How can I get a good approximation of $\sum_{s=1}^\infty x^s\ln(s)\ $?

Question

How can I get a good approximation of the sum

$$\sum_{s=1}^\infty x^s\ln(s)$$

by hand ?

If I only consider the part of the function, where it is strictly decreasing, I can bound the sum by integrals. But how can I approximate the sum upto this point ?

Answer 1

Assuming $|x|<1$, by Frullani's theorem we have $$ \log(s) = \int_{0}^{+\infty}\frac{e^{-z}-e^{-sz}}{z}\,dz \tag{1}$$ hence $$ \sum_{s\geq 1}x^s \log(s) = \int_{0}^{+\infty}\sum_{s\geq 1}\frac{(e^{-z}-e^{-sz})x^s}{z}\,dz = \frac{x^2}{1-x}\color{purple}{\int_{0}^{+\infty}\frac{1-e^{-z}}{z(e^z-x)}\,dz}\tag{2}$$ where the purple integral, for a fixed value of $x\in(-1,1)$, is not difficult to approximate through integration by parts, Cauchy-Schwarz or both. It is interesting to notice that such integral representation also provides an analytic continuation of the LHS of $(2)$: $$ \lim_{x\to -1^+}\sum_{s\geq 1}x^{s}\log(s) = \frac{1}{2}\int_{0}^{+\infty}\frac{1-e^{-z}}{z(e^z+1)}\,dz = \log\sqrt{\frac{\pi}{2}}\tag{3} $$ in agreement with Wallis product.

Answer 2

$|x|<1$

$$f(x):=\sum_{s=2}^\infty x^s\ln(s)$$

$$f(x)+f(-x)=2x^{-2}\sum_{s=2}^\infty x^{2s}\ln(2s)$$

$$=2x^{-2}\sum_{s=2}^\infty x^{2s}\left(\ln(2)+\ln(s)\right)$$

$$f(x)+f(-x)=2\ln(2)\frac{x^2}{1-x^2}+2\frac{f(x^2)}{x^2}$$

So as long as we can approximate it for $x>0$, this relationship allows us to calculate for $x<0$, and vice versa.

Note that for all $x>0$

$$\frac{x^2}{1-x}=\sum_{s=2}^\infty x^s<f(x)$$

It also seems to be fairly good of an approximation as well that is fairly easy. Try it for a few values?

Answer 3

You may at least improve convergence by the following manipulation: With $f(x) =\sum_{s\geq 1} x^s \log(s)$ we have: $$(1-x) f(x) = \sum_{s\geq 2} x^s \log(s) - \sum_{s\geq 1} x^{s+1} \log(s) = \sum_{s\geq 1} x^{s+1} \log(1+\frac{1}{s}) $$ You may then use $\log(1+\frac{1}{s}) = \frac{1}{s} - \frac{1}{2s^2} + ...$ to derive $$ f(x) = \frac{x}{1-x} \sum_{s\geq 1} x^s \sum_{k\geq 1} \frac{(-1)^{k+1}}{k s^k} = \frac{x}{1-x} \sum_{k\geq 1} \frac{(-1)^{k+1}}{k} \left( \sum_{s\geq 1} \frac{ x^s}{s^k} \right)$$ $$ f(x) = \frac{x}{1-x} \left( -\log(1-x) +\sum_{k\geq 2} \frac{(-1)^{k+1}}{k} \sum_{s\geq 1} \frac{ x^s}{s^k} \right)$$ This isolates the principal singular behavior and you may then find methods (there are more or less explicit but not very useful formulae) for approximating the sums $\sum_{s\geq 1} x^s/s^k$, $k\geq 2$.