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How can I find out whether $\sum_{k=1}^{\infty}{\frac{k+1}{2^k}}$ converges? I have split it into

$$ \sum_{k=1}^{\infty}{\frac{k+1}{2^k}} = \sum_{k=1}^{\infty}{\frac{k}{2^k}} + \sum_{k=1}^{\infty}{\frac{1}{2^k}}, $$

and applied the geometric series to the second part of the sum. But how do I deal with the first one to find the limit? I have found some similiar idea:

$$ 4= \sum_{k=1}^{\infty}{\frac{k}{2^{k-1}}} = \sum_{k=1}^{\infty}{\frac{2k}{2^k}}. $$

Is there a way to apply that idea for my case? How can I calculate the limit of / the convergence of $\sum_{k=1}^{\infty}{\frac{k}{2^k}}$?

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3 Answers 3

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Because for all $k\geq10$ we have $2^k>k^3$ and $$\sum_{k=10}^{+\infty}\frac{1}{k^2}$$ converges.

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If $S=\sum_{k=1}^{\infty}{\frac{k}{2^k}}$

then $2S=\sum_{k=1}^{\infty}{\frac{k}{2^{k-1}}}=\sum_{k=0}^{\infty}{\frac{k+1}{2^k}} = 1+\sum_{k=1}^{\infty}{\frac{k+1}{2^{k}}}= 1+\sum_{k=1}^{\infty}{\frac{k}{2^{k}}}+\sum_{k=1}^{\infty}{\frac{1}{2^{k}}} = 2+S$

and so $S=2$ and $\sum_{k=1}^{\infty}{\frac{k+1}{2^{k}}}=3$

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Hint:

$\sum_{k=1}^{\infty} x^{1-k} = \frac{x}{x-1} $ where $|x| >1$

We differentiate above to get

$\sum (1-k)x^{-k} = \frac{-1}{(x-1)^2}$

or

$-\sum kx^{-k} + \sum x^{-k} = \frac{-1}{(x-1)^2}$

put $x=2$ in order to get the value of $\sum k2^{-k}$.

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