Evaluate the following limit
$$
\lim_{n\to\infty} \frac{4^nn!}{(3n)^n}
$$
I've shown the limit is equal to $0$. One may for example use the ration test by which:
$$
\begin{align}
\frac{x_{n+1}}{x_n} &= \left({4\over 3}\right)^{n+1}\frac{(n+1)!}{(n+1)^{n+1}} \cdot \left({3\over 4}\right)^{n}\frac{n^n}{n!}\\
&= \frac{4}{3}\frac{n^n}{(n+1)^n}
\end{align}
$$
Now taking the limit of the fraction:
$$
{4\over 3}\lim_{n\to\infty} \left(n\over n+1\right)^n = {4\over 3e} < 1
$$
By this the sequence is converging to $0$. Another way could be Stirling's approximatiom, by which:
$$
\frac{4^nn!}{(3n)^n} \sim \frac{4^n}{(3n)^n}\cdot \sqrt{2\pi n}\cdot\left({n\over e}\right)^n = \left({4\over 3e}\right)^n\sqrt{2\pi n}
$$
Applying ration test after Stirling's approximation yields the same result. The problem is that Stirling's approximation has not been introduced yet. Also this limit comes right after the problems on proving some specific statements, among which are:
$$
\begin{align*}
\lim_{n\to\infty} x_n = x &\implies \lim_{n\to\infty}\frac{x_1 + x_2 + \cdots +x_n}{n} = x \tag 1\\
\lim_{n\to\infty} x_n = x &\implies \lim_{n\to\infty} \sqrt[n]{x_1x_2\dots x_n} = x\tag 2\\
\lim_{n\to\infty}\frac{x_{n+1}}{x_n} = x &\implies \lim_{n\to\infty} \sqrt[n]{x_n} = x\tag 3
\end{align*}
$$
Right before the problem from question section the book is asking to find the limit of:
$$
\lim_{n\to\infty} {1\over n}\sqrt[n]{(n+1)(n+2)\dots(2n)}
$$
This may be easily handled by applying $(3)$. My assumption is that the author expects me to use one of those proofs I've done before, however I don't see how any of them may be applied. Also please note that proving Cesaro-Stolz is following this limit.
I would appreciate if someone could point me to a way to use $(1), (2)$ or $(3)$ for evaluating the limit. Or possibly suggest other approaches to find that limit from question section.
Thank you!
Best Answer
Here is one way to prove $(3)$ from $(1)$. The proof below is pretty standard.
First observe that in $(3)$ the sequence $x_n$ must be positive. Let $a_n=\log x_n$ and $$b_n=a_{n+1}-a_n=\log\frac{x_{n+1}}{x_n}$$ Then by assumption $b_n\to \log x$. By $(1)$ we have $$\frac{b_1+b_2+\dots+b_n}{n}\to\log x$$ ie $$\frac{a_{n+1}-a_{1}}{n}\to\log x$$ which further implies that $a_n/n=\log\sqrt[n]{x_n}\to \log x$. Thus we have $\sqrt[n] {x_n} \to x$.
Proving $(2)$ from $(1)$ is even easier. One just needs to apply $(1)$ on $\log x_n$.
Cesaro-Stolz is a generalization of $(1)$ and the proof is similar to that of $(1)$.