In thinking about my question [here](https://mathoverflow.net/questions/441982) for the Linial arrangement, the following limit arose:
$$ \lim\_{n\to\infty}\frac{(n-1)\sum\_{k=0}^n {n\choose k}(k+1)^{n-2}}
 {\sum\_{k=0}^n {n\choose k}(k+1)^{n-1}}. $$
Is this limit finite? What is its value? It might be around $2.27\cdots$.



A limit involving the quotient of two sums
First of all, it seems like the value of the limit is more like $1.27...$, and not $2.27...$.


Using some heuristics outlined below it is possible to find the limit:
$$
a=1.278464542761..., 
$$
where $a$ is the root of $(x-1)e^x=1$ (can be expressed in terms of Lambert W-function).


The approach is standard, based on saddle point approximation. First, one notes that the main contribution to both sums come from large values of $k\gtrapprox n/2$. When both $k$ and $n$ are large then one has the asymptotics <https://en.wikipedia.org/wiki/Binomial_coefficient#Bounds_and_asymptotic_formulas> :
$$
\binom{n}{k}=\sqrt{\frac{n}{2\pi k(n-k)}}\frac{n^n}{k^k(n-k)^{n-k}}.
$$
Second, replace the sums with Riemann integral. Using saddle point approximation one can find that the main contribution to the sums come from values of $k$ that give maximum to the function (here the contibution from square root factors are omitted, but they can be taken into account without altering the final result):
$$
e^{(n-\epsilon)\ln(x+1)-x\ln x-(n-x)\ln (n-x)}
$$
where $\epsilon$ is 1 ir 2. This gives equation for $k\_m$:
$$
\frac{x\_m}{n-x\_m}=e^{\frac{n-\epsilon}{x\_m+1}},
$$
which is asymptotically the same as
$$
\frac{1}{z-1}=e^{z}, \quad z=n/x\_m.
$$
Then the ratio of two sums is approximated around the saddle point as
$$
\frac{(n-1)g(x\_m)I}{(x\_m+1)g(x\_m)I}\to a
$$
where $g=\binom{n}{x\_m}(x\_m+1)^{n-2}$, and $I$ is the Gaussian integral around the saddle point, which is the same both for the numerator and denominator sums.


Of course, this analysis can be made rigorous using all the usual big O notation, etc...


