An interesting sum to infinity

Is there any simple way of computing the following sum?

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

Solution 1:

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.

Solution 2:

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.