TITLE: Evaluating $\sum\limits_{n=1}^{\infty} \frac1{n}\left(\frac{np}{p+n}\right)^{n+1}$ for $0<p<1$
QUESTION [15 upvotes]: 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 $0<p<1$. Any ideas? Thanks.

REPLY [2 votes]: 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
$$
\sum_{n=1}^{\infty }\frac{1}{n}\left(p^{n+1}\right) =
p \sum_{n=1}^{\infty } \frac{1}{n}\left(p^{n}\right) =p\int \sum_{n=1}^{\infty} \ p^{n-1} dp =p\int \frac{1}{1-p} dp = -p\ln(1-p)
$$