asymptotics on series 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!
 A: 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 $x$.
A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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 $\ds{x \to \infty}$.

