I came across the following problem, known as Knuth's Series which originally was an American Mathematical Monthly problem.
Prove that $\sum_{n=1}^\infty \left(\frac{n^n}{n!e^n}-\frac{1}{\sqrt{2\pi n}}\right)=-\frac{2}{3}-\frac{\zeta\left(\frac{1}{2}\right)}{\sqrt{2\pi}}.$
It seems interesting. We are trying to compute a particular sum of the error term in Stirlings approximation. The immediate simple approaches don't seem to work.
Attempt: Why $\zeta\left(\frac{1}{2}\right)$: By partial summation we know that $\sum_{n=1}^M \frac{1}{n^s}= \frac{M^{1-s}}{1-s}+\zeta(s)+O\left(M^{-s}\right)$ for $s>0$, $s\neq 1$. This tells us where the $\frac{\zeta\left(\frac{1}{2}\right)}{\sqrt{2\pi}}$ comes from since
$\sum_{n=1}^M \frac{1}{\sqrt{2\pi n}}=\sqrt{\frac{2M}{\pi}}+\frac{\zeta\left(\frac{1}{2}\right)}{\sqrt{2\pi}}+o(1).$
Now all that remains is to prove that $\sum_{n=1}^M \frac{n^n}{n!e^n}=\sqrt{\frac{2M}{\pi}} -\frac{2}{3}+o(1).$
I am kinda stuck here, as this series seems strange to deal with. Thanks!