special functions - Bounds on geometric sum

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.

Answer

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
$$\dfrac{\partial}{\partial p} x^{-p} = -x^{-p} \log(x)$$