Show that the $\sum_{x=1}^{\infty} \frac{-(1-p)^x}{x\log(p)} = 1, x=1,2..., 0<p<1$

Question

I think I'm missing a summation identity or not understanding how to leverage log rules in this equation. If anyone could help that'd be appreciated.

$\sum_{x=1}^{\infty} \frac{-(1-p)^x}{x\log(p)} = 1, x=1,2..., 0<p<1$

so far I've got to $\frac{-1}{\log(p)}\sum_{x=1}^{\infty} \frac{(1-p)^x}{x}$

since (1-p)<1 I feel like I should use the identity $\sum_{x=1}^{\infty} (1-p)^x=\frac{1}{1-(1-p)}=\frac{1}{p}$

and so I'm left with $\frac{-1}{p\log(p)}*\sum_{x=1}^{\infty} \frac{1}{x}$

Obviously that sum goes to 0, which would make the entire sum go to 0. Since it equals 1 I am clearly missing something. Any ideas on where I went wrong? Thanks.

Answer 1

$$S=\sum_{n=1}^{\infty} \frac{-(1-p)^n}{n \ln p}=\frac{-1}{\ln p} \sum_{n=1}^{\infty} \frac{(1-p)^n}{n}= \frac{1}{\ln p} \ln (1-(1-p))= 1.$$ We have used $$\ln(1-z)=-\sum_{k=1}^{\infty} \frac{z^k}{k}.$$

Answer 2

\begin{eqnarray*} \frac{\partial}{\partial p}\sum_{x=1}^\infty\frac{(1-p)^x}x&=&\sum_{x=1}^\infty(1-p)^{x-1}=\frac1p\;. \end{eqnarray*}

Since the sum is $0$ at $p=1$, it follows that it is $\log p$.