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(B) Let $S = \sum\limits_{n = 1}^\infty \sum\limits_{k = 1}^{n - 1} \frac{k}{2^{n+k}} = \sum\limits_{n = 1}^\infty \frac{1}{2^n} \sum\limits_{k = 1}^{

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$\frac{4}{9}$

Vedclass · Answer

(B) Let $S = \sum\limits_{n = 1}^\infty \sum\limits_{k = 1}^{n - 1} \frac{k}{2^{n+k}} = \sum\limits_{n = 1}^\infty \frac{1}{2^n} \sum\limits_{k = 1}^{n - 1} \frac{k}{2^k}$.We know that $\sum\limits_{k = 1}^{n - 1} k x^k = \frac{x(1 - nx^{n-1} + (n-1)x^n)}{(1-x)^2}$.For $x = \frac{1}{2}$,$\sum\limits_{k = 1}^{n - 1} \frac{k}{2^k} = \frac{1/2(1 - n(1/2)^{n-1} + (n-1)(1/2)^n)}{(1/2)^2} = 2(1 - \frac{n}{2^{n-1}} + \frac{n-1}{2^n}) = 2 - \frac{2n}{2^{n-1}} + \frac{n-1}{2^{n-1}} = 2 - \frac{n+1}{2^{n-1}}$.Substituting this back,$S = \sum\limits_{n = 1}^\infty \frac{1}{2^n} (2 - \frac{n+1}{2^{n-1}}) = \sum\limits_{n = 1}^\infty (\frac{1}{2^{n-1}} - \frac{n+1}{2^{2n-1}})$.$S = \sum\limits_{n = 1}^\infty \frac{1}{2^{n-1}} - \sum\limits_{n = 1}^\infty \frac{n+1}{2^{2n-1}} = 2 - 2 \sum\limits_{n = 1}^\infty \frac{n+1}{4^n}$.Using $\sum\limits_{n=1}^\infty nx^n = \frac{x}{(1-x)^2}$ and $\sum\limits_{n=1}^\infty x^n = \frac{x}{1-x}$,we have $\sum\limits_{n=1}^\infty (n+1)x^n = \frac{x}{(1-x)^2} + \frac{x}{1-x} = \frac{x + x(1-x)}{(1-x)^2} = \frac{2x-x^2}{(1-x)^2}$.For $x = 1/4$,$\sum\limits_{n=1}^\infty (n+1)(1/4)^n = \frac{2(1/4) - (1/16)}{(3/4)^2} = \frac{1/2 - 1/16}{9/16} = \frac{7/16}{9/16} = \frac{7}{9}$.Thus,$S = 2 - 2(\frac{7}{9}) = 2 - \frac{14}{9} = \frac{4}{9}$.

$\sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^{n - 1} {\frac{k}{{{2^{n + k}}}}} } $ is equal to

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