Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How it is possible to approximate:

$$\sum_{i=1}^{NR}{i\cdot \left( \dfrac{1}{1-p} \right)^i} $$

share|cite|improve this question
Yes, sorry. $p \in [0,1]$ and $NR$ positive integer, bounded to a given value, 10000 for example. – Claudio Fiandrino May 17 '12 at 8:25
up vote 7 down vote accepted

Let $x = \dfrac1{1-p}$ and let $n = NR$. Then we are interested in the sum $\displaystyle \sum_{i=1}^{n} i x^i$.

\begin{align} \sum_{i=1}^{n} i x^i & = x \left(\sum_{i=1}^{n} i x^{i-1} \right)\\ & = x \left( \sum_{i=1}^{n} \frac{d x^i}{dx} \right)\\ & = x \frac{d}{dx} \left( \sum_{i=1}^{n} x^i\right)\\ & = x \frac{d}{dx} \left( x\left(\frac{x^n - 1}{x-1} \right) \right)\\ & = x \left( \frac{nx^{n+1} - (n+1)x^n + 1}{(x-1)^2} \right) \end{align}

Replacing $x$ by $\dfrac1{1-p}$ and $n$ by $NR$, we get that the solution is $$\left( \frac{NR}{p} - \frac1{p^2} + \frac1p \right) \left( \dfrac1{1-p}\right)^{NR} + \frac1{p^2} - \frac1p$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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