Finding limit using inequalities: $\liminf \frac{a_{n+1}}{a_n} \le \liminf (a_n)^ {1/n}\le\limsup (a_n)^ {1/n}\le \limsup \frac{a_{n+1}}{a_n}$ [duplicate]

Solution 1:

We will prove the right-hand side inequality $\limsup_{n\to \infty}(a_n)^{1/n}\le \limsup_{n\to \infty}\frac{a_{n+1}}{a_n}$ since we can prove the left-hand side inequality following analogously. We will assume that $a_n\ge 0$ for all $n$ and that the $\limsup_{n\to \infty}\frac{a_{n+1}}{a_n}=L<\infty$.

If $L=\limsup_{n\to \infty}\frac{a_{n+1}}{a_n}$, then for all $\epsilon>0$, there exists a number $N>0$ such that whenever $n>N$

$$\frac{a_{n+1}}{a_n}\le \sup_{m\ge n}\frac{a_{m+1}}{a_m}\le L+\epsilon \tag 1$$

From $(1)$ we have for $n>N$

$$\frac{a_n}{a_N}=\underbrace{\left(\frac{a_{N+1}}{a_N}\right)\left(\frac{a_{N+2}}{a_{N+1}}\right)\left(\frac{a_{N+3}}{a_{N+2}}\right)\cdots \left(\frac{a_{n}}{a_{n-1}}\right)}_{n-N\,\,\text{terms}}\le (L+\epsilon)^{n-N}\tag 2$$

From $(2)$, we can write

$$(a_n)^{1/n}\le (L+\epsilon)^{1-N/n}(a_N)^{1/n} \tag 3$$

Taking the $\limsup_{n\to \infty}$ on both sides of $(3)$ reveals that for all $\epsilon>0$

$$\limsup_{n\to \infty}(a_n)^{1/n}\le L+\epsilon \tag4$$

Since $\epsilon$ is arbitrary, then $(4)$ implies

$$\limsup_{n\to \infty}(a_n)^{1/n}\le L=\limsup_{n\to \infty}\frac{a_{n+1}}{a_n}$$

which proves the right-hand side inequality.

Let $a_n=\frac{n^n}{n!}$. Then we see that

$$\begin{align} \lim_{n\to \infty}\frac{a_{n+1}}{a_n}&=\lim_{n\to \infty}\frac{(n+1)^{n+1}}{n^n}\frac{n!}{(n+1)!}\\\\ &=\lim_{n\to \infty}\left(1+\frac1n\right)^n\\\\ &=e\tag 5 \end{align}$$

Note that $(5)$ implies that $\liminf_{n\to \infty}\frac{a_{n+1}}{a_n}=\limsup_{n\to \infty}\frac{a_{n+1}}{a_n}=e$.

Since $\liminf_{n\to \infty}\frac{a_{n+1}}{a_n}\le \liminf_{n\to\infty}(a_n)^{1/n}\le \limsup_{n\to\infty}(a_n)^{1/n}\le \limsup_{n\to \infty}\frac{a_{n+1}}{a_n}$, then we have

$$e=\liminf_{n\to\infty}(a_n)^{1/n}\le \limsup_{n\to\infty}(a_n)^{1/n}=e$$

which implies that

$$\lim_{n\to\infty}(a_n)^{1/n}=\lim_{n\to \infty}\frac{n}{(n!)^{1/n}}=e$$

And we are done!