$\lim_{n\rightarrow\infty}\sum_{r=0}^{n-1}\sum_{k=0}^{n-r}(-1)^k\frac{2n+3}{k!(2n+3-k)!} (1+n-r-k)^{2n+1}=0$

Question

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

Answer 1

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.

Answer 2

This is not a proof at all.

I have the feeling that I saw this question or a very similar one during the last days (and I was unable to do anything - I was hoping that some hypergeometric functions could appear somewhere).

Using a CAS, I made the calculations up to $n=1000$ (this took a real long time !).

From the results, what it seems to me is that $$S_n=\sum_{r=0}^{n-1}\sum_{k=0}^{n-r}(-1)^k\binom{2n+3}{k} \frac{(1+n-r-k)^{2n+1}}{(2n+2)!}\approx \frac {a_n}{n^{3/2}}$$ and that (more or less) $\lim_{n\to \infty } \, a_n=\frac12$. $$\left( \begin{array}{cc} n & S_n & \frac 1 {2n^{3/2}} \\ 100 & 0.000481006 & 0.000500000\\ 200 & 0.000171396 & 0.000176777\\ 300 & 0.000093540 & 0.000096225\\ 400 & 0.000060836 & 0.000062500\\ 500 & 0.000043565 & 0.000044721\\ 600 & 0.000033158 & 0.000034021 \\ 700 & 0.000026323 & 0.000026998 \\ 800 & 0.000021551 & 0.000022097 \\ 900 & 0.000018065 & 0.000018519 \\ 1000 & 0.000015427 & 0.000015811 \end{array} \right)$$

The big question is : how to prove it ? I shall very interested to see a rigorous proof.