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

How do I proceed? Please help me in this regard.

Thank you very much.

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  • $\begingroup$ You should start from $n=2$ to avoid problems. $\endgroup$ – tommy1996q Nov 9 '18 at 17:15
  • $\begingroup$ @tommy1996q Why? Note that the logarithm only depends on $x$, not on $n$. $\endgroup$ – MSobak Nov 9 '18 at 17:16
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    $\begingroup$ Yeah what I have written is perfectly fine. Otherwise the sum would be $e^{\log (\log x)} = \log x$. Which is obvious. $\endgroup$ – Dbchatto67 Nov 9 '18 at 17:18
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    $\begingroup$ Don't be impressed by the expression $\log(\log x)$, replace it by $t$. $\endgroup$ – Yves Daoust Nov 9 '18 at 19:34
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    $\begingroup$ 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. $\endgroup$ – user90369 Nov 9 '18 at 21:16
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There is no closed form expression to this series. The general term is extremely quickly decreasing. With $t=\log(\log(x))$,

$$1+t+\frac{t^2}4+\frac{t^3}{216}+\frac{t^4}{331776}+\frac{t^5}{24883200000}+\cdots$$

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    $\begingroup$ Do you have an argument that there is no closed form expression? $\endgroup$ – Henning Makholm Nov 9 '18 at 19:40
  • $\begingroup$ @HenningMakholm: firm conviction. (Note that Alpha has nothing to say about it.) I am also pretty sure that the question was wrongly reproduced. $\endgroup$ – Yves Daoust Nov 9 '18 at 19:43
  • $\begingroup$ @HenningMakholm: other argument is that this function is so close to a polynomial, that it would take the usual functions terrible contortions to match it. $\endgroup$ – Yves Daoust Nov 9 '18 at 19:53
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    $\begingroup$ I agree that it looks very probably true, but I had hoped that "so close to a polynomial" or something like it corresponded to a known rigorous criterion. $\endgroup$ – Henning Makholm Nov 9 '18 at 19:56
  • $\begingroup$ @HenningMakholm: so had I. If one could find a differential equation, then one could use the Liouville theorem. $\endgroup$ – Yves Daoust Nov 9 '18 at 20:07

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