I am confused over $\mathbb E\left(\frac{1}{1+X}\right)$ where $X$ is geometric distribution with parameter $p$. The book wants me to prove that $\mathbb E(\frac{1}{1+X})=\log\left((1-p)^{\frac{p}{p-1}}\right)$
Here's what I did.
$$\mathbb E\left(\frac{1}{1+X}\right)=\sum_{k=0}^\infty \Pr(X=k) \frac{1}{1+k}=\sum \frac{p^k (1-p)}{1+k}=\frac{1-p}{p}\sum \frac{p^{k+1}}{1+k}=\frac{1-p}{p} \left(-\log(1-p)\right)=\log\left((1-p)^{\frac{p-1}{p}}\right)\text.$$
I don't see a problem with my calculation and I checked with wolfram alpha... it showed that I am right.
Could anyone help?