# Thread: convergence

1. ## convergence

$\displaystyle \mathop {\lim }\limits_{n \to \infty } \sqrt[{n^2 }]{{\prod\limits_{i = 1}^n {\left( \begin{gathered} n \hfill \\ i \hfill \\ \end{gathered} \right)} }}$

2. Originally Posted by mms
$\displaystyle \mathop {\lim }\limits_{n \to \infty } \sqrt[{n^2 }]{{\prod\limits_{i = 1}^n {\left( \begin{gathered} n \hfill \\ i \hfill \\ \end{gathered} \right)} }}$
I get the answer to be $\displaystyle \sqrt e\approx 1.6487$.

Start by checking that $\displaystyle \int_0^1\!\!\!x\ln x\,dx = -1/4$ (easy integration by parts). Approximating this integral by a Riemann sum, you see that $\displaystyle \frac1n\sum_{k=1}^n\frac kn\ln\Bigl(\frac kn\Bigr)\sim -\frac14$ (for large n). Therefore $\displaystyle \frac1{n^2}\sum_{k=1}^nk\ln k\sim -\frac14 + \frac{(n+1)\ln n}{2n}$ .....(*).

Next,
\displaystyle \begin{aligned}\prod_{k=1}^n{n\choose k} &= \frac n1\cdot\frac{n(n-1)}{1\cdot2}\cdot \frac{n(n-1)(n-2)}{1\cdot2\cdot3}\cdots \\ &= n^{n-1}(n-1)^{n-3}(n-2)^{n-5}\cdots 2^{-n+3} = \prod_{k=1}^nk^{2k-n-1}.\end{aligned}

Take logs to see that
\displaystyle \begin{aligned}\ln\left(\sqrt[n^2 ]{\prod_{k = 1}^n {n\choose k}}\right) &= \sum_{k=1}^n\frac{2k-n-1}{n^2}\ln k \\ &\sim -\frac12 + \frac{n+1}n\ln n - \frac{n+1}{n^2}\sum_{k=1}^n\ln k\qquad\text{(from (*))} \\ &= -\frac12 + \frac{n+1}n\ln n - \frac{n+1}{n^2}\ln(n!) \\ &\sim -\frac12 + \frac{n+1}n\ln n - \frac{n+1}n(\ln n-1)\quad\ \text{(from Stirling's formula)}.\end{aligned}
This converges to 1/2 as n→∞, which leads to the result stated at the beginning.