Finding sum of the infinite series.

Find the sum of the series $$\sum_{n=0}^{\infty} \left ( \frac {\log (\log x)} {n!} \right )^n.$$

Thank you very much.

• You should start from $n=2$ to avoid problems. – tommy1996q Nov 9 '18 at 17:15
• @tommy1996q Why? Note that the logarithm only depends on $x$, not on $n$. – MSobak Nov 9 '18 at 17:16
• Yeah what I have written is perfectly fine. Otherwise the sum would be $e^{\log (\log x)} = \log x$. Which is obvious. – Dbchatto67 Nov 9 '18 at 17:18
• Don't be impressed by the expression $\log(\log x)$, replace it by $t$. – Yves Daoust Nov 9 '18 at 19:34
• Your question has nothing to do with $\log x$ . Define $f(x):=\sum\limits_{n=0}^\infty \left(\frac{x}{n!}\right)^n$ , for which no closed form is known, then the result is $f(\log\log x)$ . But that's trivial. – user90369 Nov 9 '18 at 21:16

There is no closed form expression to this series. The general term is extremely quickly decreasing. With $$t=\log(\log(x))$$,
$$1+t+\frac{t^2}4+\frac{t^3}{216}+\frac{t^4}{331776}+\frac{t^5}{24883200000}+\cdots$$