4
$\begingroup$

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. :)

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – Sangchul Lee Jun 6 '11 at 8:21
  • $\begingroup$ Terminological aside to OP: one doesn't "solve" infinite sums, one $\it evaluates$ them. $\endgroup$ – Gerry Myerson Jun 6 '11 at 13:28
  • $\begingroup$ 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! $\endgroup$ – Julian Jun 6 '11 at 19:08
5
$\begingroup$

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 $$S=\frac{\theta}{(\theta-1)^2}$$

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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, $$ \sum\limits_{i = 1}^\infty {\frac{i}{{\vartheta ^i }}} = \frac{\vartheta }{{(\vartheta - 1)^2 }}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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