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I'm currently frequenting a course in Combinatorics and I've come to this problem where I'm supposed to compute the following limit:

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

where $t$ is a real number, using Stirling's formula:

$$n! \sim \sqrt{2\pi n} \left (\frac{n}{e} \right) ^{n} $$

I've already used Stirling's formula to simplify the binomial coefficient a little bit, but I'm still far away from achieving a reasonable answer.

Any help will be appreciated.

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    $\begingroup$ What did you do so far? $\endgroup$ May 8, 2021 at 17:11
  • $\begingroup$ I just applied Stirling's formula to the binomial coefficient. I don't know what to do with the sum. $\endgroup$
    – Albert
    May 8, 2021 at 17:13
  • $\begingroup$ Probably worth estimating the largest binomial coefficient, $\binom{n}{\lfloor n/2\rfloor}.$ $\endgroup$ May 8, 2021 at 17:17

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First when $t\geq 0.$

Since: $$\binom{n}{\lfloor n/2\rfloor}^t\leq\sum_{k=0}^n\binom nk ^t\leq(n+1)\binom n{\lfloor n/2\rfloor}^t$$

and $(n+1)^{1/n}\to 1,$ we get, by the squeeze theorem, that your limit is the same as:

$$\left(\lim_{n\to\infty}\binom n{\lfloor n/2\rfloor}^{1/n}\right)^t$$

Use Stirling to approximate $\binom n{\lfloor n/2\rfloor}.$ You will get a slightly different estimate for when $n=2m$ and $n=2m-1,$ but the $n$th root will wash out that difference.


When $t<0,$ it is much easier, because $$1=\binom n0^t\leq \sum_{i=0}^n \binom{n}i^t\leq (n+1)$$

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  • $\begingroup$ Thank you for your answer. I forgot that theorem from Calculus so I had to check it out! By the way, in the first case, where $t \geq 0$, I guess the final result is $2^{t}$ right? $\endgroup$
    – Albert
    May 8, 2021 at 18:13
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    $\begingroup$ Yes, that is correct. @Albert $\endgroup$ May 8, 2021 at 18:16

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