The numerators diverge much faster this time

WARNING: PLEASE DO CALVIN'S PROBLEM BEFORE READING THIS SOLUTION. THIS SOLUTION USES THE SOLUTION TO CALVIN'S PROBLEM

Let's use some clever manipulation: the series in the problem is equal to

$\large \sum_{k=0}^{\infty}\dfrac{{(k+1)}^2}{k!}$

Now we expand the numerator:

$\large \sum_{k=0}^{\infty}\dfrac{k^2+2k+1}{k!}$

Breaking up the numerator into three separate sums:

$\large \sum_{k=0}^{\infty}\dfrac{k^2}{k!}+2\sum_{k=0}^{\infty}\dfrac{k}{k!}+\sum_{k=0}^{\infty}\dfrac{1}{k!}$

The third sum is equal to $e$ , as given in the problem. The first term of the other two summations is zero, which allows us to change the index to begin at 1.

$\large \sum_{k= \color{#D61F06}{1}}^{\infty}\dfrac{k^2}{k!}+2\sum_{k=\color{#D61F06}{1}}^{\infty}\dfrac{k}{k!}+e$

Simplifying fractions:

$\large \sum_{k= 1}^{\infty}\dfrac{k}{(k-1)!}+2\sum_{k=1}^{\infty}\dfrac{1}{(k-1)!}+e$

The second series is equal to $e$ , as given in the problem. The first series is equal to $x$ , the solution to Calvin's problem, which is $2e$

$2e + 2e + e = \boxed{5e}$

Split by partial fraction , $\displaystyle \frac{k^2}{(k-1)!} = \frac{1}{(k-3)!}+\frac{3}{(k-2)!}+\frac{1}{(k-1)!}$

Summing both sides from $k=1$ to $\infty$ , we have $\displaystyle \sum_{k=1}^{\infty}\frac{k^2}{(k-1)!}= \sum_{k=1}^{\infty} ( \frac{1}{(k-3)!}+\frac{3}{(k-2)!}+\frac{1}{(k-1)!}) = 5e$

Subtracting x and e we get $x-e= \dfrac{1}{1!} +\dfrac{2}{2!} + \dfrac{3}{3!} + \dfrac{4}{4!} + \cdots$ Simplifying the fractions, $x-e= \dfrac{1}{1!} +\dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots$ , which is equal to e. So we have $x-e=e \implies x=2e$ . Now let's set $\sum_{k=1}^{\infty} \dfrac{k^2}{(k-1)!} = y$ It follows that $y-4e=\sum_{k=1}^{\infty} \dfrac{k^2-2k}{(k-1)!} = \sum_{k=1}^{\infty} \dfrac{(k-1)^2}{(k-1)!} - \dfrac{1}{(k-1)!} = \sum_{k=2}^{\infty} \dfrac{k-1}{(k-2)!} - \sum_{k=1}^{\infty} \dfrac{1}{(k-1)!} = 2e-e=e$ Therefore, $y-4e=e \implies y=5e$