无穷级数对数

如何计算 $\sum_{i = 0}^{n}\ln\binom{n}{i}\Big/n^2$?

Colliot

如何计算 $$\sum_{i = 0}^{n}\ln\binom{n}{i}\Big/n^2$$?

我只能这么估计:

$$\binom{n}{i}<\left(\dfrac{n\cdot e}{k}\right)^k$$ $$\sum_{i = 0}^{n}\ln\binom{n}{i}\Big/n^2<\dfrac{1}{n}\dfrac{k}{n}\left(\ln(\dfrac{n}{k}\cdot e)\right)\rightarrow\int_0^1t\ln\dfrac{e}{t}dt=\dfrac{3}{4}$$

但数值计算的结果似乎是 $\dfrac{1}{2}$。

怎么才能证明这点呢?


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考虑到$$\binom{n+1}{i}\big/\binom{n}{i}=\dfrac{n+1}{n+1-i}$$ 由Stolz定理有$$\begin{align} &\ \ \ \ \ {\lim_{n \to \infty}}\sum_{i = 0}^{n}\ln\binom{n}{i}\Big/n^2\\ &={\lim_{n \to \infty}}\sum_{i = 0}^{n}\ln\dfrac{n+1}{n+1-i}\Big/(2n+1)\\ &={\lim_{n \to \infty}}\ln\dfrac{(n+1)^{n+1}}{(n+1)!}\Big/(2n+1)\\ &={\lim_{n \to \infty}}\ln(\dfrac{n+1}{n})^n\Big/2=\dfrac{1}{2} \end{align}$$ (练习了一下刚学的LaTeX)

Edited at 07/03/2018

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