A series problem by Knuth 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!
 A: Here is a series of hints to summarise the approach that appeared in the American Math Monthly. I have divided them by horizontal bars so hopefully it can ensure you only read one at a time. Unfortunately, this particular solution does not appear to be very general.

Prove the following separately first: 
$$ \sum_{k=1}^{\infty} \biggl( \frac{k^k}{k!e^k} - \frac{ (1/2)_{k-1} }{\sqrt{2} (k-1)!} \biggr) = \frac{-2}{3} $$ $$ \sum_{k=1}^{\infty} \biggl( \frac{1}{\sqrt{2\pi k}} - \frac{ (1/2)_{k-1} }{\sqrt{2} (k-1)!} \biggr) = \frac{ \zeta (1/2) }{\sqrt{2\pi}} $$ where the rising factorial is defined: $ (a)_0 = 1 \mbox{  and  } (a)_m = a(a+1)(a+2) \cdots (a+m-1) $

Abel's theorem comes in handy: If $\sum_{k=0}^{\infty} a_k $ converges, then $$\sum_{k=0}^{\infty} a_k = \lim_{x\to 1^{-} } \sum_{k=0}^{\infty} a_k x^k $$

Some power series (both for $ |z| < 1 $) : $$ \mathrm{W}(z) = \sum_{k=1}^{\infty} \frac{k^{k-1} z^k}{k!e^k} $$ $$ \sum_{k=1}^{\infty} \frac{ (1/2)_{k-1} }{(k-1)!} z^{k-1} = \frac{1}{\sqrt{1-z}} $$ where $\mathrm{W}(z) $ satisfies $ \mathrm{W} \exp(-\mathrm{W}) = z/e $ (See Lambert W Function), and the second series comes from Newton's Binomial Expansion.
