Square/factorial

$= \large \displaystyle\sum_{n = 0}^{\infty} \dfrac{ n^2}{n!}$

$= \displaystyle\sum_{n = 0}^{\infty} \dfrac{ n^2}{n ( n - 1)!}$

$= \displaystyle\sum_{n = 1}^{\infty} \dfrac{ n + 1 - 1}{(n - 1)!}$

$= \displaystyle\sum_{n = 1}^{\infty} \left( \dfrac{n-1}{(n-1)!} + \dfrac{1}{(n-1)!} \right)$

$= \displaystyle\sum_{n = 2}^{\infty} \dfrac{1}{(n - 2)!} + \displaystyle\sum_{n = 1}^{\infty} \dfrac{1}{(n - 1)!}$

$= \left(\dfrac{1}{0!}+ \dfrac{1}{1!} + \dfrac{1}{2!} + \ldots + \infty\right) + \left( \dfrac{1}{0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \ldots + \infty \right)$

$\large \Rightarrow e \ + \ e \ = \color{#3D99F6}{2}e \Rightarrow \boxed{\color{#3D99F6}{\lambda} = 2}$

$e^x = \large \displaystyle\sum_{n = 0}^{\infty} \dfrac{ x^n}{n!} \\ \dfrac{d(e^x)}{dx} = e^x = \large \displaystyle\sum_{n = 0}^{\infty} \dfrac{ n x^{n-1}}{n!} \\ \dfrac{d^2(e^x)}{dx^2} = e^x = \large \displaystyle\sum_{n = 0}^{\infty} \dfrac{ (n^2-n) x^{n-2}}{n!}$

For $x = 1$ we have:

$e = \large \displaystyle\sum_{n = 0}^{\infty} \dfrac{1}{n!} \\ e = \large \displaystyle\sum_{n = 0}^{\infty} \dfrac{ n }{n!} \\ e = \large \displaystyle\sum_{n = 0}^{\infty} \dfrac{ (n^2-n)}{n!}$

Manipulating the last equation we obtain:

$e = \large \displaystyle\sum_{n = 0}^{\infty} \dfrac{ (n^2)}{n!} - \large \displaystyle\sum_{n = 0}^{\infty} \dfrac{ (n)}{n!} \\ e = \large \displaystyle\sum_{n = 0}^{\infty} \dfrac{ (n^2)}{n!} - e \\ 2e = \large \displaystyle\sum_{n = 0}^{\infty} \dfrac{ (n^2)}{n!}$

Thus, $\lambda = 2$ .

$\begin{aligned} S & = \sum_{n=\color{#3D99F6}0}^\infty \frac {n^2}{n!} \\ & = \sum_{n=\color{#D61F06}1}^\infty \frac {n^2}{n!} \\ & = \sum_{n=\color{#3D99F6}0} ^\infty \frac {(n+1)^2}{(n+1)!} \\ & = \sum_{n=\color{#3D99F6}0} ^\infty \frac {n+1}{n!} \\ & = \sum_{n=\color{#3D99F6}0} ^\infty \frac n{n!} + \sum_{n=\color{#3D99F6}0}^\infty \frac 1{n!} \\ & = \sum_{n=\color{#D61F06}1} ^\infty \frac n{n!} + \sum_{n=\color{#3D99F6}0}^\infty \frac 1{n!} \\ & = \sum_{n=\color{#D61F06}1} ^\infty \frac 1{(n-1)!} + \sum_{n=\color{#3D99F6}0}^\infty \frac 1{n!} \\ & = \sum_{n=\color{#3D99F6}0}^\infty \frac 1{n!} + \sum_{n=\color{#3D99F6}0}^\infty \frac 1{n!} \\ & = e+e=2e \end{aligned}$

$\implies \lambda = \boxed{2}$