$X$ 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.

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Please show us your work –  leonbloy Dec 13 '12 at 16:29
Where is your series? –  Patrick Li Dec 13 '12 at 17:03
its a geometric series.... its an expectation formula. I am sorry, but I am unfarmilar with the formating of this site. The series is sigma, index k from 1 to infinity, and the following series (1/k)p (1-p)^k-1 I know I am going to get scolded about not knowing the proper notation on math exchange but I have a final tomorrow . Also, I know there is a general formula but that will not suffice, I will have to know how to derive it. the answer comes out to be -p/(1-p)log(p) I believe according to my professor. –  Olivia Irving Dec 13 '12 at 17:18

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 $a=1-p<1$: $$\sum\limits_{k=1}^\infty {p(1-p)^{k-1}\over k}={-p\ln p\over 1-p}.$$

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Thank you this clarified everything! –  Olivia Irving Dec 13 '12 at 19:12
@OliviaIrving You're welcome. –  David Mitra Dec 14 '12 at 2:50