Summations involving $\sum_k{x^{e^k}}$ I'm interested in the series
$$\sum_{k=0}^\infty{x^{e^k}}$$
I started "decomposing" the function as so:
$$x^{e^k}=e^{(e^k \log{x})}$$
So I believe that as long as $|(e^k \log{x})|<\infty$, we can compose a power series for the exponential.  For example, 
$$e^{(e^k \log{x})}=\frac{(e^k \log{x})^0}{0!}+\frac{(e^k \log{x})^1}{1!}+\frac{(e^k \log{x})^2}{2!}+\dots$$
Then I got a series for
$$\frac{(e^k \log{x})^m}{m!}=\sum_{j=0}^\infty{\frac{m^j \log{x}^m}{m!j!}k^j}$$
THE QUESTION
I believe that we can then plug in the last series into the equation to get
$$\sum_{k=0}^\infty{x^{e^k}}=\sum_{k=0}^\infty{\sum_{j=0}^\infty{ \sum_{m=0}^\infty{\frac{m^j \log{x}^m}{m!j!}k^j} }}$$
Is the order of summations correct? i.e. Must $\sum_k$ come before $\sum_j$?
Also, can we switch the order of the summations?  If so, which order(s) of summations are correct?
 A: If we consider values $\alpha,\beta$ and $t$ greater then zero with $\alpha\beta=2\pi$ then your series can be expressed in relation to several other sums under the following double exponential series identity:
$$\alpha \sum_{k=0}^\infty e^{te^{k\alpha}}=\alpha\left(\frac{1}{2}-\sum_{k=1}^\infty\frac{(-1)^{k}t^k}{k!(e^{k\alpha}-1)}\right)-\gamma-\ln(t)+2\sum_{k=1}^\infty\varphi(k\beta)$$
Where we have:
$$\varphi(\beta)=\frac{1}{\beta}\Im\left(\frac{\Gamma(i\beta+1)}{t^{i\beta}}\right)=\sqrt{\frac{\pi}{\beta\sinh(\pi \beta)}}\cos\left(\beta\log\left(\frac{\beta}{n}\right)-\beta-\frac{\pi}{4}-\frac{B_2}{1\times 2 }\frac{1}{\beta}+\cdots\right)$$
This along with a similar identity was stated without a proof on page $279$ of Ramanujan's second notebook. Though in $1994$ Bruce C Berndt and James Lee Hafner published a proof which can be found here . Unfortunately I can't access the article without paying, however just by looking at the identity I'm more then willing to bet they made use of the Poisson summation formula.
