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Is there any simple way of computing the following sum?

$$\sum_{k=1}^\infty \frac1{k\space k!}$$

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It is equal to $\displaystyle{\int_0^1\dfrac{e^t-1}{t}\,dt}$. – Jonas Meyer Jun 1 '12 at 20:51
The value is Ei$(1) -\gamma = \int_0^1 \frac{e^t-1}{t} dt$ if that's any help. – Logan Maingi Jun 1 '12 at 20:51
@Chris: I just used Steven Stadnicki's method to get that, but I didn't know whether it answered the question. – Jonas Meyer Jun 1 '12 at 20:59
@JonasMeyer Agreed - I'm not really sure that it does, since it only boils it down to the integral and then some magic happens. I'd add a proof of the evaluation to my answer, but a bit of digging about the web isn't actually turning up any details on it, and none of the usual spots seem to say anything about it. – Steven Stadnicki Jun 1 '12 at 21:12
up vote 12 down vote accepted

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.

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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.

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$\def\d{\delta} \def\e{\epsilon} \def\g{\gamma} \def\pv{\mathrm{PV}} \def\pv{\mathcal{P}} \def\pv{\mathrm{P}}$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.

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