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Consider the sum $\sum_{x=1}^{\infty} \frac{\log{x}}{z^x}$. We can assume that $z\geq1$ (and is real). Mathematica gives this sum as

-PolyLog^(1, 0)[0,1/z]

Even after reading the manual page for PolyLog I don't understand what this function is like and I certainly don't know how the sum was derived.

Are there simple upper and lower bounds for this sum?

I also tried to compute $\int_{x=1}^{\infty} \frac{\log{x}}{z^x}$ in the hope that this would shed more light but Mathematica gives

Gamma[0, Log[z]]/Log[z]

which I also didn't find helpful.

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up vote 0 down vote accepted

I assume you mean $\displaystyle \sum_{x=1}^\infty \dfrac{\ln(x)}{z^x}$. Since $z > 1$ and $\ln(x)$ grows very slowly, the terms go to $0$ rapidly, so you can get good bounds by taking a partial sum and bounds for the "tail". For $x \ge n+1$ we have $\ln(n+1) \le \ln(x) \le \ln(n+1) + \dfrac{x-n-1}{n+1}$ so $$ \sum_{x=1}^n \dfrac{\ln(x)}{z^x} + \dfrac{z \ln(n+1)}{z^{n+2}-z^{n+1}} \le \sum_{x=1}^\infty \dfrac{\ln(x)}{z^x} \le \sum_{x=1}^n \dfrac{\ln(x)}{z^x} + \dfrac{1+(z-1)(n+1)\ln(n+1)}{(n+1)(z-1)^2 z^n}$$

Some explanation for the PolyLog: by definition $$\text{PolyLog}(p,1/z) = \sum_{x=1}^\infty \dfrac{x^{-p}}{z^x}$$ Take the derivative of this with respect to $p$ (which is what the $(1,0)$ superscript refers to), evaluated at $p=0$ and you get $-1$ times your sum, because $$ \dfrac{\partial}{\partial p} x^{-p} = -x^{-p} \log(x)$$

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Thanks (edited question to clarify). As this is $E(\log(X))$ for a geometric random variable I am particularly interested to know if it is the same order as $\log{E(X)}$. – Anush Dec 11 '12 at 20:55
What do you mean by "the same order"? – Robert Israel Dec 11 '12 at 21:02
I mean within a constant multiplicative factor, ideally. – Anush Dec 11 '12 at 21:09
What is the relationship between $E(\log(X))$ and $\log{E(X)}$ for a geometric random variable? Jensen's inequality gives you one way but what can you get the other way? (Thanks for the polylog explanation.) – Anush Dec 11 '12 at 22:05
If $\mu = E[X] = 1/p$, we have $P(X \ge \mu) = (1-p)^{\lfloor \mu \rfloor}$ and $E[\log(X)] \ge \log(\mu) P(X \ge \mu)$. Note that as $p \to 0+$, $P(X \ge \mu) \to 1/e$. – Robert Israel Dec 11 '12 at 22:59

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