limit of a sequence with Lagrange inversion formula? In a recent question to the Pari/GP mailing lists, a user stated that the
limit as $N\to\infty$ of
$$\sum_{n=1}^N\dfrac{n^ne^{-n}}{n!}-\dfrac{2\sqrt{N}}{\sqrt{2\pi}}$$
is equal to $-2/3$. This seems to be an application of Lagrange inversion
and/or the expansion of Lambert's $W$ function around $-1/e$, but I have not been able to find a proof. I would be interested in such a proof.
 A: One way to obtain the -2/3 is by singularity analysis.
The first step is to construct the generating function of your sequence.
From the Taylor expansion of the Lambert $W$ function at 0, one gets that $-W(-x)$ is the generating function of the sequence $N^{N-1}/N!$ and therefore by differentiation
$$\frac{-W(-x)}{1+W(-x)}=\sum_{N\ge1}{\frac{N^N}{N!}x^N}.$$
Replacing $x$ by $x/e$ and multiplying by $1/(1-x)$ yields the desired generating function
$$F(x):=\frac{-W(-x/e)}{(1-x)(1+W(-x/e))}=
\sum_{N\ge1}{\left(\sum_{n=1}^N{\frac{n^ne^{-n}}{n!}}\right)x^N}.$$
From there, the result follows from an analysis at $x=1$. From the known expansion of $-W(-x)$ at $1/e$, one deduces
$$F(x)=\frac{\sqrt{2}}{2(1-x)^{3/2}}-\frac{2}{3(1-x)}+O\left(\frac1{\sqrt{1-x}}\right),\quad x\rightarrow1.$$
Now singularity analysis (or Darboux's method) deduces the asymptotic expansion of your sequence as
$$\frac{\sqrt{2 N}}{\sqrt{\pi}}-\frac23+O(1/\sqrt{N}).$$
With slightly more work along the same lines, one obtains a full asymptotic expansion beginning with
$${\frac {\sqrt {2N}}{\sqrt {\pi}}}-\frac23+{\frac {
\sqrt {2}}{3\sqrt {\pi N}}}-{\frac {37\sqrt {2}}{864\sqrt {\pi}N^{3/2}}}+{\frac {359\sqrt {2}
 }{64800\sqrt {\pi}N^{5/2}}}+O \left(  {N}^{-7/2} \right) 
.$$
