Amazing Factorial Summation

$\begin{aligned} \color{#3D99F6}{S} & = \frac{1 \cdot 2}{1!+2!} + \frac{2 \cdot 3}{2!+3!} + \frac{3 \cdot 4}{3!+4!} + \frac{4 \cdot 5}{4!+5!} + \cdots \\ & = \sum_{n=1}^\infty \frac{n(n+1)}{n!+(n+1)!} \\ & = \sum_{n=1}^\infty \frac{n(n+1)}{n!(n+2)} \\ & = \sum_{n=1}^\infty \frac{n+1}{(n+2)(n-1)!} \\ & = \color{#3D99F6}{\sum_{n=0}^\infty \frac{n+2}{(n+3)n!}} \end{aligned}$

Now consider the Maclaurin series of $e^x$ ,

$\begin{aligned} \sum_{n=1}^\infty \frac{x^n}{n!} & = e^x \\ \sum_{n=1}^\infty \frac{x^{n+2}}{n!} & = x^2e^x \quad \quad \small \color{#3D99F6}{\text{Integrate both sides}} \\ \sum_{n=1}^\infty \frac{x^{n+3}}{(n+3)n!} & = x^2e^x - 2xe^x + 2e^x \color{#D61F06}{- 2} \quad \quad \small \color{#D61F06}{\text{Get from putting }x=0};\quad \color{#3D99F6}{\times \frac{1}{x} \text{ both sides}} \\ \sum_{n=1}^\infty \frac{x^{n+2}}{(n+3)n!} & = xe^x - 2e^x + \frac{2e^x}{x} - \frac{2}{x} \quad \quad \small \color{#3D99F6}{\text{Differentiate both sides}} \\ \sum_{n=1}^\infty \frac{(n+2)x^{n+1}}{(n+3)n!} & = e^x + xe^x - 2e^x + \frac{2e^x}{x} - \frac{2e^x}{x^2} + \frac{2}{x^2} \quad \quad \small \color{#3D99F6}{\text{Putting }x=1} \\ \color{#3D99F6} {\sum_{n=1}^\infty \frac{n+2}{(n+3)n!}} & = e + e - 2e + 2e - 2e + 2 \\ \implies \color{#3D99F6}{S} & = \boxed{2} \end{aligned}$

Let us denote the general term by $T_{n}$ .

Then we have

$T_n = n(n+1)/n! (n+2)$ .

Adding and removing 2 from numerator ,

$T_n = n-1/n! + 2/n!(n+2)$ .

Therefore we generate last expression before telescoping.

$T_n = (1/(n-1)! - 1/n!) + 2( 1/(n+1)! - 1/(n+2)!)$

Thus after taking sum from n = 1 to $\infty$ Answer is 2