Infinite Multiple summation

Since $\frac{1}{n(n+1)(n+2)\cdots(n+p)} \; = \; \frac{1}{p}\left[ \frac{1}{n(n+1)\cdots(n+p-1)} - \frac{1}{(n+1)(n+2)\cdots(n+p)}\right]$ we deduce that $\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)\cdots(n+p)} \; = \; \frac{1}{p \times p!} \hspace{2cm} p \ge 1$ and hence the desired sum is $\sum_{p=1}^\infty \frac{1}{p \times p!} \; = \; \mathrm{Ei}(1) - \gamma \; = \; \int_0^1 \frac{e^x-1}{x}\,dx \; = \; -\int_0^1 e^x \ln x\,dx \; = \; \boxed{1.31790215}$ (the integral representations of the sum are standard results). To three decimal places, the answer should be $1.318$ .