Matematické Fórum

stuart clark · 29. 10. 2014 09:06

Calculation of $\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{k}{2^{n+k}} =$

Brano · 29. 10. 2014 11:34

$\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{k}{2^{n+k}} = \sum_{k=1}^{\infty}\frac{k}{2^{k}}\sum_{n=1}^{\infty}\frac{1}{2^{n}} = \sum_{k=1}^{\infty}\frac{k}{2^{k}}=(*)$

$\sum_{k=0}^\infty x^k=\frac{1}{1-x}$ differentiating both sides and multiplying by $x$ yields
$\sum_{k=0}^\infty kx^k=\frac{x}{(1-x)^2}$ so

$(*)=2$

stuart clark · 01. 11. 2014 07:51

Thanks ↑ Brano:

Matematické Fórum

#1 29. 10. 2014 09:06

Sum of infinite Series

#2 29. 10. 2014 11:34

Re: Sum of infinite Series

#3 01. 11. 2014 07:51

Re: Sum of infinite Series

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