Evaluate infinite sum involving n!

Question

Evaluate $\sum_{n=1}^\infty \frac{1}{n×n!}$

I really don't know where to begin with this but I'm pretty sure $e$ is involved somehow.

If it can help, $n×n!=(n+1)!-n!$

Answer 1

You have $$ \sum\limits_{1\, \le \,\,n} {\;{1 \over {n \cdot n!}}} = \sum\limits_{0\, \le \,\,n} {\;{1 \over {\left( {n + 1} \right) \cdot \left( {n + 1} \right)!}}} = \sum\limits_{0\, \le \,\,n} {\;{1 \over {\left( {n + 1} \right)^2 \cdot n!}}} $$

Then, starting from the series of $e^x$, we get $$ \eqalign{ & e^{\,x} = \sum\limits_{0\, \le \,\,n} {\;{{x^n } \over {n!}}} \cr & \int_0^x {e^{\,u} du} = \sum\limits_{0\, \le \,\,n} {\;{{x^{n + 1} } \over {\left( {n + 1} \right)n!}}} \cr & {1 \over x}\int_0^x {e^{\,t} dt} = \sum\limits_{0\, \le \,\,n} {\;{{x^n } \over {\left( {n + 1} \right)n!}}} \cr & \int_0^x {{1 \over u}\left( {\int_0^u {e^{\,t} dt} } \right)du} = \sum\limits_{0\, \le \,\,n} {\;{{x^{n + 1} } \over {\left( {n + 1} \right)^2 \cdot n!}}} \cr & {1 \over x}\int_0^x {{1 \over u}\left( {\int_0^u {e^{\,t} dt} } \right)du} = \sum\limits_{0\, \le \,\,n} {\;{{x^n } \over {\left( {n + 1} \right)^2 \cdot n!}}} \cr} $$

so that $$ \eqalign{ & f(x) = \sum_{0\, \le \,n} {{x^n \over {(n + 1)^2 \cdot n!}}} = {1 \over x} \int_0^x {1 \over u}\left( \int_0^u {e^t \, dt } \right) \, du = {1 \over x} \int_0^x {e^u - 1 \over u} \, du = \cr & = {1 \over x} \left( -\ln x + \operatorname{Ei}(x) - \lim_{t\;\to\;0+} \left( -\ln t + \operatorname{Ei}(t) \right) \right) = \cr & = {1 \over x} \left( -\ln x + \operatorname{E}_1 (x) + \lim_{t\; \to \;0+} (\ln t - \operatorname{E}_1 (t)) \right) \cr} $$ Note that the integrand function $(e^u - 1)/ u$ is continuous ( $\lim\limits_{x\; \to \;0} ( (e^x - 1)/x) = 1$ ) and positive.
The Exponential Integral function is as in MathWorld: $$ \operatorname{Ei}(x) = -\int_{-x}^\infty {{e^{-u} \over u} \, du = \int_{-\infty }^x {e^u \over u} \, du = \gamma + \ln x + \sum\limits_{1\, \le \,\,n} {x^n \over n \cdot n!}} $$

So we can conclude that $$ \bbox[lightyellow] { \sum\limits_{1\, \le \,\,n} {\;{1 \over {n \cdot n!}}} = \int_0^1 {{{e^{\,u} - 1} \over u}du} = {\rm Ei}(1) - \gamma \approx 1.3179 }$$

Answer 2

HINT:

$$e^x=\sum_{i=0}^\infty \frac{x^n}{n!}$$

And also

$$\sum_{i=0}^\infty \frac{a_nx^n}{n}=\sum_{i=0}^\infty a_n\int_0^x t^{n-1}dt$$