b) If x = 1, the series diverges for α ≤ −1, converges conditionally for α
in the interval −1 < α < 0, and converges absolutely for α ≥ 0.
Proof: If x = 1, we consider four cases as follows: (i) α ≤ −1, (ii)
−1 < −α < 0, (iii) α = 0, and (iv) α > 0 :
(i) As α ≤ −1, say a n = α(α−1)···(α−n+1)
n!
. Then
|a n | =
−α (−α + 1) · · · (−α + n − 1)
n!
32
≥ 1 for all n.