Evaluate $\sum_{n=0}^{\infty} \frac{{\left(\left(n+1\right)\ln{2}\right)}^n}{2^n n!}$ Evaluate: $$\sum_{n=0}^{\infty} \frac{{\left(\left(n+1\right)\ln{2}\right)}^n}{2^n n!}$$
I am not sure where to start.  The ${\left(n+1\right)}^n$ term is obnoxious as I can't split the fraction.  Perhaps this can be turned into a double summation by using binomial theorem on ${\left(n+1\right)}^n$?
 A: On the Wikipedia page for the Lambert W function, we find the following Maclaurin series identity:
$$\bigg(\frac{W(x)}{x}\bigg)^r =\sum_{n=0}^\infty \frac{r(n+r)^{n-1}}{n!}(-x)^n$$
Performing some elementary transformations on this series yields the following identitiy:
$$\frac{1}{1+W(x)}\bigg(\frac{W(x)}{x}\bigg)^r=\sum_{n=0}^\infty \frac{(n+r)^n}{n!}(-x)^n$$
If we plug in $r=1$ and $x=-\ln(2)/2$, and use the fact that $W(-\ln(2)/2)=-\ln(2)$, we obtain the value of the series that you’re looking for:
$$\color{green}{\frac{2}{1-\ln(2)}}=\sum_{n=0}^\infty \frac{(n+1)^n}{n!}\bigg(\frac{\ln(2)}{2}\bigg)^n$$
which, corresponding with numerical approximations, is about $6.5178$.
A: Will an upper bound do? Expand $n! \approx (\frac{n}{e})^n \sqrt{2 \pi n}$ and rewrite 
$$
(n+1)^n = n^n (1+\frac{1}{n})^n \leq n^n e
$$
A lot of terms cancel out and you get a sum
$$
\sum_{k=1}^{\infty}\frac{a^k}{\sqrt{k}}
$$
For some $|a|<1$, which is essential to convergence. This sum probably has some solution, but if you are happy with the upper-bound using integral, it is this:
$$
\int_{1}^{\infty}\frac{a^x}{\sqrt{x}} dx
$$
which has solution in the form of an complimentary error function.
