b***k
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xiaoxiaokuan (小小矿) 于 (Thu Mar 6 12:49:33 2008) 提到:
sum of n^2*(1/2)^n, n=1 to infinity
thanks
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dArtagnan (达达尼昂) 于 (Thu Mar 6 12:57:43 2008) 提到:
let the sum be S
R = S - 1/2 S = 1*(1/2) + \sum [(n+1)^2 - n^2]*(1/2)^(n+1)]
= 1*(1/2) + \sum (2n+1) (1/2)^(n+1)
and then calculate
R - 1/2 R
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xiaoxiaokuan (小小矿) 于 (Thu Mar 6 12:59:09 2008) 提到:
wow, thanks!
☆─────────── | r***w
发帖数: 35 | 2 No need to use maple, it is elementary, integrate the function f(x)=(1/2)^2x
^2 on [1,\infty], by L'Hospital's rule, you can get
\sum=[(ln2)^2+2*ln2-2]/[2*(ln2)^3]
Or, sum x^n=1/(1-x) (geometric) |
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