# Find the value of sum (n/2^n) [duplicate]

I have the series $\sum_{n=0}^\infty \frac{n}{2^n}$. I must show that it converges to 2.

I was given a hint to take the derivative of $\sum_{n=0}^\infty x^n$ and multiply by $x$ , which gives

$\sum_{n=1}^\infty nx^n$ , or $\sum_{n=0}^\infty nx^n$.

Clearly if I take $x=\frac{1}{2}$ , the series is $\sum_{n=0}^\infty \frac{n}{2^n}$. How do I proceed from here?

## marked as duplicate by Martin Sleziak, Claude Leibovici calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 18 '16 at 9:37

Notice that if $|x|<1$ then the original series converges with
$$\sum_{n=0}^\infty x^n \;\; =\;\; \frac{1}{1-x}.$$
Computing the derivative and plugging in $x=\frac{1}{2}$ should hopefully seem easier now.