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I calculated (with the help of Maple) that the following infinite sum is equal to the fraction on the right side.

$$ \sum_{i=1}^\infty \frac{i}{\vartheta^{i}}=\frac{\vartheta}{(\vartheta-1)^2} $$

However I don't understand how to derive it correctly. I've tried numerous approaches but none of them have worked out so far. Could someone please give me a hint on how to evaluate the infinite sum above and understand the derivation?

Thanks. :)

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I suggest you to consider the geometric series $1 + x + x^2 + x^3 + \cdots = (1 - x)^{-1}$. Differentiating both sides will give you a result that leads to the given identity. – Sangchul Lee Jun 6 '11 at 8:21
Terminological aside to OP: one doesn't "solve" infinite sums, one $\it evaluates$ them. – Gerry Myerson Jun 6 '11 at 13:28
Thanks to all you guys :) @Gerry Myerson Thanks :P My command of English isn't that bad but math terms are seriously giving me a hard time! – Julian Jun 6 '11 at 19:08
up vote 5 down vote accepted

let $$S=\sum_{i=1}^\infty\frac{i}{\theta^i}, (\theta>1),$$ then $$ \begin{align} S-\frac{1}{\theta}S&=\sum_{i=1}^\infty\frac{i}{\theta^i}-\sum_{i=1}^\infty\frac{i}{\theta^{i+1}}\\ &=\sum_{i=1}^\infty\frac{i}{\theta^i}-\sum_{i=2}^\infty\frac{i-1}{\theta^{i}}\\ &=\frac{1}{\theta}+\sum_{i=2}^{\infty}\frac{1}{\theta^i}\\ &=\frac{1}{\theta}+\frac{1}{\theta^2-\theta}, \end{align} $$ which yields $$S=\frac{\theta}{(\theta-1)^2}$$

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Several good methods have been suggested. Here's one more. $$\eqalign{\sum{i\over\theta^i}&={1\over\theta}+{2\over\theta^2}+{3\over\theta^3}+{4\over\theta^4}+\cdots\cr&={1\over\theta}+{1\over\theta^2}+{1\over\theta^3}+{1\over\theta^4}+\cdots\cr&\qquad+{1\over\theta^2}+{1\over\theta^3}+{1\over\theta^4}+\cdots\cr&\qquad\qquad+{1\over\theta^3}+{1\over\theta^4}+\cdots\cr&\qquad\qquad\qquad+{1\over\theta^4}+\cdots\cr&={1/\theta\over1-(1/\theta)}+{1/\theta^2\over1-(1/\theta)}+{1/\theta^3\over1-(1/\theta)}+{1/\theta^4\over1-(1/\theta)}+\cdots\cr}$$ which is a geometric series which you can sum to get the answer.

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Hint: Consider the expectation (first moment) of the geometric distribution. Specifically, letting $1 - p = 1/\vartheta $ in the derivation of the formula ${\rm E}(Y)=(1-p)/p$ (in that link) gives exactly what you are looking for.

Elaborating. It is shown in the Wikipedia link how to derive the equality $$ \sum\limits_{k = 0}^\infty {(1 - p)^k pk} = \frac{{1 - p}}{p}, \;\; 0 < p \leq 1. $$ Letting $1-p = 1/\vartheta $, so that $p=(\vartheta - 1)/\vartheta$, this gives $$ \sum\limits_{k = 1}^\infty {k\frac{1}{{\vartheta ^k }}} = \frac{{1 - p}}{{p^2 }} = \frac{1}{\vartheta }\frac{{\vartheta ^2 }}{{(\vartheta - 1)^2 }} = \frac{\vartheta }{{(\vartheta - 1)^2 }}. $$ That is, $$ \sum\limits_{i = 1}^\infty {\frac{i}{{\vartheta ^i }}} = \frac{\vartheta }{{(\vartheta - 1)^2 }}. $$

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