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How do I evaluate this (find the sum)? It's been a while since I did this kind of calculus.

$$\sum_{i=0}^\infty \frac{i}{4^i}$$

Rose
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  • Hint: Look at $f(x) = \sum \frac{ x^i }{4^i}$. Can you work out what $f(x)$ is in a simpler form? Now...what would $f'(x)$ be? And what would $f'(1)$ be? – John Hughes Sep 03 '15 at 03:18
  • HINT: Let $S = \sum_{i\geq 0} \frac{i}{4^i}$. Compute $4S$. – GAVD Sep 03 '15 at 03:20
  • It's not a geometric series. – zhw. Sep 03 '15 at 03:49
  • This is a duplicate of a duplicate of a … of a duplicate of a duplicate. – Akiva Weinberger Sep 03 '15 at 04:06
  • Some posts which are very similar: http://math.stackexchange.com/questions/87030/proof-by-induction-sum-limits-i-0n-i-2i-1-n1-2n-1 http://math.stackexchange.com/questions/11464/how-to-compute-the-formula-sum-r-1d-r-cdot-2r http://math.stackexchange.com/questions/90637/what-is-the-limit-of-sum-limits-n-1-inftyn2-3n http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn (@columbus8myhw Which question would you suggest as the duplicate target?) – Martin Sleziak Sep 03 '15 at 05:46

3 Answers3

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For $-1<x<1$, the series $\sum_{i=0}^{\infty}x^i$ converges absolutely to $\frac{1}{1-x}$ $$\sum_{i=0}^{\infty}x^i=\frac{1}{1-x}$$ Then \begin{align*} \sum_{i=0}^{\infty}ix^{i-1} &= \frac{d}{dx}\left(\frac{1}{1-x}\right)\\ &=\frac{1}{(1-x)^2}\\ \sum_{i=0}^{\infty}ix^i&=\frac{x}{(1-x)^2} \end{align*} Now, by plugging $x=1/4$ into the last equation, we have $$\sum_{i=0}^{\infty}\frac{i}{4^i}=\frac{1/4}{(1-1/4)^2}=\frac{4}{9}$$

Ángel Mario Gallegos
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Another approach is to write

$$\begin{align} \sum_{i=0}^{\infty}\frac{i}{4^i}&=\sum_{i=1}^{\infty}\frac{1}{4^i}\left(\sum_{j=1}^{i}1\right)\\\\ &=\sum_{j=1}^{\infty}\sum_{i=j}^{\infty}\frac{1}{4^i}\\\\ &=\sum_{j=1}^{\infty}\frac{1}{4^j}\frac{1}{1-\frac14}\\\\ &=\frac{1/4}{(1-\frac14)^2}\\\\ &=\frac49 \end{align}$$

Mark Viola
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Let me try.

Set $$S = \sum_{i\geq 0}\frac{i}{4^i}.$$

Then we have $$4S = \sum_{i \geq 0} \frac{i+1}{4^i} = \sum_{i\geq 0}\frac{i}{4^i} + \sum_{i\geq 0}\frac{1}{4^i} = S + \frac{1}{1-\frac{1}{4}}$$

So, $3S = \frac{4}{3}$. It implies that $$S = \frac{4}{9}.$$

GAVD
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