# How to find the sum of this series? [duplicate]

The series is $1\cdot\frac{1}{2} + 2\cdot\frac{1}{4} + 3\cdot\frac{1}{8} + \cdots$

Or in other words

$$\sum_{n=1}^{\infty}\frac{n}{2^n}$$

What kind of series is this and how to find the sum? Thanks....

## marked as duplicate by Chris Culter, Mark Viola sequences-and-series 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(); } ); }); }); Sep 26 '17 at 0:54

• Hint. Differentiate the power series for $1/(1-x)$, then multiply by $x$. – Ethan Bolker Sep 26 '17 at 0:20
• That sounds like a great idea! Why don't you do it and I'll give you a lovely green check mark :) – Ben S. Sep 26 '17 at 0:25
• I think the third term should have $\frac18$. – marty cohen Sep 26 '17 at 0:29
• @martycohen fixed thanks. – Ben S. Sep 26 '17 at 0:30
• Nice accepted answer so I won't post mine. It's a good technique to remember. – Ethan Bolker Sep 26 '17 at 12:17

If $s(a) =\sum_{n=0}^{\infty} na^n$ for $|a| < 1$, then
$\begin{array}\\ as(a) &=\sum_{n=0}^{\infty} na^{n+1}\\ &=\sum_{n=1}^{\infty} (n-1)a^{n}\\ &=\sum_{n=1}^{\infty} na^{n}-\sum_{n=1}^{\infty} a^{n}\\ &=s(a)-\dfrac{a}{1-a}\\ \text{so}\\ s(a) &=\dfrac{a}{(1-a)^2}\\ \end{array}$
• Thank you. I am a little confused by one thing. You have $\sum_{n=1}^{\infty} a^{n} = a/(1-a)$ but I thought that it was $\sum_{n=0}^{\infty} a^{n}$ that summed to that? – Ben S. Sep 26 '17 at 0:40
• $\sum_{n=0}^{\infty} a^{n} = 1/(1-a)$. $\sum_{n=k}^{\infty} a^{n} = a^k/(1-a)$. – marty cohen Sep 26 '17 at 2:05