Tough and tricky Summations.

$\begin{aligned} S & = \sum_{n=0}^\infty \frac 1{n!} \left[\sum_{\color{#3D99F6}k=0}^{\color{#3D99F6}n}({\color{#3D99F6}k+1})\int_0^1 2^{-({\color{#3D99F6}k+1})x} dx \right] \\ & = \sum_{n=0}^\infty \frac 1{n!} \left[\sum_{\color{#D61F06}k=1}^{\color{#D61F06}n+1}{\color{#D61F06}k}\int_0^1 2^{-{\color{#D61F06}k}x} dx \right] \\ & = \sum_{n=0}^\infty \frac 1{n!} \left[ \sum_{k=1}^{n+1} k \left[-\frac {2^{-kx}}{k \ln 2} \right]_0^1 \right] \\ & = \sum_{n=0}^\infty \frac 1{n! \ln 2} \sum_{k=1}^{n+1}\left(1 - 2^{-k}\right) \\ & = \frac 1{\ln 2} \sum_{n=0}^\infty \frac 1{n!} \left(n+1 - \frac 12 \left(\frac {1- 2^{-(n+1)}}{1-\frac 12} \right) \right) \\ & = \frac 1{\ln 2} \sum_{n=0}^\infty \frac 1{n!} \left(n+ 2^{-(n+1)}\right) \\ & = \frac 1{\ln 2}\left(\sum_{\color{#3D99F6}n=0}^\infty \frac n{n!} + \frac 12 \sum_{n=0}^\infty \frac {2^{-n}}{n!} \right) \\ & = \frac 1{\ln 2}\left(\sum_{\color{#D61F06}n=1}^\infty \frac n{n!} + \frac {\sqrt e}2 \right) \\ & = \frac 1{\ln 2}\left(\sum_{\color{#D61F06}n=1}^\infty \frac 1{(n-1)!} + \frac {\sqrt e}2 \right) \\ & = \frac 1{\ln 2}\left(\sum_{\color{#3D99F6}n=0}^\infty \frac 1{n!} + \frac {\sqrt e}2 \right) \\ & = \frac {e+\frac {\sqrt e}2}{\ln 2} \approx \boxed{5.11} \end{aligned}$