Interesting Limit (8)

$\sum_{k=0}^n \log \binom{n}{k} = \log \prod_{k=0}^n \binom{n}{k} = \log \frac{(n!)^{n+1}}{(2! \times ... \times n!)^2} = \log \left( 2^{3-n} \times 3^{5-n} \times ... \times n ^{2n - 1 - n} \right) = \sum_{k=2}^{n} (2k - 1 - n) \log k$

Diving by $n^2$ we get three series: $\sum_{k=2}^n \frac{2k \log k}{n^2} - \sum_{k=2}^n \frac{\log k}{n^2} - \sum_{k=2}^n \frac{\log k}{n}$ First, notice that $\sum_{k=2}^n \frac{\log k}{n^2} = \frac{\log(n!)}{n^2} \leq \frac{\log(n^n)}{n^2} = \frac{\log n}{n} \to 0$ so the middle term vanishes. Now, remembering Euler's summation formula in the form $\sum_{k=2}^n f(k) = \int_2^n f(x) dx + O(f(n))$ , if we apply this to the functions $f(x) = 2x \log x$ and $f(x) = \log(x)$ we notice that the error terms $\frac{f(n)}{n^2}$ vanish in both cases. Thus we can say that in the limit: $\lim_{n \to \infty} \sum_{k=2}^n \frac{2k \log k}{n^2} - \sum_{k=2}^n \frac{\log k}{n} = \lim_{n \to \infty} \frac{1}{n^2} \int_2^n 2x \log x dx - \frac{1}{n} \int_2^n \log x dx = \lim_{n \to \infty} -\frac{1}{2} + \log n + \frac{2 - \log 4}{n^2} - \log n + 1 - \frac{2 - \log 4}{n} = -\frac{1}{2} + 1 = \frac{1}{2}$

Note: What has actually been proven is the asymptotic formula $\sum_{k=0}^n \log \binom{n}{k} = \frac{n^2}{2} + O(n \log n)$

Relevant wiki: Stolz–Cesàro theorem

$\begin{aligned} L&=\lim_{n\to\infty}\frac{\displaystyle\sum_{k=0}^n\ln\binom nk}{n^2}&\small\color{#3D99F6}\text{Using Stolz-Cesàro theorem} \\&=\lim_{n\to\infty}\frac{\displaystyle\sum_{k=0}^{n+1}\ln\binom {n+1}k-\sum_{k=0}^n\ln\binom nk}{(n+1)^2-n^2} \\&=\lim_{n\to\infty}\frac{\displaystyle\sum_{k=0}^n\ln\dfrac{\binom {n+1}k}{\binom nk}+\ln\binom {n+1}{n+1}}{2n+1} \\&=\lim_{n\to\infty}\frac{\displaystyle\sum_{k=0}^n\ln\frac{n+1}{n-k+1}}{2n+1} \\&=\lim_{n\to\infty}\frac{(n+1)\ln (n+1)-\displaystyle\sum_{k=1}^{n+1}\ln k}{2n+1}&\small\color{#3D99F6}\text{Using Stolz-Cesàro theorem again} \\&=\lim_{n\to\infty}\frac{(n+1)\ln (n+1)-n\ln n-\ln (n+1)}{(2n+1)-(2n-1)} \\&=\lim_{n\to\infty}\frac{\ln\left(\left(1+\dfrac 1n\right)^n\right)}2 \\&=\frac{\ln e}2=\boxed{0.5} \end{aligned}$