Infinite square factorial summation

The given series converges absolutely and the convergence can be verified using ratio test. The given sum can be written as,

$S=\frac{1^2}{3!}+\frac{2^2}{4!}+\frac{3^2}{5!}+\ldots=\sum_{k=3}^\infty\frac{(k-2)^2}{k!}\\ \implies S=\sum_{k=3}^\infty\frac{k^2-4k+4}{k!}\\ \implies S=\sum_{k=3}^\infty\frac{k^2}{k!}+\sum_{k=3}^\infty\left(-\frac{4k}{k!}+\frac{4}{k!}\right)\\ \implies S=\sum_{k=3}^\infty\frac{k}{(k-1)!}+\sum_{k=3}^\infty\left(-\frac{4}{(k-1)!}+\frac{4}{k!}\right)\\ \implies S=\sum_{k=3}^\infty\frac{(k-1)+1}{(k-1)!}+\sum_{k=3}^\infty\left(-\frac{4}{(k-1)!}+\frac{4}{k!}\right)\\ \implies S=\sum_{k=3}^\infty\frac{1}{(k-2)!}+\sum_{k=3}^\infty\frac{1}{(k-1)!}+\sum_{k=3}^\infty\left(-\frac{4}{(k-1)!}+\frac{4}{k!}\right)\\ \implies S=\sum_{k=1}^\infty\frac 1{k!}+\sum_{k=2}^\infty\frac 1{k!}+\sum_{k=3}^\infty\left(-\frac{4}{(k-1)!}+\frac{4}{k!}\right)$

Now, recall the Maclaurin expansion of $e^x$ which converges $\forall~x$ . For the case of $x=1$ , we get the series for $e$ which is,

$e=\sum_{k=0}^\infty\frac 1{k!}=1+\sum_{k=1}^\infty\frac 1{k!}=2+\sum_{k=2}^\infty\frac 1{k!}$

Also, note that the last sum (the one on the extreme right in $S$ ) is a telescoping sum and as such, the sum $S$ can be evaluated as,

$S=(e-1)+(e-2)+\left(-\frac{4}{(3-1)!}\right)=e-1+e-2-2=2e-5$

$S=2e-5\approx 0.436564\implies 1000S\approx 436.564\implies \left\lfloor 1000S\right\rfloor=\boxed{436}$

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Note: The "sum-breaking" done in 3rd and 6th lines of the calculation is valid because the sums obtained after "breaking" are also convergent sums (can be easily verified using convergence tests).

Let us use the expansion of the exponential function around zero, which is given by, for all $x \in \mathbb{R}$ $e^x = \sum_{n \geq 0} \dfrac{x^n}{n!}$ Dividing both sides by $x^2$ , we find, for all $x \in \mathbb{R} / \{0\}$ $\dfrac{e^x}{x^2} = \sum_{n \geq 0} \dfrac{x^{n-2} }{n!}.$ Then, as this sum is absolutely convergent, we can take the derivative with respect to $x$ at both sides of the equation and interchange the derivative with the sum, as follows $\dfrac{d}{dx} \left( \dfrac{e^x}{x^2} \right) = \sum_{n \geq 0} \dfrac{d}{dx} \left( \dfrac{x^{n-2} }{n!} \right) = \sum_{n \geq 0} \dfrac{(n-2) x^{n-3} }{n!}.$ After that, we multiply by $x$ at both sides of the equation and then, take another derivative, at both sides, in respect to $x$ (we can still interchange derivative with sum, because the sum is still absolutely convergent), $\dfrac{d}{dx} \left( x \dfrac{d}{dx} \left( \dfrac{e^x}{x^2} \right) \right) = \sum_{n \geq 0} \dfrac{d}{dx} \left( \dfrac{(n-2) x^{n-2} }{n!} \right) = \sum_{n \geq 0} \dfrac{(n-2)^2 x^{n-3} }{n!}.$ Then, we can make $x=1$ and rewrite the right-hand side of the above equation as $4 + 1 + \sum_{n \geq 1} \dfrac{n^2}{(n+2)!},$ and solve the left-hand side of the equation, which give us $\left[ \dfrac{d}{dx} \left( x \dfrac{d}{dx} \left( \dfrac{e^x}{x^2} \right) \right) \right]_{x=1} = 2e.$ Finally, we can conclude that our desired sum $\sum_{n \geq 1} \dfrac{n^2}{(n+2)!} = 2e - 5.$