An interesting sum to infinity Is there any simple way of computing the following sum?
$$\sum_{k=1}^\infty \frac1{k\space k!}$$
 A: The exponential integral function can be written as:
$$
\mathrm{Ei}(x) = \gamma + \log|x| + \sum_{k=1}^{\infty} \frac{x^k}{k\; k!}
$$
Plug $x = 1$ to get:
$$
\sum_{k=1}^{\infty} \frac{1}{k\; k!} = \mathrm{Ei}(1) - \gamma
$$
Where $\gamma$ is Euler–Mascheroni constant and $\mathrm{Ei}(1)$ is given by A091725.
A: $\def\d{\delta}
\def\e{\epsilon}
\def\g{\gamma}
\def\pv{\mathrm{PV}}
\def\pv{\mathcal{P}}
\def\pv{\mathrm{P}}$We show another way to get the integral representation of the sum and explain its relation to the exponential integral.
Let
$$S(x) = \sum_{k=1}^\infty \frac{x^k}{k k!}.$$
The sum we are interested in is $S(1)$, but, as is often the case, it is easier to get the sum for any $x>0$.
(There is a straightforward extension to $x<0$.)
Notice that
$$\begin{eqnarray*}
S'(x) &=& \sum_{k=1}^\infty \frac{x^{k-1}}{k!} \\
&=& \frac{1}{x}\left( \sum_{k=0}^\infty \frac{x^{k}}{k!} - 1\right) \\
&=& \frac{e^x-1}{x}.
\end{eqnarray*}$$
Therefore,
$\displaystyle S(x) = \int_a^x dt\, \frac{e^t-1}{t}.$
To find $a$ just notice that $S(0) = 0$, so $a=0$,
$$S(x) = \int_0^x dt\, \frac{e^t-1}{t}.$$
The argument of the integral is perfectly well-behaved at $t=0$, so
$$\begin{eqnarray*}
S(x) &=& \lim_{\e\to 0} \int_\e^x dt\, \frac{e^t-1}{t} \\
&=&  \lim_{\e\to 0} \left(
    \int_\e^x dt\, \frac{e^t}{t} - \int_\e^x dt\,\frac{1}{t}
    \right) \\
&=& \lim_{\e\to 0} \left(
    \pv \int_{-\infty}^x dt\,\frac{e^t}{t} - \pv \int_{-\infty}^\e dt\,\frac{e^t}{t}
    -\log x + \log \e
    \right) \\
&=& \lim_{\e\to 0} \left(
    \mathrm{Ei}(x) - \mathrm{Ei}(\e) - \log x + \log \e
    \right) \\
&=& \lim_{\e\to 0} \left(
    \mathrm{Ei}(x) - (\g + \log \e) - \log x + \log \e
    \right) \\
&=& \mathrm{Ei}(x) - \g - \log x.
\end{eqnarray*}$$
(See below for a derivation of $\mathrm{Ei}(\e) = \g + \log \e + O(\e)$.) 
Therefore, 
$$\sum_{k=1}^\infty \frac{x^k}{k k!} = \mathrm{Ei}(x) - \g - \log x$$
and so 
$$\sum_{k=1}^\infty \frac{1}{k k!} = \mathrm{Ei}(1) - \g.$$
Some details
Above we use the definition of the exponential integral
$$\mathrm{Ei}(x) = \pv \int_{-\infty}^x dt\,\frac{e^t}{t},$$
where $\pv\int$ stands for the Cauchy principal value, and the series expansion for $\mathrm{Ei}(x)$ for small $x$, which we derive now.
Split the integral, 
$$\begin{eqnarray*}
\mathrm{Ei}(x) &=&  \lim_{\d\to0}\left[
\underbrace{\int_{-\infty}^{-\d} dt\,\frac{e^t}{t}}_{I_1} 
+ \underbrace{\int_{\d}^{x} dt\,\frac{e^t}{t}}_{I_2} 
\right].
\end{eqnarray*}$$
For $I_1$, let $t=-s$ and integrate by parts,
$$I_1 = \log\d - \int_\d^\infty ds\, e^{-s}\log s.$$
For $I_2$, Taylor expand $e^t$ and integrate, 
$$I_2 = \log x - \log\d + O(x).$$
Thus, 
$$\begin{eqnarray*}
\mathrm{Ei}(x) &=& \lim_{\d\to0}\left[
\left(\log\d - \int_\d^\infty ds\, e^{-s}\log s\right)
+\left(\log x - \log\d + O(x)\right)
\right] \\
&=& \g + \log x + O(x),
\end{eqnarray*}$$
where we recognize the integral representation of the Euler-Mascheroni constant, $\g = -\int_0^\infty ds\,e^{-s}\log s$.
Notice that if we kept the higher order terms in the expansion for $I_2$ we would find
$$\mathrm{Ei}(x) = \g + \log x + \sum_{k=1}^\infty \frac{x^k}{k k!},$$
the correct expansion for the exponential integral for $x>0$.
In fact, this immediately gives our sum, 
$$\sum_{k=1}^\infty \frac{1}{k k!} = \mathrm{Ei}(1) - \g.$$
This is the approach of @Ayman Hourieh. 
A: First of all, consider the power series for $e^x$, $\displaystyle\sum_{k=0}^{\infty}\frac{x^k}{k!}$.  Now subtract off the constant term and divide by $x$: $\displaystyle{\frac{e^x-1}{x} = \sum_{k=1}^{\infty}\frac{x^{k-1}}{k!}}$.  Now integrate: $\displaystyle{\int_0^x \frac{e^t-1}{t} dt = \sum_{k=1}^{\infty}\frac{x^k}{k\cdot k!}}$ (note that the lower limit is dictated by the constant term).  Finally, evaluate at $x=1$; the value of your sum is the value of the definite integral $\displaystyle{\int_0^1 \frac{e^t-1}{t} dt }$.  Wolfram Alpha evaluates this to $\mathrm{Ei}(1)-\gamma$, so there's probably no better closed form than that.
