Evaluating $\sum_{i=1}^{\infty}\frac{(i\ln 2)^i}{2^ii!}$

Question

I seek the proof of the evaluation to the sum $$\sum_{i=1}^{\infty}\frac{(i\ln 2)^i}{2^ii!} = \frac{1}{1-\ln2}-1 \approx 2.25889.$$

It is almost a power series, if not for the $i$ in the bracket.

I have no idea how to proceed. I tried integrating and differentiating the summand (as it is apparently a common technique), but the $i^i$ term causes problems.

It might be related to the fact that the sum is equal to the geometric sum $\sum_{i=1}^{\infty}\ln^i2$.

Any solutions would be greatly appreciated.

Answer 1

Use the following formula: $$\mathrm{W}(-x)=-\sum_{n=1}^\infty\frac{n^{n-1}}{n!}x^n, \quad x\in(-1/e,1/e)$$ (for reference, see Derivation of the series for $W(x)$) here $W$ is Lambert W function, then $$xW'(-x)=\sum_{n=1}^\infty\frac{n^{n}}{n!}x^n.$$ Take $x=\frac{\ln2}{2}\in(-1/e,1/e)$, we will get $$ \sum_{n=1}^{\infty}\frac{(n\ln 2)^n}{2^n\cdot n!}=\frac{\ln2}{2}W'\left(-\frac{\ln2}{2}\right).$$ Use $$W\left(\frac{-\log(x)}{x}\right)=\log({x}^{-1}),\quad W'(z)=\frac{1}{z+e^{W(z)}},$$ ( see $W\left(\frac{-\log(x)}{x}\right)=\log({x}^{-1})$) we can get $$\sum_{n=1}^{\infty}\frac{(n\ln 2)^n}{2^n\cdot n!} =\frac{\ln2}{2}\frac{1}{-(\ln2)/2+e^{W(-(\ln2)/2)}} =\frac{\ln2}{2}\frac{1}{\frac12-(\ln2)/2} =\frac{\ln2}{1-\ln 2}.$$

For the reference of $W'(z)=\frac{1}{z+e^{W(z)}}$, we can refer Lambert W function

Answer 2

This is too long for a comment.

For a quick approximation of the partial sums $$a_i=\frac{i^i \,\left(\frac{\log (2)}{2}\right)^i}{i!}$$

Let $$\alpha=\log \left(\frac{1}{2} e \log (2)\right)$$

Using the asymptotics $$a_i=\frac{e^{\alpha \,i}}{\sqrt{2 \pi } \sqrt{i}}\Bigg(1-\frac{1}{12\, i}+\frac{1}{288\, i^2}+\frac{139}{51840\,i^3}-\frac{571}{2488320 \,i^4}+O\left(\frac{1}{i^5}\right)\Bigg)$$ and $$\sum_{i=1}^p \frac { e^{\alpha \,i}}{i^{n+\frac{1}{2}} }=\text{Li}_{n+\frac{1}{2}}\left(e^{\alpha }\right)-e^{\alpha (p+1)}\,\,\Phi \left(e^{\alpha },n+\frac{1}{2},p+1\right)$$ where appear the polylogarithm function and the Lerch transcendent function.