sumlimits_{n = 1}^infty {sumlimits_{k = 1}^{ | vedclass

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(B) Let $S = \sum_{n=1}^\infty \sum_{k=1}^{n-1} \frac{k}{2^{n+k}}$.We can rewrite the sum as $S = \sum_{n=1}^\infty \frac{1}{2^n} \sum_{k=1}^{n-1} \fr

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$\frac {4}{9}$

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(B) Let $S = \sum_{n=1}^\infty \sum_{k=1}^{n-1} \frac{k}{2^{n+k}}$.We can rewrite the sum as $S = \sum_{n=1}^\infty \frac{1}{2^n} \sum_{k=1}^{n-1} \frac{k}{2^k}$.Using the formula for an arithmetico-geometric series,$\sum_{k=1}^{n-1} k x^k = \frac{x(1-x^{n-1})}{(1-x)^2} - \frac{(n-1)x^n}{1-x}$. For $x = \frac{1}{2}$,this simplifies to $\sum_{k=1}^{n-1} \frac{k}{2^k} = 2 - \frac{n+1}{2^{n-1}}$.Substituting this back into the expression for $S$:$S = \sum_{n=1}^\infty \frac{1}{2^n} \left( 2 - \frac{n+1}{2^{n-1}} \right) = \sum_{n=1}^\infty \left( \frac{2}{2^n} - \frac{n+1}{2^{2n-1}} \right)$.$S = \sum_{n=1}^\infty \frac{1}{2^{n-1}} - \sum_{n=1}^\infty \frac{n+1}{2^{2n-1}}$.The first part is a geometric series: $\sum_{n=1}^\infty \frac{1}{2^{n-1}} = 1 + \frac{1}{2} + \frac{1}{4} + \dots = \frac{1}{1 - 1/2} = 2$.The second part is $\sum_{n=1}^\infty \frac{n+1}{2^{2n-1}} = 2 \sum_{n=1}^\infty \frac{n+1}{4^n} = 2 \left( \sum_{n=1}^\infty \frac{n}{4^n} + \sum_{n=1}^\infty \frac{1}{4^n} \right)$.Using $\sum_{n=1}^\infty n x^n = \frac{x}{(1-x)^2}$,for $x = 1/4$,we get $\frac{1/4}{(3/4)^2} = \frac{4}{9}$.Using $\sum_{n=1}^\infty x^n = \frac{x}{1-x}$,for $x = 1/4$,we get $\frac{1/4}{3/4} = \frac{1}{3}$.So,the second part is $2 \left( \frac{4}{9} + \frac{1}{3} \right) = 2 \left( \frac{4+3}{9} \right) = \frac{14}{9}$.Finally,$S = 2 - \frac{14}{9} = \frac{18-14}{9} = \frac{4}{9}$.

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

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