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.
Best Answer
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 $$1=\binom n0^t\leq \sum_{i=0}^n \binom{n}i^t\leq (n+1)$$