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What would be the simplest way to prove that $\sum\limits_{i=1}^{\infty}\dfrac{i}{2^i}$ converges?

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closed as off-topic by Najib Idrissi, drhab, Yiorgos S. Smyrlis, Asaf Karagila, M. Vinay Sep 20 '14 at 9:33

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Another possible way : consider $$\sum\limits_{i=1}^{\infty}{i}{x^i}=x\sum\limits_{i=1}^{\infty}{i}{x^{i-1}}=x \frac{d}{dx}\Big(\sum\limits_{i=1}^{\infty}{x^i}\Big)=x \frac{d}{dx}\Big(\frac{x}{1-x}\Big)=\frac{x}{(1-x)^2}$$ Now, replace $x$ by $\frac{1}{2}$ to get $2$.

The same procedure would apply to $$\sum\limits_{i=1}^{\infty}\dfrac{i}{a^i}=\frac{a}{(a-1)^2}$$

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  • $\begingroup$ This is precisely what occurred to me. +1 for you. $\endgroup$ – MPW Sep 20 '14 at 6:37
  • $\begingroup$ Perfect! Thank you so much. $\endgroup$ – crgolden Sep 20 '14 at 8:01
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We have $$\lim_{n\to\infty} \left(\frac{n}{2^n}\right)^{1/n}=\lim_{n\to\infty}\frac{n^{1/n}}{2}=\frac{1}{2},$$ so by the Root Test our series converges. The Ratio Test also works nicely.

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One way is this:

  1. Prove that for all $i\ge 1$, $i<(1.9)^i$.

  2. Comparison test with $\sum \frac{(1.9)^i}{2^i}=\sum (\frac{1.9}{2})^i$, a convergent geometric series.

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