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!

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.
