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!

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\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 $\ds{x \to \infty}$.

