Sum/Product

$n+$ can be simplified to $\frac{n(n+1)}{2}$ . Plugging this into the summation cancels out an $n$ in the numerator and denominator, leaving $\sum_{n=2}^∞$ $\frac{n+1}{2((n-1)!)}$ . Notice that this summation is equivalent to a similar summation from $n=1$ to infinity if 1 is added to $n$ in both the numerator and denominator. Doing this gives $\sum_{n=1}^∞$ $\frac{n+2}{2(n!)}$ . This summation can be split into two summations: $\frac{1}{2}$ $\sum_{n=1}^∞$ $\frac{1}{(n-1)!}$ and $\sum_{n=1}^∞$ $\frac{1}{n!}$ . The constant $e$ is equal to $\sum_{n=1}^∞$ $\frac{1}{(n-1)!}$ . Using this, the former summation is equal to $\frac{e}{2}$ and the latter summation is equal to $e-1$ . Adding these together yields $\frac{3e-2}{2}$ , or about 3.077.