I want to compute the limit $$\lim_{i\rightarrow\infty}\frac{1}{(2i+1)!}\sum_{k=0}^i(-1)^{k+1}\binom{2i+3}{k}(i-k+2)^{2i+1}$$
Context: from this article, I'm trying to compute the limit $$\lim_{i\rightarrow\infty}\frac{1}{(2i+1)!}[A_{1,0}(i,i)-A_{0,1}(i+1,i-1)]$$ where $A_{p,q}(i,j)$ is given by the formula (1.1) of the article.
I tried to simplify the limit above by putting the $(2i+1)!$ inside the sum and then cancel some factorial from the binomial: $$\sum_{k=0}^i(-1)^{k+1}\frac{1}{(2i+1)!}\binom{2i+3}{k}(i-k+2)^{2i+1}= \sum_{k=0}^i(-1)^{k+1}\frac{(2i+3)(2i+2)}{k!(2i+3-k)!}(i-k+2)^{2i+1}.$$
Since $(2i+3)(2i+2)$ does not depend on $k$, we can remove them from the sum: $$(2i+3)(2i+2)\sum_{k=0}^i(-1)^{k+1}\frac{(i-k+2)^{2i+1}}{k!(2i+3-k)!}.$$
So it all boils down to $$\lim_{i\rightarrow\infty}(2i+3)(2i+2)\sum_{k=0}^i(-1)^{k+1}\frac{(i-k+2)^{2i+1}}{k!(2i+3-k)!}.$$
The product outside the sum is of the order $4i^2$. If I could show that the part inside the sum is of an order such that the product goes to zero I would be happy. But the alternating signs of the sums really annoys me.
Extra: According to Mathematica, it should converge to zero:
f[i_] := N[(Sum[(-1)^(k + 1) Binomial[2 i + 3,
k] (i - k + 2)^(2 i + 1), {k, 0, i}])/((2 i + 1)!)]
Array[f, 100]
Output= {2.16667, 0.0416667, 0.268056, 0.185995, 0.150008, \
0.123581, 0.104117, 0.0892726, 0.0776475, 0.0683426, 0.0607579, \
0.054479, 0.0492115, 0.0447412, 0.0409088, 0.0375939, 0.0347037, \
0.0321657, 0.0299226, 0.0279285, 0.0261465, 0.0245463, 0.0231028, \
0.0217955, 0.020607, 0.0195227, 0.0185303, 0.0176191, 0.0167803, \
0.016006, 0.0152894, 0.0146248, 0.0140069, 0.0134314, 0.0128942, \
0.0123919, 0.0119214, 0.0114799, 0.011065, 0.0106745, 0.0103065, \
0.00995912, 0.00963083, 0.00932019, 0.00902589, 0.00874677, \
0.00848173, 0.00822982, 0.00799014, 0.00776186, 0.00754426, \
0.00733663, 0.00713836, 0.00694886, 0.00676759, 0.00659407, \
0.00642785, 0.00626849, 0.00611561, 0.00596884, 0.00582785, \
0.00569232, 0.00556197, 0.00543652, 0.00531571, 0.00519932, \
0.00508711, 0.00497887, 0.00487443, 0.00477358, 0.00467616, \
0.00458202, 0.00449099, 0.00440293, 0.00431771, 0.00423521, \
0.00415531, 0.00407788, 0.00400283, 0.00393005, 0.00385946, \
0.00379095, 0.00372444, 0.00365986, 0.00359712, 0.00353615, \
0.00347689, 0.00341926, 0.00336321, 0.00330867, 0.00325559, \
0.00320392, 0.0031536, 0.00310458, 0.00305682, 0.00301027, \
0.00296489, 0.00292063, 0.00287747, 0.00283537}
This solution depends on finding an asymptotic expression for
$$ S(n) := \sum_{k=1}^n(-1)^{n+k}\binom{2n}{n+k}k^{2n-3}. $$
The sum of the problem's first line can be written as, with $n=i+2,$ via simple series manipulations, $$ \sum_{k=0}^i(-1)^{k+1}\binom{2i+3}{k}(i-k+2)^{2i+1}=-\sum_{k=1}^n(-1)^{n+k}\binom{2n-1}{n-k}k^{2n-3} + (-1)^n\binom{2n-1}{n-1} .$$
To get to the form $S(n),$ recognize that $\binom{2n-1}{n-k}=(n+k)/(2n)\binom{2n}{n+k}$ and that the sum with even powers of $k$ in the summand is identically zero. It should also be recognized that the last term of the previous equation, when divided by (2i+1)!, goes rapidly to zero by Stirling's formula.
Now I've worked out a similar problem on Math Overflow 294590, and I won't repeat it, but using those techniques (with $f(x)=Li_{-(2n-3)}(x)$) one derives,
$$S(n)=A(2n-3,n-3)-2\,A(2n-3,n-2)+A(2n-3,n-1)$$ where the $A(n,k)$ are the Eulerian numbers. This formula can be recognized as a finite difference approximation to
$$S(n)\approx\frac{d^2}{dk^2}\,A(2n-3,n+k-2)\big|_{k=0}.$$
Now go look up the asymptotics for the Eulerian numbers near the central part (where they are strongly peaked) to find that
$$A(2n-3,n+k-2)\sim (2n-3)!\sqrt{\frac{3}{\pi(n-1)}}\exp{(-3k^2/(n-1))}.$$
Therefore we have
$$S(n) \sim \, (2n-3)! \Big( \frac{-6}{n-1} \sqrt{\frac{3}{\pi(n-1)}} \Big). $$
From this it is clear (remember $i=n+2$) to see that limit of the proposer's first equation goes to 0 as $n^{-3/2}.$