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Finding $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{n^2}\sum^{n}_{r=0}\ln\binom{n}{r}$.

My Try: Using $\text{A.M G.M H.M}$

$$ \frac{1}{n+1}\sum^{n}_{r=0}\binom{n}{r}\geq \sqrt[n+1]{\prod^{n}_{r=0}\binom{n}{r}}\geq \frac{n+1}{\sum^{n}_{r=0}\frac{1}{\binom{n}{r}}}$$

$$\bigg(\frac{2^n}{n+1}\bigg)^{\frac{1}{n}}\geq \bigg(\prod^{n}_{r=0}\binom{n}{r}\bigg)^{\frac{1}{n(n+1})}\geq \bigg(\frac{n+1}{\sum^{n}_{r=0}\frac{1}{\binom{n}{r}}}\bigg)^{\frac{1}{n}}$$

Can anyone please explain is my process is right. if not how can i solve it .Help me please. Thanks

DXT
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  • I think the question might be hinting towards relation between Reimann sum and integrals – Rohan Shinde Jun 06 '18 at 03:35
  • https://math.stackexchange.com/questions/2972852/product-of-binomials/2972969#2972969 – DXT Mar 14 '19 at 08:49
  • https://math.stackexchange.com/questions/4531304/a-limit-that-has-binomials-lim-n-to-infty-sqrtn2-n-prod-i-0n-frac – DXT Apr 01 '23 at 06:32

1 Answers1

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Using the Stolz-Cesaro Theorem, we find that

$$\begin{align} \lim_{n\to \infty}\frac{\sum_{k=0}^n\log\left(\binom{n}{k}\right)}{n^2}&\overbrace{=}^{\text{SCT}}\lim_{n\to \infty}\frac{\sum_{k=0}^{n+1}\log\left(\binom{n+1}{k}\right)-\sum_{k=0}^n\log\left(\binom{n}{k}\right)}{(n+1)^2-n^2}\\\\ &=\lim_{n\to \infty}\frac{\sum_{k=1}^n \left(\log\left(\binom{n+1}{k}\right)-\log\left(\binom{n}{k}\right) \right)}{2n+1}\\\\ &=\lim_{n\to \infty}\frac{\sum_{k=0}^n \log(n+1)-\log(n+1-k)}{2n+1}\\\\ &=\lim_{n\to \infty} \frac{(n+1)\log(n+1)-\sum_{k=1}^n\log(k)}{2n+1}\\\\ &=\lim_{n\to \infty} \frac{(n+1)\log(n+1)-n\log(n)-\sum_{k=1}^n\log(k/n)}{2n+1}\\\\ &=\lim_{n\to \infty}\frac{\log(n+1)+n\log\left(1+\frac1n\right)-n\left(\frac1n \sum_{k=1}^n\log(k/n)\right)}{2n+1}\\\\ &=\frac12 \end{align}$$

Mark Viola
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