求收斂區間(n次方和x次方)

2010-06-18 7:52 am

∑(n=1，∞) (5^n/n^2)(x-3)^2n收斂區間?
謝謝各位

更新1:

To 天助: Thank you your reply.Can you explain using the reason of the Root test?thank you.

更新2:

thank you.

更新3:

To 天助: Can you answer the radius of convergence,thank you?

更新4:

thank you

更新5:

結案前謝謝天助的回答

回答 (1)

天助

2010-06-18 8:00 am

✔ 最佳答案

By the Root test:
a(n)= 5^n (x-3)^(2n) / n^2
lim(n->) |a(n)|^(1/n) (lim(n->∞) n^(1/n)=1)
= 5|x-3|^2
so, if |x-3|^2 < 1/5, then the series is conv.
if |x-3|^2 > 1/5, then the series is div.
if |x-3|^2= 1/5, then
series=Σ[n=1~∞] 1/n^2 is a p-series p=2, thus it is conv.
Ans: the interval of conv. is |x-3|^2 <= 1/5, or 3-1/√5 <= x <= 3+1/√5

2010-06-18 00:16:38 補充：
We can use the Ratio test too.
lim(n->∞) |a(n+1)/a(n)|= 5|x-3|

2010-06-18 00:17:04 補充：
(Sorry!) lim(n->∞) |a(n+1)/a(n)|= 5|x-3|^2

2010-06-22 10:49:50 補充：
the radius of conv. = half length of the interval of conv.= 1/√5

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