This question already has an answer here:

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

Or in other words


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


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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ Hint. Differentiate the power series for $1/(1-x)$, then multiply by $x$. $\endgroup$ – Ethan Bolker Sep 26 '17 at 0:20
  • $\begingroup$ That sounds like a great idea! Why don't you do it and I'll give you a lovely green check mark :) $\endgroup$ – Ben S. Sep 26 '17 at 0:25
  • $\begingroup$ I think the third term should have $\frac18$. $\endgroup$ – marty cohen Sep 26 '17 at 0:29
  • $\begingroup$ @martycohen fixed thanks. $\endgroup$ – Ben S. Sep 26 '17 at 0:30
  • $\begingroup$ Nice accepted answer so I won't post mine. It's a good technique to remember. $\endgroup$ – Ethan Bolker Sep 26 '17 at 12:17

Without calculus:

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} $

  • $\begingroup$ 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? $\endgroup$ – Ben S. Sep 26 '17 at 0:40
  • $\begingroup$ $\sum_{n=0}^{\infty} a^{n} = 1/(1-a)$. $\sum_{n=k}^{\infty} a^{n} = a^k/(1-a)$. $\endgroup$ – marty cohen Sep 26 '17 at 2:05
  • $\begingroup$ Aha. Thanks again. $\endgroup$ – Ben S. Sep 26 '17 at 2:14

Not the answer you're looking for? Browse other questions tagged or ask your own question.