$\lim_{n\rightarrow\infty}\sum_{r=0}^{n-1}\sum_{k=0}^{n-r}(-1)^k\frac{2n+3}{k!(2n+3-k)!} (1+n-r-k)^{2n+1}=0$ I'm trying to prove that
$$\sum_{r=0}^{n-1}\sum_{k=0}^{n-r}(-1)^k\frac{2n+3}{k!(2n+3-k)!}
(1+n-r-k)^{2n+1}$$
converges to zero as $n\rightarrow\infty$. The conjecture came from the computations on Mathematica:
f[n_] := Sum[
  Sum[(-1)^k (2 n + 
      3) (1 + n - r - k)^(2 n + 1)/(k! (2 n + 3 - k)!), {k, 0, 
    n - r}], {r, 0, n - 1}]
ListLinePlot[{Table[f[n], {n, 100}]}]

Plot for 1≤n≤100
And the motivation for the problem is this article where I evaluate
$$\sum_{r=0}^{n-1}\nu(\mathbf{A}_{0,1}(n-r,n+r))-\nu(\mathbf{A}_{1,0}(n-1-r,n+1+r))=$$
Current trials: I can rewrite the sum as
$$\sum_{r=0}^{n-1}\sum_{k=0}^{n-r}(-1)^k\binom{2n+3}{k}
\frac{(1+n-r-k)^{2n+1}}{(2n+2)!}$$
and remove the $(2n+2)!$ from the inner sum since it doesn't depend on $k$:
$$\sum_{r=0}^{n-1}\frac{1}{(2n+2)!}\sum_{k=0}^{n-r}(-1)^k\binom{2n+3}{k}
(1+n-r-k)^{2n+1}$$
Then if I fix $r$, the inner sum is asymptotic to $(1+i-r)^{2i+1}$ (the factor $k=0$ prevails over the rest). However, I cannot just sum the asymptotes since asymptotic functions doesn't behave well with summation.
Any suggestion will be great (it doesn't need to be a complete solution)!
Thanks in advance
 A: The first asymptotic term is 
$$ S_n \sim \frac{1}{2}\sqrt{\frac{3}{\pi}} \, (n+1)^{-3/2} .$$
The proof depends on series manipulations to get an expression in which an inner sum has a closed form, and the resulting single sum has a known (to me) asymptotic form. From the proposer's form
$$ (2n+1)!\,S_n = \sum_{r=0}^{n-1}\sum_{k=0}^{n-r} (-1)^k \binom{2n+3}{k} (1+n-r-k)^{2n+1} $$
let $r\to r+n$ in the summand, take the negative of the outer summation index, and furthermore shift the outer index by one; i.e.,
$$ (2n+1)!\,S_n = \sum_{r=2}^{n+1}\sum_{k=0}^{r} (-1)^k \binom{2n+3}{k} (r-k)^{2n+1}.$$
Now let $k \to r-k$ and notice the inner sum can then be extended to infinity:
$$ (2n+1)!\,S_n = \sum_{r=2}^{n+1}\sum_{k=1}^{\infty} (-1)^{r+k} \binom{2n+3}{r-k} k^{2n+1}.$$
We can exchange summations and perform explicitly the inner sum.  With Mathematica and some simplification we find, letting $N = n+1, $
$$ \sum_{r=2}^{N} (-1)^r \binom{2N+1}{r-k} = (-1)^N\binom{2N}{N+k}
+ \frac{(2N)!}{(1-k)!(2N+k-1)!}.$$
The last term is non-zero only for $k=1$.  Therefore we have
$$ (2n+1)!\,S_n = \sum_{k=1}^{N} (-1)^{k+N}k^{2N-1}\binom{2N}{N+k} -1.$$  When solving for $S_n,$ the last term is exponentially small and can be dropped.  Using the technique I mentioned in MSE 2720893, one can show how such an expression is related to the Eulerian numbers, and use their known asymptotics to find
$$ \sum_{k=1}^{N} (-1)^{k+N}k^{2N-1}\binom{2N}{N+k} \sim (2N-1)! \sqrt{\frac{3}{\pi\,N}}.$$  A factorial identity and algebra completes the proof.
