$e$ game

$\begin{aligned}S_1 & = \displaystyle\sum_{k=1}^{2}\frac{k^2}{3k}\\& = \frac{1}{3}\displaystyle\sum_{k=1}^{2}k \\& = \boxed{1}\end{aligned}$ $\begin{aligned} S_2 & = \displaystyle\sum_{k=0}^{\infty} \frac{(k-1)^2}{k!} \\ & = \displaystyle\sum_{k=0}^{\infty} \frac{k^2-2k+1}{k!} \\ & = \displaystyle\sum_{k=0}^{\infty} \frac{k^2}{k!}-\frac{2k}{k!}+\frac{1}{k!} \\& = \displaystyle\sum_{k=0}^{\infty} \frac{k}{(k-1)!}-\frac{2}{(k-1)!}+\frac{1}{k!} \\& = \displaystyle\sum_{k=0}^{\infty} \frac{k-1+1}{(k-1)! }-\frac{2}{(k-1)!}+\frac{1}{k!} \\&= \displaystyle\sum_{k=0}^{\infty}\frac{k-1}{(k-1)!}+\frac{1}{(k-1)!}-\frac{2}{(k-1)!}+\frac{1}{k!} \\& = \displaystyle\sum_{k=0}^{\infty}{\color{#D61F06}\frac{1}{(k-2)!}-\frac{1}{(k-1)!}}+\frac{1}{k!} = (e-e) +\displaystyle\sum_{k=0}^{\infty}\frac{1}{k!} \\& = \displaystyle\sum_{k=0}^{\infty} \frac{1}{k!} \\& = \boxed{e}\end{aligned}$ $\begin{aligned} S_3 & = \displaystyle\sum_{k=0}^{\infty} \frac{(2k+3)^2}{5k!} \\ & = \frac{1}{5}\displaystyle\sum_{k=0}^{\infty} \frac{4k^2+12k+9}{k!} \\ & = \frac{1}{5}\displaystyle\sum_{k=0}^{\infty} \frac{4k^2}{k!}+\frac{12k}{k!}+\frac{9}{k!} \\& = \frac{1}{5}\displaystyle\sum_{k=0}^{\infty} \frac{4k}{(k-1)!}+\frac{12}{(k-1)!}+\frac{9}{k!} \\& = \frac{1}{5}\displaystyle\sum_{k=0}^{\infty} \frac{4(k+1)}{(k-2)! }+\frac{12}{(k-1)!}+\frac{9}{k!} \\& =\frac{1}{5}(4(2e)+12e+9e) \\& = \boxed{\frac{29e}{5}}\end{aligned}$

$\begin{aligned} S_1 & = \sum_{k=1}^2 \frac {k^2}{3k} = \sum_{k=1}^2 \frac {k}{3} = \frac 13 + \frac 23 = 1 \color{#D61F06} \ne e \end{aligned}$

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$\begin{aligned} S_2 & = \sum_{k=0}^\infty \frac {(k-1)^2}{k!} = \sum_{k=0}^\infty \frac {k^2-2k+1}{k!} = {\color{#3D99F6}\sum_{k=0}^\infty \frac {k^2}{k!}} - 2{\color{#D61F06} \sum_{k=0}^\infty \frac k{k!}} + \sum_{k=0}^\infty \frac 1{k!} = {\color{#3D99F6}s_2} - 2{\color{#D61F06}s_1} + s_0 \end{aligned}$

• $\begin{aligned} s_0 & = \sum_{k=0}^\infty \frac 1{k!} = e \end{aligned}$
• $\begin{aligned} s_1 & = \sum_{\color{#3D99F6}k=0}^\infty \frac k{k!} = \sum_{\color{#D61F06}k=1}^\infty \frac k{k!} = \sum_{\color{#3D99F6}k=0}^\infty \frac {k+1}{(k+1)!} = \sum_{\color{#3D99F6}k=0}^\infty \frac 1{k!} = e \end{aligned}$
• $\begin{aligned} s_2 & = \sum_{\color{#3D99F6}k=0}^\infty \frac {k^2}{k!} = \sum_{\color{#D61F06}k=1}^\infty \frac {k^2}{k!} = \sum_{\color{#3D99F6}k=0}^\infty \frac {(k+1)^2}{(k+1)!} = \sum_{\color{#3D99F6}k=0}^\infty \frac {k+1}{k!} = s_1+s_0 = 2e \end{aligned}$

$\implies S_2 = s_2-2s_1+s_0 = 2e-2e+e\color{#3D99F6} = e$ . Therefore, the answer is $\boxed{\displaystyle \sum_{k=0}^\infty \frac {(k-1)^2}{k!}}$ .

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$\begin{aligned} S_3 & = \sum_{k=0}^\infty \frac {(2k+3)^2}{5k!} = \sum_{k=0}^\infty \frac {4k^2+12k+9}{5k!} = \frac {4s_2+12s_1+9s_0}5 = \frac {8+12+9}5e = \frac {29}5 e \color{#D61F06} \ne e \end{aligned}$

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Generalisation: $\displaystyle s_n = \sum_{k=0}^{n-1} {n-1 \choose k} s_k$