# Application of Stirling's formula

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.

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

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

• 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? May 8, 2021 at 18:13
• Yes, that is correct. @Albert May 8, 2021 at 18:16