$\lim_{n\to\infty} \sqrt[n]{n!}$ and $\lim_{n\to\infty}\frac{1}{n} \sqrt[n]{n!}$ with differentiation and integration tools.

Question

Find the following limits: $$\lim_{n\to\infty} \sqrt[n]{n!} \hspace{3cm} \lim_{n\to\infty}\frac{1}{n} \sqrt[n]{n!}$$

This exercise are in my homework of Real Analysis about Riemann-Stietljes integration. However, I don't find the relation of these limits with the integration with Riemaniann sums. In this exercise I've already demonstrated that:

$$\int_{0}^{a}\exp(x)dx = \int_{*0}^{a} \exp(x)dx = \int_0^{*a}\exp(x)dx = \exp(a)-1.$$

Where we have the following definitions:

$$P = \lbrace x_0,\ldots,x_n\rbrace = \text{Partition of the interval }[0,a],$$ $$m_j = \inf\lbrace f(x):x\in[x_{j-1},x_j]\rbrace, \hspace{3mm} M_j= \sup\lbrace f(x):x\in[x_{j-1},x_j]\rbrace,$$ $$s(f,P) = \sum_{j=1}^n m_j (x_j-x_{j-1}), \hspace{3mm}S(f,P) = \sum_{j=1}^{n} M_j(x_j-x_{j-1}),$$ $$\int_{*0}^a f(x)dx =\sup\lbrace s(f,P):P \text{ partition of [0,a]} \rbrace,$$ $$\int_0^{*a} f(x)dx=\inf\lbrace S(f,P):P \text{ partition of [0,a]} \rbrace.$$

I think taht I have to used the result already demonstrated but I'm not sure. I appreciate any help or hint :)

Answer 1

I'd suggest you start with the second problem by taking the logarithm and simplifying; the key simplification you need is $\log(n!)=\sum_{k=1}^n \log(k)$. Once you understand the second one, the first one is straightforward.

Incidentally this trick of starting off with the logarithm is the usual procedure for studying infinite products and similar expressions.

Answer 2

Note that $$\log(\sqrt[n]{n!}) = \frac{1}{n}\sum_{i=1}^n \log i.$$

We have $$\sum_{i=1}^n \log i \geq \int_1^{n-1} \log x \, dx,$$ since $\log (i+1) \geq \int_i^{i+1} \log x , dx.$$

Note that $$\int_1^{n-1} \log x \, dx = (n-1)(\log(n-1)-1) + 1.$$ Thus, $$\frac{1}{n}\sum_{i=1}^n\geq \frac{1}{n}\int_1^{n-1} \log x \, dx$$ which will diverge by our above closed form. Hence the first limit is infinity (using continuity of the logarithm).

The second limit is $1/e,$ which we compute as follows. Note $$\log(\sqrt[n]{n!}/n) = -\log n + \frac{1}{n}\sum_{i=1}^n \log i.$$

We bound that sum as $$\frac{n-1}{n}(\log (n-1) - 1) + \frac{1}{n} \leq \frac{1}{n}\sum_{i=1}^n \log i \leq (\log n -1) + \frac{1}{n},$$ so when you subtract $\log n$ and let $n\rightarrow \infty,$ you (by the squeeze theorem) find that the limit approaches $-1.$ But then using continuity of the logarithm, the original limit is $e^{-1} = 1/e.$