If $S_n$ denotes the sum of the products of the first $n$ natural numbers taken two at a time, then find $$\sum_{n=0}^\infty\frac{S_n}{(n+1)!}$$
$$ S_n=\sum_{1\leq i<j\leq n} a_ia_j=\frac{1}{2}\bigg[(\sum a_i)^2-\sum a_i^2\bigg]\\ =\frac{1}{2}\bigg[\frac{n^2(n+1)^2}{4}-\frac{n(n+1)(2n+1)}{6}\bigg]$$ Then $$\sum_{n=0}^\infty\frac{S_n}{(n+1)!}=\frac{1}{24}\bigg[\sum_{n=0}^\infty\frac{3}{(n-3)!}+\sum_{n=0}^\infty\frac{8}{(n-2)!}\bigg]\\ =T_0+T_1+T_2+\frac{1}{24}\bigg[\sum_{n=3}^\infty\frac{3}{(n-3)!}+\sum_{n=3}^\infty\frac{8}{(n-2)!}\bigg]\\ =0+0+2+\frac{1}{8}\Big[1+\frac{1}{1!}+\dots\Big]+\frac{1}{3}\Big[\frac{1}{1!}+\frac{1}{2!}+\dots\Big]\\=2+\frac{e}{8}+\frac{1}{3}\big[e-1\big]=2+\frac{11e}{24}-\frac{1}{3}=\frac{11e}{24}+\frac{5}{3} $$
But my reference gives the solution $\frac{11e}{24}$, so what is going wrong with my attempt ?
You are almost correct, just pay attention to the starting index of each sum at each step: $$\begin{align} \sum_{n=0}^\infty\frac{S_n}{(n+1)!}&=\frac{1}{24}\sum_{n=1}^\infty\frac{3n(n+1)-2(2n+1)}{(n-1)!}\\ &=\frac{1}{24}\sum_{n=1}^\infty\frac{3(n-1)(n-2)+8(n-1)}{(n-1)!}\\ &=\frac{1}{24}\sum_{n=3}^\infty\frac{3}{(n-3)!}+\frac{1}{24}\sum_{n=2}^\infty\frac{8}{(n-2)!}\\ &=\frac{e}{8}+\frac{e}{3}=\frac{11e}{24}. \end{align}$$