# Summation of an infinite sequence: $\sum\limits_{k=1}^\infty \frac{k}{2^k}$ [duplicate]

I am wondering how to sum this infinite sequence: $\displaystyle\sum\limits_{k=1}^\infty \frac{k}{2^k}$. According to wolfram alpha this equals to $2$, but I would like to know why.

I'll never pass an opportunity to link: mathdl.maa.org/images/cms_upload/268948749035.pdf (which gives the value of ${1\over2}\sum_{k=1}^\infty {k\over2^k}$). –  David Mitra Dec 17 '11 at 12:27
@David: I like $$\begin{array}{} \frac12&+&\frac14&+&\frac18&+&\frac1{16}&+&\dots&=&1\\ &&\frac14&+&\frac18&+&\frac1{16}&+&\dots&=&\frac12\\ &&&&\frac18&+&\frac1{16}&+&\dots&=&\frac14\\ &&&&&&\frac1{16}&+&\dots&=&\frac18\\ &&&&&&&&\vdots&&\vdots\\ \hline \frac12&+&\frac24&+&\frac38&+&\frac4{16}&+&\dots&=&2 \end{array}$$ –  Brian M. Scott Dec 17 '11 at 12:37