Evaluating $\sum\limits_{n=1}^{\infty} \frac1{n}\left(\frac{np}{p+n}\right)^{n+1}$ for $0I'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 $0<p<1$. Any ideas? Thanks.
 A: If $0<p<1$ 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 
$0<p<1$, then
$$
\sum_{n=1}^{\infty }\frac{1}{n}\left(p^{n+1}\right) =
p \sum_{n=1}^{\infty } \frac{1}{n}\left(p^{n}\right) =p\int \sum_{n=1}^{\infty} \ p^{n-1} dp =p\int \frac{1}{1-p} dp = -p\ln(1-p)
$$
A: 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.
