求此題收斂區間在那(n次方,n階層)

Prelemma(Stirling formula): n!~ √(2πn) *(n/e)^n -----(1)
(2) lim(n->∞) (n!/n^n)^(1/n)= 1/e

By the Root test:
a(n)= 3^n* (x-1)^n * n!/n^n
lim(n->∞)|a(n)|^(1/n)=lim(n->∞) 3|x-1|*(n!/n^n)^(1/n)= 3|x-1|/e (by (2))
so that,
if |x-1|< e/3, then the series is conv.
if |x-1|> e/3, then the series is div.
if |x-1|=e/3, then
series=Σ[n=1~∞] (-1)^n [n! (e/n)^n] ~ Σ[n=1~∞] (-1)^n *√(2πn) (by(1))
while lim(n->∞)√(2πn)≠ 0,
so that, the series is div.

Ans: the interval of conv. is |x-1|<e/3 or 1-e/3 < x < 1+e/3

2010-06-18 10:53:56 補充：
1. Stirling's formula is a famous formula, he express n! as power fn.
2. most terms of a(n)=3^n*(x-1)^n *n!/n^n are power fn. of power n except n!,
so thinking about it by the Root test,
We can solve it by the Ratio test,
lim(n -> ∞) |a(n+1)/a(n)|= 3|x-1|*lim(n -> ∞) [n/(n+1)]^n= 3|x-1|/e

2010-06-18 10:54:13 補充：
3. lim(n-> ∞) (n!/n^n)^(1/n)
=exp{ lim(n-> ∞)Σ[k=1~n] (1/n)ln(k/n) }
=exp[∫(0~1) ln(x) dx]= exp(-1)= 1/e

2010-06-22 10:50:27 補充：
the radius of conv. = half length of the interval of conv.= e/3

2010-06-22 10:53:39 補充：
The major part of terms of the series is power fn. so we can use the root or ratio test.

2010-06-27 02:29:08 補充：
本題若不用Stirling formula,求收斂半徑猶可,但收斂區間的端點就不易判別了

2010-06-27 17:39:25 補充：
求收斂半徑時,沒有用到Stirling formula,只用普通的Root test,及 e的定義
Stirling formula只用在 |x-1|=e/3時(即收斂區間的端點)
此時須更詳細的分析n!與n^n的關係

To Heaven help heart clear:

本題若不用Stirling formula,求收斂半徑猶可,但收斂區間的端點就不易判別了

→看來太早結案了,不太清楚您說的意思,可以請您列算式說明嗎?謝謝

不好意思了