# Alternating shifted central binomial sum with Cauchy weights

My question is how can one show that

$$\lim_{n \to \infty} \frac{1}{\binom{2n}{n}}\sum_{k=1}^n (-1)^k \binom{2n}{n+k}\frac{x^2}{x^2+\pi^2k^2} = \frac{1}{2}\Big(\frac{x}{\sinh(x)}-1\Big)$$

I find the proposed identity nice because it feels like we can look at the presence of the fraction in $$x$$ on the left hand side, for large $$x$$, as a perturbation to the textbook sum of alternating binomial coefficients with a little rearrangement.

I'm also generally interested in techniques to handle alternating sums of shifted central binomial sums with different weights, which prompted my question at MSE 2824529.

Skbmoore posted a proof of a neat and useful identity at 2827591 (and Tired commmented a slick proof) that $$\,\,\,\,\,\sum_{k=1}^n (-1)^{k+1} \binom{2n}{n+k} k^s = \binom{2n}{n} \sin(\pi s/2) \int_0^\infty \frac{dx \, \,x^s}{\sinh{\pi x}} \frac{n!^2}{(n+ix)!(n-ix)!}.$$

Likely an apt modification of that formula will net us the asymptotics above, but I'm not quite clever enough with modifying the integrand to give me something nice in the limit of $$s$$ to zero.

• This question relates the $\sinh$ product formula $$\sinh (x)=x \prod _{k=1}^{\infty } \left(\frac{x^2}{\pi ^2 k^2}+1\right)$$ to the equivalent infinite summation. I get as far as trying to prove $$\frac{1}{2} \left(\frac{1}{\prod _{k=1}^n \left(\frac{x^2}{\pi ^2 k^2}+1\right)}-1\right)=\sum _{k=1}^n \frac{(-1)^k x^2 \left(\prod _{i=1}^k (-i+n+1)\right)}{\left(\pi ^2 k^2+x^2\right) \left(\prod _{i=1}^k (i+n)\right)}$$ Feb 2, 2019 at 19:23

Here is an approach:

We consider the following sum in the limit $$n\rightarrow\infty$$ (which we denote by $$S$$) :

$$S_n=\frac{1}{\binom{2n}{n}}\sum_{k=0}^n(-1)^k\binom{2n}{n+k}\frac{1}{x^2+\pi^2 k^2}$$

We split the range of summation at some $$\delta$$ such that $$1<<\delta< to get $$S_n=S_{2,n}+S_{1,n}$$ and observe that $$k< in $$[0,\delta]$$ so we can expand the binomial around small $$k/n$$ which means that

$$S_{1,n}=\sum_{k=0}^{\delta}(-1)^k\left(1+O(k^2/n^3)\right)\frac{1}{x^2+\pi^2 k^2}$$ since $$|\sum_{k=0}^{\delta}(-1)^k\frac{k^2}{x^2+\pi^2 k^2}|<\sum_{k=0}^{\delta}\frac{k^2}{x^2+\pi^2 k^2}<\sum_{k=0}^{\delta}1=\delta$$ we can bound the $$O(k^2/n^3)$$ term as follows

$$S_{1,n}=\sum_{k=0}^{\delta}(-1)^k\frac{1}{x^2+\pi^2 k^2}+o(\delta/n^3)$$ Taking limits we get that $$S_{1,n}$$ approaches a constant to determined later

$$S_1=\sum_{k=0}^{\infty}(-1)^k\frac{1}{x^2+\pi^2 k^2}\,\,\,\, (*)$$

Next we show that the tail sum $$S_{2,n}$$ approaches zero so that the limit in question is indeed given by $$(*)$$. To this end, observe that on $$[\delta,n]$$ we have $$\pi k >> x$$ so we can expand the fraction around large $$k$$. We find

$$S_{2,n}=\frac{1}{\binom{2n}{n}}\sum_{k=\delta}^n(-1)^k\binom{2n}{n+k}\left(\frac{1}{\pi^2 k^2}+O(1/k^4)\right)$$ since the binomial coefficient has a maximum at $$k=0$$ we can bound $$\frac{\binom{2n}{n+k}}{\binom{2n}{n}}\leq1$$ so

$$|S_{2,n}|<\frac{1}{\pi^2}\sum_{k=\delta}^n\frac{1}{k^2}\sim \frac{1}{\pi^2} \int_{\delta}^n\frac{dk}{k^2}\sim \frac{1}{\pi^2\delta}$$ so, choosing $$\delta$$ to be a (slowly enough) increasing sequence of $$n$$ we have indeed that

$$S_2=0\,\,\,\,(**)$$

Note that the $$O(1/k^4)$$ are vanishing even faster and can therefore also be neglected. Putting anything $$(*)$$ and $$(**)$$ together we get

$$S=S_1=\frac{1}{2x^2}+\frac{1}{2\sinh(x)x}$$

were the last equality follows from Mittag-Lefflers Theorem (or the product expansion of sine)

• Nice :). Could you comment why Delta should be slowly increasing? Slowly with respect to what? Feb 4, 2019 at 20:40
• @Thomas such that $\delta(n)\rightarrow \infty$ as well as $\delta(n)/n\rightarrow 0$ as $n \rightarrow \infty$. You can take $\delta(n)=\sqrt{n}$ for example.. Feb 4, 2019 at 20:42
• @Thomas this method is called asymptotic matching in case you want to take a deeper dive Feb 4, 2019 at 20:45
• Thanks for the answer! I'll also look more into asymptotic matching. I'm curious whether it would work for MSE 2824529 or MSE 2825442 as well. Those questions are already answered, but I appreciate new techniques. Feb 4, 2019 at 22:41