Evaluating $\sum\limits_{n=1}^{\infty} \frac1{n}\left(\frac{np}{p+n}\right)^{n+1}$ for $0<p<1$

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.

Solution 1:

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 $$ \sum_{n=1}^{\infty }\frac{1}{n}\left(p^{n+1}\right) = p \sum_{n=1}^{\infty } \frac{1}{n}\left(p^{n}\right) =p\int \sum_{n=1}^{\infty} \ p^{n-1} dp =p\int \frac{1}{1-p} dp = -p\ln(1-p) $$