Infinite Factorial Summation with Sum of Squares

The sum of the first n squares can be written as $\frac{n(n+1)(2n+1)}{6}$ so the sum can be written as $\sum_{n=1}^\infty \frac{n(n+1)(2n+1)}{6(n+1)!}$ The (n+1) and n terms cancel with the factorial and leave behind $\sum_{n=1}^\infty \frac{2n+1}{6(n-1)!}$

We make a substitution here for k = n-1, making the sum $\sum_{k=0}^\infty \frac{2k+3}{6k!}$ and split up into two separate sums, $\sum_{k=0}^\infty \frac{2k}{6k!}$ and $\sum_{k=0}^\infty \frac{3}{6k!}$ . It is known that $\sum_{k=0}^\infty \frac{1}{k!}$ is equal to e , so the second sum $\sum_{k=0}^\infty \frac{(3)}{6k!}$ equates to $\frac{e}{2}.$

The first sum is equivalent to $0 +\sum_{k=1}^\infty \frac{k}{3k!} = \sum_{k=1}^\infty \frac{1}{3(k-1)!}$ or $\frac{e}{3}$

Thus our answer is $S = \frac{e}{3} + \frac{e}{2} = \frac{5e}{6}$ , so $\lfloor100S\rfloor = 226$