What is $\lim\limits_{n\to\infty}\frac {1}{n^2}\sum\limits_{k=0}^{n}\ln\binom{n}{k} $?

Question

It was originally asked on another website but nobody has been able to prove the numerical result. The attempts usually go by Stirling's approximation or try to use the Silverman-Toeplitz theorem.

Answer 1

As already noted by @SimpleArt, the expression equals

$$\tag 1 (n+1)\ln n!-2\sum_{k=1}^n\ln k!.$$

The sum on the right equals

$$\sum_{k=1}^n\sum_{j=1}^{k}\ln j = \sum_{j=1}^{n}(n—j+1)\ln j= (n+1)\ln n! - \sum_{j=1}^{n} j\ln j.$$

Therefore $(1)$ equals

$$\tag 2 -(n+1)\ln n! +2 \sum_{j=1}^{n} j\ln j.$$

Now $\ln n! = \sum_{k=1}^{n} \ln k,$ and this sum can be compared nicely with $\int_1^n \ln x \, dx$ to get

$$\ln n! = n\ln n - n +O(\ln n).$$

Similarly, $\sum_{j=1}^{n} j\ln j $ can be compared to $\int_1^\infty x \ln x \, dx$ to give

$$\sum_{j=1}^{n} j\ln j = \frac{n^2\ln n}{2}-\frac{n^2}{4} +O(n\ln n).$$

Putting this altogether, we see some nice cancellation shows $(2)$ equals $n^2/2 + O(n\ln n).$ Dividing by $n^2$ then gives $1/2$ for the desired limit.

Answer 2

By Stolz Cezaro $$\lim\limits_{n\to\infty}\frac {1}{n^2}\sum\limits_{k=0}^{n}\ln\binom{n}{k}=\lim\limits_{n\to\infty}\frac {1}{2n+1} \left(\sum\limits_{k=0}^{n+1}\ln\binom{n+1}{k}- \sum\limits_{k=0}^{n}\ln\binom{n}{k} \right)\\ =\lim\limits_{n\to\infty}\frac {1}{2n+1} \sum\limits_{k=0}^{n}\left(\ln\binom{n+1}{k}- \ln\binom{n}{k} \right)\\ =\lim\limits_{n\to\infty}\frac {1}{2n+1} \sum\limits_{k=0}^{n}\ln\left(\frac{(n+1)!}{k!(n+1-k)!} \frac{k!(n-k)!}{n!} \right)\\ =\lim\limits_{n\to\infty}\frac { \sum\limits_{k=0}^{n}\ln\left(\frac{(n+1)}{(n+1-k)} \right)}{2n+1}\\ =\lim\limits_{n\to\infty}\frac { \ln\left(\prod\limits_{k=0}^{n}\frac{(n+1)}{(n+1-k)} \right)}{2n+1}\\ =\lim\limits_{n\to\infty}\frac { \ln\left(\frac{(n+1)^n}{(n+1)!} \right)}{2n+1}\\ $$

Applying stolz Cezaro again, we get $$ =\lim\limits_{n\to\infty}\frac { \ln\left(\frac{(n+2)^{n+1}}{(n+2)!} \right)-\ln\left(\frac{(n+1)^n}{(n+1)!} \right)}{2} \\ =\lim\limits_{n\to\infty}\frac { \ln\left(\frac{(n+2)^{n+1}}{(n+2)!} \frac{(n+1)!}{(n+1)^n} \right)}{2} \\ =\lim\limits_{n\to\infty}\frac { \ln\left(\frac{(n+2)^{n}} {(n+1)^n} \right)}{2}\\ =\lim\limits_{n\to\infty}\frac { \ln\left(1+\frac{1}{n+1} \right)^n}{2}\\ =\frac{\ln(e)}{2}=\frac{1}{2} $$

Answer 3

Here is as exact an answer as you want, from my answer here: Prove that $\prod_{k=1}^{\infty} \big\{(1+\frac1{k})^{k+\frac1{2}}\big/e\big\} = \dfrac{e}{\sqrt{2\pi}}$

$$\prod_{k=0}^n \binom{n}{k} \sim C^{-1}\frac{e^{n(n+2)/2}}{n^{(3n+2)/6}(2\pi)^{(2n+1)/4}} \exp\big\{-\sum_{p\ge 1}\frac{B_{p+1}+B_{p+2}}{p(p+1)}\frac1{n^p}\big\}\text{ as }n \to \infty $$ where $$\begin{align} C &= \lim_{n \to \infty} \frac1{n^{1/12}} \prod_{k=1}^n \big\{k!\big/\sqrt{2\pi k}\big(\frac{k}{e}\big)^k\big\}\\ &\approx 1.04633506677...\\ \end{align} $$ and the $\{B_p\}$ are the Bernoulli numbers, defined by $$\sum_{p \ge 0} B_p\frac{x^p}{p!} = \frac{x}{e^x-1} .$$

Taking logs,

$$\sum_{k=0}^n \ln\binom{n}{k} \sim \ln C+n(n+2)/2-(3n+2)\ln n/6-(2n+1)\ln(2\pi)/4 -\sum_{p\ge 1}\frac{B_{p+1}+B_{p+2}}{p(p+1)}\frac1{n^p}\\ \text{ as }n \to \infty $$

Dividing by $n^2$,

$$\dfrac1{n^2}\sum_{k=0}^n \ln\binom{n}{k} \to \dfrac12 \text{ as }n \to \infty $$