#1: Initial revision
by
Klein Moretti
·
2023-10-16T17:29:16Z (about 1 year ago)
Calculation of limit
How to prove
$$\lim\limits_{n\to\infty}\frac{n^{\frac{m-1}2}\sum_{k=0}^m\left(C_n^k\right)^m}{2^{nm}}=\left(\frac2\pi\right)^{\frac m2}\sqrt{\frac\pi{2m}}.$$