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Let $X$ be a binomial distributed with parameters $n$ and $p=1/k$. Then we have $E[X]=n/k$. But what is $E[\dfrac{1}{X}\mid X>0]$? Is there a nice closed formula getting the exact value or approximation?

Paul S.
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    Related: http://math.stackexchange.com/questions/154060/sum-with-binomial-coefficients-sum-k-1m-frac1km-choose-k – leonbloy Nov 08 '13 at 00:49

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I don't know a closed formula for this, but you could simply calculate it as $E[\frac{1}{X}|X>0]=\sum\limits_{i=1}^n i^{-1}\frac{B(i;n,p)}{1-B(0;n,p)}$

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I know this is old but I wanted to save others time. There is no known closed form solution, but bounds can be placed. (You've already seen that it can be found numerically)

Jensen's inequality provides a lower bound. $$E[1/X]\ge E[X]$$ I never found an upper bound but Wikipedia does offer a citation for establishing the rough order of an upper bound: $$E[1/(1+X)] \approx O(1/np)+o(1/n)$$ https://en.wikipedia.org/wiki/Inverse_distribution#Reciprocal_of_binomial_distribution https://en.wikipedia.org/wiki/Jensen%27s_inequality

Jojker
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