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I'm trying to find the following limit for $t\geq1$

$$\lim_{n\to\infty} \sqrt[n]{\sum_{k =0}^n \binom{n}{k}^t}.$$

I suspect this can be solved using generating functions but did not make any headway.

Looking at this thread, Combinatorial proof of summation of $\sum\limits_{k = 0}^n {n \choose k}^2= {2n \choose n}$. I'm guessing the answer to be $t$. As pointed out by Gary, the limit is $4$ when $t = 2$ and the appropriate guess would be $2^t$.

David
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    For $t=2$, you would get $4 \neq t$ since $\binom{2n}{n} \sim \frac{{4^n }}{{\sqrt {\pi n} }}$. The general answer should be $2^t$ based on https://math.stackexchange.com/questions/976777/an-asymptotic-expression-of-sum-of-powers-of-binomial-coefficients – Gary Mar 14 '21 at 10:49

1 Answers1

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Assume that $n$ is even (we can deal with odd cases as well, but it's a little messier), writing $n=2m$. Then, since for all $0\leq k\leq 2m$ we have $\binom{2m}{k}\leq \binom{2m}{m}$, we get $$ \binom{2m}{m}^t\leq \sum_{k=0}^{2m} \binom{2m}{k}^t \leq (2m+1)\binom{2m}{m}^t \tag{1} $$ and, using standard estimates for the Binomial coefficient (Stirling's inequality) we have $$ \binom{2m}{m} \operatorname*{\sim}_{m\to\infty} \frac{C}{\sqrt{m}}\cdot 2^{2m} \tag{2} $$ for some constant $C>0$. Combining (1) and (2) and taking the $2m$-th root, we get by the squeeze theorem that, as $m\to\infty$, the limit is $\boxed{2^t}$.

Clement C.
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