Inspired by Paul Patawaran

$\begin{aligned} S & = \sum_{n=1}^\infty \frac 1{n(n+1)(n+1)!} \\ & = \sum_{n=1}^\infty \left(\frac 1n - \frac 1{n+1}\right) \frac 1{(n+1)!} \\ & = \sum_{n=1}^\infty \left(\frac 1{n(n+1)n!} - \frac 1{(n+1)(n+1)!}\right) \\ & = \sum_{n=1}^\infty \left(\frac 1{nn!} - \frac 1{(n+1)n!} - \frac 1{(n+1)(n+1)!}\right) \\ & = \sum_{n=1}^\infty \left(\frac 1{nn!} - \frac 1{(n+1)(n+1)!}\right) - \color{#3D99F6} \sum_{n=1}^\infty \frac 1{(n+1)n!} & \small \color{#3D99F6} \text{See note.} \\ & = 1 - \color{#3D99F6}(e-2) \\ & = \boxed{3-e} \end{aligned}$

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Note: Consider the Maclaurin series of $e^x$ .

$\small \begin{aligned} \sum_{n=0}^\infty \frac {x^n}{n!} & = e^x \\ \int \sum_{n=0}^\infty \frac {x^n}{n!} \ dx & = x + \frac {x^2}{2!} + \frac {x^3}{3!} + \frac {x^4}{4!} + ... \\ \sum_{\color{#3D99F6}n=0}^\infty \frac {x^{n+1}}{(n+1)n!} & = e^x - 1 \\ \frac x{1(0!)} + \sum_{\color{#D61F06}n=1}^\infty \frac {x^{n+1}}{(n+1)n!} & = e^x - 1 \\ \implies \sum_{n=1}^\infty \frac {x^{n+1}}{(n+1)n!} & = e^x - 1 - x & \small \color{#3D99F6} \text{Putting }x = 1 \\ \sum_{n=1}^\infty \frac 1{(n+1)n!} & = e - 2 \end{aligned}$