標題: Power Series [打印本頁] 作者: kingwinner 時間: 10-4-6 05:55 PM
標題: Power Series
1) Evaluate
∞
∑ 1 / [n! (n+2)]
n=0
(hint: sum an appropriate power serires)
2) Evaluate
∞
∑ 1 / [n(n+1) 5^n]
n=1
(hint: start with a geometric series)
請高手賜教, thx!作者: Gnoehc 時間: 10-4-6 06:00 PM
Consider
for the first one.
Consider
For the second one. Since two power serieses converge uniformly, termwise integration applies.作者: kingwinner 時間: 10-4-7 04:41 AM
但問題係我唔知要點 integrate 先 work...
1) Integrating term by term, e^x = ∑ x^(n+1) / (n+1)!
2) Integrating term by term, ∑ x^(n+1) / (n+1)
但都唔係我想要既 series...
pls help...
[ 本帖最後由 kingwinner 於 10-4-7 06:34 AM 編輯 ]作者: Naozumi 時間: 10-4-7 07:57 AM
你真係太死板啦
xe^x = ∑ x^(n+1) / n!
跟住兩邊都求definite integral 咪得lor作者: kingwinner 時間: 10-4-7 11:21 AM
thx, Q1 識做了, 但 Q2仲係唔識...
Q2)
∞
∑ x^n = 1/(1-x) for |x|<1
n=0
Take the indefinite integral of both sides...
∞
∑ x^(n+1) /(n+1) + C= -ln(1-x)
n=0
Put x=0 => C=0
∞
∑ x^(n+1) /(n+1) = -ln(1-x)
n=0
Divide both sides by x^2
∞
∑ x^(n-1) /(n+1) = - ln(1-x) / x^2
n=0
∞
∑ x^(n-1) /(n+1) = - [ln(1-x) / x^2] - 1/x
n=1
跟住點做?
thx
[ 本帖最後由 kingwinner 於 10-4-7 11:52 AM 編輯 ]作者: Gnoehc 時間: 10-4-9 01:21 AM
It is not necessary to find the closed form of the integral (Though not hard)
If you are patient enough, you have finally (I just compute it through computer)