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As a follow up of my previous question here is an improved version of the series:

$$\sum_{i=1}^{2NR}{i\cdot \left( \dfrac{1}{1-p} \right)^i}\, \left(1-\dfrac{2R-(2NR-i)\Delta}{2R} \right) $$


  • $p\in [0,1]$
  • $R\in [0,1]$
  • $N$ positive integer
  • $\Delta$ infinitesimal

Notice that for $i=2NR$ the last term of the series, that is actually a probability, is 0 and it is correct.

My question is: it is possible to get a closed form expression?

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up vote 2 down vote accepted

Note that a little algebra will transform your sum into $$\sum_{i=1}^{2NR}\left( Ki +\frac{\Delta}{2R}i^2\right)\left(\frac{1}{1-p}\right)^i,$$ where $K$ is an easily computed constant.

The answer to the previous post tells you how to find $\sum_1^{2NR} ix^i$. Multiply that sum by $K$. The only remaining task is to find $\sum_1^{2NR}i^2x^i$.

This is done using more or less the same idea as the preceding answer by Marvis. Let $f(x)=1+x+x^2+\cdots+x^n$. There is an explicit closed form for $f(x)$. Note that $$xf'(x)=\sum_{i=0}^n ix^{i}.$$ If $g(x)=xf'(x)$, then by the same trick $$xg'(x)= \sum_{i=0}^n i^2x^{i}.$$

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Thanks: the only problem was exactly how to treat the probability term. :) – Claudio Fiandrino May 22 '12 at 14:33

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