# geometric sum with probabilities

I have the following sum to consider:

$$p_i/(1+r)^i$$

for i = 1 to 10

My question is, whether this can be somehow evaluated using some sort of mathematical insight without explicitly computing each value? p_i are given and r is const.

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Do the $p_i$ take systematic values somehow? If not, I don't believe there is a mathematical insight to be had. –  Daan Michiels May 10 '12 at 19:24
It's just a polynomial evaluation. If $f(z)=\sum p_i z^i$ then you are looking for $f(\frac{1}{1+r})$. Unless you know more about the $p_i$, then you can't say much more than this. –  Thomas Andrews May 10 '12 at 19:34
pi are pretty much random... –  Bober02 May 10 '12 at 19:40
If $r$ is known to be small, you might be able to estimate it with a power series. At minimum, you'd need $|r|<1$, or no power series will converge, but ideally you'd have an even stricter bound on $r$, like $|r|<\frac{1}{10}$ –  Thomas Andrews May 10 '12 at 19:44
Without an explicit form Daan is right there is no way to get the numerator separated from the denominator to get the geometric part out. Is r an arbitary constant, a positive constant or apositive costant <1. Clearly r must be restricted not to be -1. Now if pi is constant then since Σ 1/(1+r)^i =[ 1-(1+r)^11]/-r summing from 1 to 10 the result would be C [(1+r)^11 -1]/r where pi=C for each i. –  Michael Chernick May 10 '12 at 19:45