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I have problem understanding this when I am reading about the generating function of random walk. $f_0(n)$ is the probability of "first" return after n steps in random walk.

If $\sum_{n=1}^{\infty}f_0(n) = 1$, i.e. the walker is certain to return, why is $\sum_{n=1}^{\infty}nf_0(n) = \infty$?

Intuitively, $\sum_{n=1}^{\infty}nf_0(n)$ is infinity as $n$ gets to infinity so the sum should blow up. But if $f_0(n)$ goes like $1/n^3$ i think this may not be true. However, $1/n^3$ does not satisfy $\sum_{n=1}^{\infty}f_0(n) = 1$. So the constraint $\sum_{n=1}^{\infty}f_0(n) = 1$ posts a limitation on the functional property of $f_0(n)$.

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    What is $f_0(n)$? – 2017-01-23
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    The theorem is false. Let $f_0(n) = e^{-n}$, and $S$ be the sum over all $n$. Then $\frac{1}{S} \sum_{n=1}^{\infty} e^{-n} = 1$ but $\sum_{n=1}^{\infty} ne^{-n}$ is easily seen to converge. – 2017-01-23
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    Even easier, let f0 (n) = 1 if n = 1, 0 otherwise. – 2017-01-23
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    f0(n) is the probability of "first" return occurs after n steps. I miss the "first" word before. So I think it has some special property. – 2017-01-23
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    Well we don't know what that property is, so maybe you want to find it and put it in the post. – 2017-01-23

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