This question already has an answer here:

In the following series:

$ \displaystyle\sum\limits_{n=1}^{\infty}n*\frac{1}{2^n}$

I've found that the series converges to 2 by looking it up but how would one calculate the summation? You can't use the formula for a geometric sum because the series' terms don't differ by a common ratio. I'm sure I'm just missing something but it's been a while since I've done series. Thanks!


marked as duplicate by Martin R, mathematics2x2life, Brian Borchers, Xander Henderson, user99914 Mar 21 '18 at 17:33

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.

  • $\begingroup$ ooh that is a duplicate. Thanks! $\endgroup$ – Anthony Mar 21 '18 at 16:27

When you list out the terms according to the given general term by plugging in different values of $n$, you would find that it is an Arithmetico - Geometrtic Progression (AGP).

Let $$S = \displaystyle\sum\limits_{n=1}^{\infty}n*\frac{1}{2^n}$$ $$\implies S = \frac12 + \frac{2}{2^2} + \frac{3}{2^3} + .....$$

$$\implies \frac{S}{2} = \frac{1}{2^2} + \frac{2}{2^3} + \frac{3}{2^4} + $$ [After shifting the terms on the RHS by one place to the right.]

Subtracting; $$\implies \frac{S}{2} = \frac12 + \frac{1}{2^2} + \frac{1}{2^3} + ....$$ which yields $$\frac{S}{2} = 1$$
and thus $$S = 2$$

  • $\begingroup$ What are the rules again when subtracting infinite sums? Is it okay to do it as long as you're not cancelling stuff out? $\endgroup$ – Anthony Mar 21 '18 at 16:28
  • $\begingroup$ @Anthony: IIRC it's a legitimate move when the series are absolutely convergent. In this case, both $S$ and $S/2$ can be seen to be absolutely convergent by applying the ratio test. $\endgroup$ – Michael Seifert Mar 21 '18 at 18:12
  • $\begingroup$ @Anthony I forgot to mention the fact that $\frac{S}{2}$ has been written after one right shift, which leads to the subsequent computation. Edited the post now. $\endgroup$ – Saksham Mar 22 '18 at 7:58

Let $$ f(x) = \sum_{n = 0}^\infty x^n = \frac{1}{1-x}, $$ assuming that $|x| < 1$. Then we have $$ x f'(x) = x \sum_{n = 0}^\infty n x^{n-1} = \sum_{n = 0}^\infty n x^n, $$ but also $$ x f'(x) = \frac{x}{(1-x)^2}. $$ Thus, $$ \sum_{n = 0}^\infty n x^n = \frac{x}{(1-x)^2} $$ for all $|x| < 1$. In particular, with $x = 1/2$, the series sums to 2, as you found.

  • $\begingroup$ Very cool. Thanks! $\endgroup$ – Anthony Mar 21 '18 at 15:19

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