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I'm trying to find the value of the following sum

$\sum_{n=1}^{\infty} \frac1{n}\left(\frac{np}{p+n}\right)^{n+1}$

where $0. Any ideas? Thanks.

  • 0
    @Patrick, your argument shows that the sum is $\le p^2/(1-p)$. By a similar one, it is also $\ge p^2/(1+p)$.2011-09-24

2 Answers 2

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If $0 then $\dfrac{np}{n+p}<\dfrac{np}{n}=p$. Thus $ \sum_{n=1}^{\infty } \frac{1}{n}\left(\frac{np}{p+n}\right)^{n+1}< \sum_{n=1}^{\infty } \frac{1}{n}\left(p^{n+1}\right) = . . .(1) . . .= -p \ln(1-p) $ (1): If $0, then $ \sum_{n=1}^{\infty }\frac{1}{n}\left(p^{n+1}\right) = p \sum_{n=1}^{\infty } \frac{1}{n}\left(p^{n}\right) =p\int \sum_{n=1}^{\infty} \ p^{n-1} dp =p\int \frac{1}{1-p} dp = -p\ln(1-p) $

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    thanks didier, but I think my answer was correct and you did brief it.2011-09-24
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I can't comment so that's why i'm giving an answer but isn't it true that $\frac 1n \left( \frac {np}{p+n} \right)^{n+1} = \frac 1n \frac {n^{n+1}p^{n+1}}{(p+n)^{n+1}} = \frac {n^np^{n+1}}{(p+n)^{n+1}} = \left( \frac {n^{-1}p}{p+n} \right)^{n+1}$ so couldn't you just sum $\sum _{n=1}^{\infty } \left( \frac {n^{-1}p}{p+n} \right)^{n+1}$ using $\frac {a}{1-r}$ with $a = \left( \frac {p}{p+n} \right)^2$ and $r = \frac {n^{-1}p}{p+n}$ which yields $\sum _{n=1}^{\infty } \left( \frac {n^{-1}p}{p+n} \right)^{n+1} = \frac {\left( \frac {p}{p+n} \right)^2}{1-\frac {n^{-1}p}{p+n}} = \frac{p^2}{(p+n)^2} \frac {p+n}{p+n- n^{-1}p}$? or have i missed something? hope that helps.

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    Don't do that .2011-10-01