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Calculation of limit

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How to prove $$\lim\limits_{n\to\infty}\frac{n^{\frac{m-1}2}\sum_{k=0}^m\left(C_n^k\right)^m}{2^{nm}}=\left(\frac2\pi\right)^{\frac m2}\sqrt{\frac\pi{2m}}.$$

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It doesn't seem right. For m=1 the LHS the fraction inside the limit is (1*(1+n)/2^n) = n/2^n, that g... (2 comments)
It doesn't seem right. For m=1 the LHS the fraction inside the limit is (1*(1+n)/2^n) = n/2^n, that g...
Arpad Horvath‭ wrote about 1 year ago · edited about 1 year ago

It doesn't seem right. For m=1 the LHS the fraction inside the limit is (1*(1+n))/2^n = n/2^n, that goes to 0, but the RHS is sqrt(2/pi)*sqrt(pi/2) = 1.

Peter Taylor‭ wrote about 1 year ago

More generally, when $n > m$ the sum has no support so the LHS is zero.