limsup of fraction ratio to limsup of the square of ratio Let $({a_n})_{n\geq 0}$ be non decreasing in $n$. How to show that
$$\limsup \frac{a_n}{n} = \limsup \frac{a_{n^2}}{n^2} ?$$
 A: 1) As $a_ {n+1}\geq a_n$ for all $n$, we have that $a_n\to L\in\mathbb{R}$, or $a_n\to +\infty$. I leave to you the first case.
2) We can suppose that $a_n\geq 0$ for all $n$. Note that we have always $\displaystyle \limsup \frac{a_n}{n} \geq \limsup \frac{a_{n^2}}{n^2}$. Hence if $\displaystyle \limsup \frac{a_n}{n} =0$ or if $\displaystyle \limsup \frac{a_{n^2}}{n^2}=+\infty$, we are done.
3) Hence we can suppose that $\displaystyle\limsup \frac{a_{n^2}}{n^2}=L\in ]0,+\infty[$. Let $\varepsilon>0$. There exists $N$ such that for $n\geq N$, we have $\displaystyle \frac{a_{n^2}}{n^2} \leq L+\varepsilon$, hence $a_{n^2}\leq (L+\varepsilon)n^2$.
4) Let $n\geq N^2$. There exists $q$ such that $q\geq N$ and $q^2\leq n<(q+1)^2$. As $a_m$ is non decreasing, we get that $$a_n\leq a_{(q+1)^2}\leq (L+\varepsilon)(q+1)^2\leq (L+\varepsilon)(n+2\sqrt{n}+1)$$
Now it is easy to see that $\displaystyle \limsup \frac{a_n}{n} \leq L+\varepsilon$ and thus that $\displaystyle \limsup \frac{a_n}{n} \leq L$, and we are done.
A: Consider the sequences $\{{\frac{a_n}{n}}_{}\}_{n=0}^{\infty}$ and $\{{\frac{a_{n^2}}{n^2}}_{}\}_{n=0}^{\infty}.$ Notice that every single term you see in the later sequence appears in the first one. Thus $\limsup \frac{a_n}{n}\geq\limsup \frac{a_{n^2}}{{n^2}}$. On the other hand if let$\limsup\frac{a_n}{n}=A\not=\infty$ the we can find a subsequence of $\{\frac{a_n}{n}\}$ that tends to $A$ as $n$ goes to infinity.
But then we can find a subsequence of $\{\frac{a_{n^2}}{n^2}\}$( by $"\epsilon-n$"definition) that tends to $A$.done!
