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I believe, based on numerical evidence, that $$\sum_{n=2}^\infty \frac{x^n}{(\log n) n!} \sim \frac{\exp(x)}{\log(x)}$$ as $x\to\infty$. However, I am not sure how to prove this. What would be a good way to approach this problem?

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2 Answers 2

up vote 7 down vote accepted

Consider the related finite sum $$ s(x)=\sum\limits_n\frac{x^n}{n!}\mathbf 1_{x/2\leqslant n\leqslant 2x}=\mathrm e^x\cdot\mathbb P(x/2\leqslant N_x\leqslant 2x), $$ where $N_x$ is a Poisson random variable with parameter $x$. A large deviation estimates ensures that, for every $x$ large enough, $\mathbb P(x/2\leqslant N_x\leqslant 2x)=1-p(x)$ where $0\lt p(x)\leqslant\mathrm e^{-cx}$ for some positive $c$.

The series $f(x)$ to be estimated is $f(x)=g(x)+h(x)$ where $g(x)$ sums the terms such that $x/2\leqslant n\leqslant2x$ and $h(x)$ sums the other terms. Note that $0\leqslant h(x)\leqslant\mathrm e^x p(x)/\log2$ and $s(x)/\log(2x)\leqslant g(x)\leqslant s(x)/\log(x/2)$, hence $$ \mathrm e^{x}\frac{1-\mathrm e^{-cx}}{\log x+\log2}\leqslant f(x)\leqslant\frac{\mathrm e^x}{\log x-\log2}+\mathrm e^x\frac{\mathrm e^{-cx}}{\log2}. $$ Finally, $$ f(x)=\frac{\mathrm e^x}{\log x}\,\left(1+O\left(\frac1{\log x}\right)\right). $$ Note: With some more care, one can replace the $O\left(\frac1{\log x}\right)$ error term by $O\left(x^{a-1/2}\right)$, for every positive $a$.

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Here is a possible way but is not a complete solution. (This is a bit too long for a comment and hence have community wikied it)

From Stirling, we have that $$n! \sim \sqrt{2 \pi n} \left(\dfrac{n}{e} \right)^n$$ Equivalently, $\exp(x) \sim \sqrt{2 \pi} \dfrac{x^{x+1/2}}{x!}$. Hence, $$\exp(x) \sim \sum_{n= x- \sqrt{x}}^{x+\sqrt{x}} \dfrac{x^n}{n!}$$ (The above step needs proof.)

Hence, $$\sum_{n= x- \sqrt{x}}^{x+\sqrt{x}} \dfrac{x^n}{n! \log (n)} \sim \sum_{n= x- \sqrt{x}}^{x+\sqrt{x}} \dfrac{x^n}{n! \log (x)} \sim \dfrac{\exp(x)}{\log(x)}$$ where we have used the fact that $\log(x \pm \sqrt{x}) \sim \log(x)$.

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For each positive $a$, the sum from $x-a\sqrt{x}$ to $x+a\sqrt{x}$ is equivalent to a multiple of $e^x$, not to $e^x$. –  Did Nov 5 '12 at 9:09
    
@did Isn't that absorbed in $\sqrt{2 \pi}$? Anyway let me get back to it tomorrow morning to try and fix this. –  user17762 Nov 5 '12 at 9:21

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