Find the sum of the series

We know that:

$\large \displaystyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

Differentiating with respect to $x$ and multiplying by $x$ three times one gets:

$\large \displaystyle xe^x = \sum_{n=0}^{\infty} \frac{nx^n}{n!}$

$\large \displaystyle (x^2+x)e^x = \sum_{n=0}^{\infty} \frac{n^2x^n}{n!}$

$\large \displaystyle (x^3+3x^2+x)e^x = \sum_{n=0}^{\infty} \frac{n^3x^n}{n!}$

Making $x=1$ :

$\color{#20A900} \boxed{ \large \displaystyle e = \sum_{n=0}^{\infty} \frac{n}{n!} }$ $\color{#20A900} \boxed{ \large \displaystyle 2e = \sum_{n=0}^{\infty} \frac{n^2}{n!} }$ $\color{#20A900} \boxed{ \large \displaystyle 5e = \sum_{n=0}^{\infty} \frac{n^3}{n!} }$

Then:

$\large \displaystyle S = \sum_{n=1}^{\infty} \sum_{k=1}^n \frac{k^2}{n!}$

$\large \displaystyle S = \frac16 \sum_{n=1}^{\infty} \frac{n(n+1)(2n+1)}{n!}$

$\large \displaystyle S = \frac16 \sum_{n=0}^{\infty} \frac{n(n+1)(2n+1)}{n!}$

$\large \displaystyle S = \frac16 \left [ 2 \sum_{n=0}^{\infty} \frac{n^3}{n!} + 3\sum_{n=0}^{\infty} \frac{n^2}{n!} + \sum_{n=0}^{\infty} \frac{n}{n!} \right ]$

$\large \displaystyle S = \frac16 \left [ 2 \cdot 5e + 3 \cdot 2e + e \right]$

$\color{#20A900} \boxed{ \large \displaystyle S = \frac{17e}{6} }$

Thus:

$\color{#3D99F6} \large \displaystyle a=17, b = 6, \boxed{a+b=23}$