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Define $$ f(x)=\sum_{k=1}^\infty \frac{x^k}{k\cdot k!}. $$ Is there a way to find the asymptotics of $f(x)$ as $x\rightarrow \infty$? What I suspect is $$ f(x)\sim \frac{e^x}{x}, $$ because \begin{align*} f(x)&=\sum_{k=1}^\infty \frac{x^k}{k\cdot k!}\\ &\geq \sum_{k=1}^\infty \frac{x^k}{(k+1)\cdot k!}\\ &=\frac{1}{x}\sum_{k=1}^\infty \frac{x^{k+1}}{(k+1)!}\\ &=\frac{1}{x}(e^x-1-x)\sim\frac{e^x}{x}, \end{align*} but this is only a lower bound, how to get a similar upper bound? Thanks a lot!

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    what is the series for $f'(x)?$ – 2017-01-17
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    Well there is $f(x)=\sum_{k=1}^\infty \frac{x^k}{k\cdot k!}\leq\sum_{k=1}^\infty \frac{x^k}{k!}=e^x-1\sim e^x$ – 2017-01-17
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    $f'(x)=\frac{e^x-1}{x},$ but it doesn't help, because the anti-derivative will in turn give such a similar series. – 2017-01-17
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    Yes, but $e^x$ has a different order than $\frac{e^x}{x},$ I actually wish to have an exact asymptotics as $\frac{e^x}{x}$ – 2017-01-17
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    You can easily estimate the error you made with that approximation. Since $\frac{1}{k} - \frac{1}{k+1} = \frac{1}{k(k+1)}$, and $\frac{1}{k+2} < \frac{1}{k} \leqslant \frac{3}{k+2}$ for $k \geqslant 1$, the error is (for $x > 0$) $$c(x)\sum_{k = 1}^{\infty} \frac{x^k}{(k+2)!} = c(x) \frac{e^x-1-x - \frac{1}{2}x^2}{x^2}$$ with $1 < c(x) < 3$ (and $c(x) \to 1$ as $x\to\infty$). Thus $$f(x) = \frac{e^x}{x} + \frac{e^x}{x^2} + O\biggl(\frac{e^x}{x^3}\biggr)$$ (where I've skipped the proof of the remainder bound, it's very similar). – 2017-01-17
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    that's right..works so well, thanks... – 2017-01-17

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Noting $$ f'(x)=\sum_{k=1}^\infty \frac{x^{k-1}}{k!}=\frac{e^x-1}{x}$$ one has, \begin{eqnarray} \lim_{x\to\infty}\frac{f(x)}{\frac{e^x}{x}}=\lim_{x\to\infty}\frac{f'(x)}{\frac{e^xx-e^x}{x^2}}=\lim_{x\to\infty}\frac{\frac{e^x-1}{x}}{\frac{e^xx-e^x}{x^2}}=\lim_{x\to\infty}\frac{e^x-1}{e^x}\frac{x}{x-1}=1 \end{eqnarray} and hence $$ f(x)\sim\frac{e^x}{x} $$ for big $x$.

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    Oh, this works so nice, thanks a lot..... – 2017-01-17
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Here it is a (very!) brute-force approach. Both $\frac{e^t-1}{t}$ and its primitive are entire functions, hence

$$ g(x)\stackrel{\text{def}}{=}\frac{1}{x\,e^x}\sum_{k\geq 1}\frac{x^k}{k\cdot k!} = e^{-x}\sum_{k\geq 0}\frac{x^k}{(k+1)(k+1)!} = \sum_{j,k\geq 0}\frac{(-1)^j x^{k+j}}{(k+1)^2 j!k!}$$ leads to: $$ g(x) = \sum_{s\geq 0}\frac{x^s}{s!}\sum_{k=0}^{s}\frac{\binom{s}{k}(-1)^{s-k}}{(k+1)^2}=\sum_{s\geq 0}\frac{x^s}{s!}\int_{0}^{1}\sum_{k=0}^{s}\binom{s}{k}(-1)^{s-k}u^k(-\log u)\,du$$ then to: $$ g(x) = \sum_{s\geq 0}\frac{x^s (-1)^s}{s!}\int_{0}^{1}(1-u)^s(-\log u)\,du = \sum_{s\geq 0}\frac{x^s (-1)^s H_{s+1}}{(s+1)!}$$ and the behaviour in a neighbourhood of the origin has no secrets anymore.
We may deal with the behavior in a left neighbourhood of $+\infty$ by noticing that, by Frullani's theorem, the Laplace transform of $\frac{e^x-1}{x}$ is given by $-\log\frac{s-1}{s}$ for $s>1$. It follows that: $$ \mathcal{L}\left(\sum_{k\geq 1}\frac{x^k}{k\cdot k!}\right)=-\frac{1}{s}\,\log\left(\frac{s-1}{s}\right)=\mathcal{L}\left(\frac{e^x-1}{x}\right)+\frac{s-1}{s}\,\log\left(\frac{s-1}{s}\right) $$ and: $$ \mathcal{L}\left(e^{-x}\sum_{k\geq 1}\frac{x^k}{k\cdot k!}\right)=\frac{1}{s+1}\log\left(1+\frac{1}{s}\right)$$ $$ \mathcal{L}\left(x\, e^{-x}\sum_{k\geq 1}\frac{x^k}{k\cdot k!}\right)=\frac{1\color{red}{-s\log\left(\frac{s}{s+1}\right)}}{s\color{blue}{(s+1)^2}}$$ where the red term $\color{red}{\to 0}$ and the blue term $\color{blue}{\to 1}$ as $s\to 0^{+}$. This proves the asymptotic behaviour of $\int_{0}^{x}\frac{e^t-1}{t}\,dt$ as $x\to +\infty$ is given by $\frac{e^x}{x}\mathcal{L}^{-1}\left(\frac{1}{s}\right)=\frac{e^x}{x}$ as wanted.

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{f}\pars{x} & \equiv \sum_{k = 1}^{\infty}{x^{k} \over k\ k!} = \sum_{k = 1}^{\infty}{x^{k} \over k!}\int_{0}^{1}t^{k - 1}\,\dd t = \int_{0}^{1}{1 \over t}\sum_{k = 1}^{\infty}{\pars{xt}^{k} \over k!}\,\dd t = \int_{0}^{1}{1 \over t}\pars{\expo{xt} - 1}\,\dd t \\[5mm] & = \int_{0}^{x}{\expo{t} - 1 \over t}\,\dd t = \int_{t\ =\ 0}^{t\ =\ x}{1 \over t}\,\dd\pars{\expo{t} - t} = {\expo{x} - x \over x} + \int_{0}^{x}{\expo{t} - t \over t^{2}}\,\dd t \\[5mm] & = {\expo{x} - x \over x} + \int_{t\ =\ 0}^{t\ =\ x} {1 \over t^{2}}\,\dd\pars{\expo{t} - {1 \over 2}\,t^{2}} = {\expo{x} - x \over x} + {\expo{x} - x^{2}/2 \over x^{2}} + 2\int_{0}^{x}{\expo{t} - t^{2}/2 \over t^{3}}\,\dd t \end{align}

It's clear that $\ds{\,\mrm{f}\pars{x} \sim {\expo{x} \over x}}$ as $\ds{x \to \infty}$.