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I came across the sum $\sum_{k=0}^n (-1)^k {2n \choose k} (1-\frac{k}{n})^2 \ln(1-\frac{k}{n})$ while working through some integrals, and I believe that it tends as $\frac{4^{n}}{n^{5/2}}$. I'm particularly interested in the coefficient on this putative dominant term in an asymptotic expansion as n tends to infinity.

I tried Taylor expanding the piece $(1-\frac{k}{n})^2 \ln(1-\frac{k}{n})$, which yields a relatively friendly series. Then my next step would be to evaluate sums of the form $\sum_{k=0}^n (-1)^k {2n \choose k} k^m$ for some arbitrary natural number m. I saw a more specialized sum with more complete bounds here. However, I'm having difficulty generalizing the accepted answer's technique with my bounds.

I found, via derivatives of the trigonometric power formula for sin^2n, that $\sum_{k=0}^n (-1)^k {2n \choose k} (1-\frac{k}{n})^{2m}=0$ for natural m < n. That might help if one expands just the logarithm.

Note that I take $x^2 \ln(x) $ to vanish at x=0 so that I can write my upper bound to n for aesthetic reasons alone.

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  • $\begingroup$ Good question but one thing missed: is $n$ tending to infinity? $\endgroup$ – Szeto Jun 19 '18 at 5:42
  • $\begingroup$ Yep, n is tending to infinity. I've added that in. $\endgroup$ – user196574 Jun 19 '18 at 5:44
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As J. D'Aurizio notes, the proposer's question is answered if the asymptotics of his eq (1) can be determined. I'll outline a proof of the following: $$ \sum_{k=1}^n (-1)^k \binom{2n}{n+k} k^2 \log{k} \sim \binom{2n}{n} \frac{\pi}{4}\Big( 7\frac{\zeta(3)}{\pi^3} + \frac{93}{n} \frac{\zeta(5)}{\pi^5} + ...\Big).$$

The $\zeta(n)$ are Riemann's zeta. The OP conjectured an asymptotic form, which is correct up to an alternating factor, and asked what the value of the leading constant is. By using the asymptotics for the central binomial,

$$ \sum_{k=0}^n (-1)^k \binom{2n}{k} (1-k/n)^2 \log{(1-k/n)} \sim \frac{4^n}{n^{5/2}}\, (-1)^n \, \frac{7\zeta(3)}{4\pi^2\sqrt{\pi}}.$$

My top equation is not so difficult to derive if one should have the identity available,

$$ (A)\,\,\,\,\,\sum_{k=1}^n (-1)^{k+1} \binom{2n}{n+k} k^s = \binom{2n}{n} \sin(\pi s/2) \int_0^\infty \frac{dx \, \,x^s}{\sinh{\pi x}} \frac{n!^2}{(n+ix)!(n-ix)!}. $$ Differentiate with respect to $s$ and set it afterwards to 2. Use the asymptotic expansion for the ratio of gamma functions to get $$ \sum_{k=1}^n (-1)^{k} \binom{2n}{n+k} k^2 \log{k} = \binom{2n}{n} \frac{\pi}{2} \int_0^\infty \frac{dx \, \,x^2}{\sinh{\pi x}} \big(1+\frac{x^2}{n} + ...\big). $$ Integrate term-by-term and the integrals are known by Mathematica in terms of $\zeta(n)$.

I don't know of a reference that has (A). My proof is long and I'll skip it unless there is suitable interest. I was really surprised at the final result because I was expecting a $\log(n)$ to be in the mix.

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  • $\begingroup$ "A" is quite the striking identity! It seems lucky that s=2 causes the sine term to vanish on product rule, was the identity designed for such? I'll be posting a whole host of curious and surprising sums over the next few days. $\endgroup$ – user196574 Jun 20 '18 at 20:28
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    $\begingroup$ proof for $A$? thx :-) $\endgroup$ – tired Jun 20 '18 at 23:25
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    $\begingroup$ tired and user196574: posted my proof as MSE 2827591. Question asks for a generalization or simplification. $\endgroup$ – skbmoore Jun 21 '18 at 18:37
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A partial answer. $$\frac{1}{n^2}\sum_{k=0}^{n}(-1)^k \binom{2n}{k}(n-k)^2 = \delta(n-1)$$ hence it is enough to find the asymptotic behaviour of $$\frac{(-1)^n}{n^2}\sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k k^2\log k\tag{1}$$ where for any $n\geq 2$ we have $$ \sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k = \frac{1}{2}\binom{2n}{n} \tag{O}$$ $$ \sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k k = -\frac{n}{4n-2}\binom{2n}{n} \tag{A}$$ $$ \sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k k^2 = 0 \tag{B}$$ $$ \sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k k^3 = \frac{n^2}{2(2n-1)(2n-3)}\binom{2n}{n}. \tag{C}$$ We have $H_k=\log k+\gamma+O\left(\frac{1}{k}\right)$ and $$ -\frac{\log(1-z)}{1-z}=\sum_{n\geq 1} H_n z^n\tag{D} $$ $$ \sin^{2n}(x)=\frac{1}{4^n}\binom{2n}{n}+\frac{(-1)^n}{2^{2n-1}}\sum_{k=0}^{n-1}(-1)^k\binom{2n}{k}\cos(2(n-k)x)\tag{E}$$ hence the asymptotic behaviour of the sum in which $\log u$ is replaced by $H_u$ can be found through the following approach:

  1. Apply $\frac{d^2}{dx^2}$ to $(E)$;
  2. Evaluate $(D)$ at $z=e^{2ix}$ and $z=e^{-2ix}$;
  3. Write the wanted combinatorial sum, where $\log k$ has been replaced by $H_k$,
    as a contour integral, by exploiting Parseval's theorem and the previous points;
  4. Estimate such integral through the saddle point/Laplace method.

On the other hand, what we really need to find is the derivative at $m=2$ of $\sum_{k=0}^{n}\binom{2n}{n+k}(-1)^k k^m$.

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    $\begingroup$ Could you explain further the replacement of log(k) by the harmonic number? It seems to me that the 1/k^3 and higher corrections might have a non-negligible contribution to the sums. For example, for very large n it seems that we have approximately $\sum_{k=1}^n {2n \choose n+k} (-1)^k \frac{1}{k} = -\ln(2) {2n \choose n} $ which would make such a term of order $4^n / \sqrt{n}$ after Stirling's approx and would arise from an O(1/k^3) correction to the harmonic number approximation. $\endgroup$ – user196574 Jun 19 '18 at 20:29
  • $\begingroup$ I've made another question addressing sums of the form $\sum_{k=0}^n {2n \choose n+k} (-1)^k k^m$ and their large-n form. I've quoted some of your results O through C there since I find them really interesting! Specifically how all seem to depend on n similarly in the large-n limit. $\endgroup$ – user196574 Jun 19 '18 at 22:05
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    $\begingroup$ @user196574: you are right, I assumed that the contribution provided by $\frac{1}{k},\frac{1}{k^3}$ were negligible, but that might not be the case. I am going to fix the answer and clarify it is just a partial one. $\endgroup$ – Jack D'Aurizio Jun 19 '18 at 22:10

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