Is there any simple way of computing the following sum?
$\sum_{k=1}^\infty \frac1{k\space k!}$
Is there any simple way of computing the following sum?
$\sum_{k=1}^\infty \frac1{k\space k!}$
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.
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.
$\def\d{\delta} \def\e{\epsilon} \def\g{\gamma} \def\pv{\mathrm{PV}} \def\pv{\mathcal{P}} \def\pv{\mathrm{P}}$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.