TITLE: Apparent Contradiction by WolframAlpha in Calcuation of $\sum_{k=1}^\infty k^{k-1} e^{-k} / k!$
QUESTION [6 upvotes]: Consider the infinite sum
$$S = \sum_{k=1}^\infty \frac{k^{k-1} e^{-k}}{k!}.$$
Note that a basic version of Stirling's formula says that $k! \ge (k/e)^k \sqrt{2\pi k}$, and this tells us that
$$S_N = \sum_{k=1}^N \frac{k^{k-1} e^{-k}}{k!} \le \sum_{k=1}^N \frac{k^{k-1} e^{-k}}{k^k e^{-k} \sqrt{2\pi k}} = \frac{1}{\sqrt{2\pi}} \sum_{k=1}^N k^{-3/2},$$
and so $S_N \to S$ with $S$ a convergent series. Moreover, a stronger form of Stirling's formula says that $k! \ge (k/e)^k \sqrt{2\pi k} e^{1/(12k + 1)}$. Plugging this in similarly gives
$$S \le \frac{1}{\sqrt{2\pi}} \sum_{k=1}^\infty k^{-3/2} e^{-1/(12k+1)}.$$
WolframAlpha tells us here that this sum has numerical value approximately $0.95$, and specifically $S<1$. However, when asking WolframAlpha what $S$ is directly here, it says that $S = 1$.
As far as I can see, these two claims by WolframAlpha cannot be consistent. That is, unless I've made a mistake somewhere... Note that I haven't used an approximation for all $k$ which is only valid for large $k$: Wikipedia claims (via this paper) that the above stronger form holds for all positive integers $k$, not just sufficiently large ones.
If someone could enlighten me, I'd be most appreciative!

Note that this SE question along with the fact that $z \mapsto z e^{-z}$ is uniquely maximised at $z = 1$ tells us that $S = 1$. Indeed, taking $x = 1/e$ there gives us that $S$ satisfies $S e^{-S} = e^{-1}$, and the only $S$ satisfying this is $S = 1$, by the maximisation property mentioned above.
Thus it does appear that WolframAlpha is getting stung by rounding rounding errors. Thanks everyone for the comments.

REPLY [2 votes]: Lambert's W function has series $$W(x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}\,k^{k-1}}{k!}x^k \tag 1$$
which has radius of convergence $1/e$.  The series in $(1)$ can be obtained using Lagrange's Inversion Theorem.
Since we have $-W(-1/e)=1$, then from $(1)$ we see that
$$\sum_{k=1}^\infty \frac{(-1)^{k-1}\,k^{k-1}}{k!}\left(-\frac1e\right)^k=1$$
from which we recover the identity
$$\sum_{k=1}^\infty \frac{k^{k-1}e^{-k}}{k!}=1$$