$X$ is a Geometric random variable find the expectation of $1/X$ Let $X$ be a geometric random variable with parameter $p$, find the expectation of $E[1/X]$.
I need help simplifying the series. 
 A: You need to find the value of $\sum\limits_{k=1}^\infty {p(1-p)^{k-1}\over k}$. Towards this end, let's find a formula for the sum of the series $\sum\limits_{k=1}^\infty {a^{k-1}\over k}={1\over a}\sum\limits_{k=1}^\infty {a^k\over k}$ when $0<a<1$. Note that this series indeed converges since it's dominated by a convergent geometric series.
But how to find its sum? 
A hint here is to apply the Integration Theorem for power series to an appropriate series (that is a series that produces the series under scrutiny after integrating term by term).

I'll attempt to do this without making any errors:
Let's consider the series $\sum\limits_{k=1}^\infty {a^{k-1} }$. We will think of this series as a power series in the variable $a$. For a fixed $a\in(0,1)$, term by term integration of this series gives
$$\tag{1}
\int_0^a\sum\limits_{k=1}^\infty {t^{k-1} } \,dt
=\sum\limits_{k=1}^\infty \int_0^a {t^{k-1} } \,dt
=\sum\limits_{k=1}^\infty   {a^{k } \over k}.
$$
But, for $0<t\le a$,  $\sum\limits_{k=1}^\infty {t^{k-1} }={1\over 1-t} $; so the integral on the left hand side of $(1)$ is
$$
\int_0^a {1\over 1-t}\,dt=-\ln(1-a).
$$
We now have
$$
\sum\limits_{k=1}^\infty {a^{k-1}\over k}
={1\over a}\sum\limits_{k=1}^\infty {a^{k}\over k}={-\ln(1-a)\over a};
$$
and so, with $ a=1-p<1$:
$$
\sum\limits_{k=1}^\infty {p(1-p)^{k-1}\over k}={-p\ln p\over 1-p}.
$$
