Application of Stirling's formula

Question

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.

Answer 1

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.

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