TITLE: Suppose $\lim \sup_{n \to \infty}a_n \le \rho$. Show $\lim \sup_{n \to \infty} a_n^{{(n-m)}/{n}} \le \rho$.
QUESTION [6 upvotes]: Suppose that $\{a_n\}$ is a sequence of positive numbers with $\lim \sup_{n \to \infty} a_n\le \rho$. Show $\lim \sup_{n \to \infty} a_n^{{(n-m)}/{n}} \le \rho$, where $m \in \Bbb N$.
I was thinking about doing something along the lines of
$\lim \sup_{n \to \infty} a_n^{{(n-m)}/{n}}=\lim \sup_{n \to \infty}a_n (a_n^{-m})^{\frac {1}n}$
Then proving that $(a_n^{-m})^{\frac {1}n}$ goes to $1$ as you take the limit supremum. Please let me know if I am completely off base with this logic.
EDIT: 
$$(a_n^{-m})^{\frac {1}n} = \frac 1 {\underbrace{a_n^{1/n} \cdots a_n^{1/n}}_{m \text{ times}}}$$
Because $a_n^{1/n}$ converges to 1, $\lim_{n \to \infty} a_n^{1/n} =  \lim \sup_{n \to \infty} a_n^{1/n}=1$.
Thus, $\lim \sup_{n \to \infty} a_n^{{(n-m)}/{n}}=\lim \sup_{n \to \infty} a_n\le \rho $.
Is this correct? If not, where did I mess up? If so, this is not the most elegant solution, is there a better one?

REPLY [3 votes]: The problem is that $(a^{-m}_n)^{\frac 1n}$ might not be even bounded. An example is that $a_n = \frac{1}{n^{\frac{n}{m}}}$. In this case we have $(a^{-m}_n)^{\frac 1n} = n$. This observation suggests that one should not break the term into two. 
Indeed, the claim can be proved directly. Let $\epsilon >0$. Then there is $N_1 \in \mathbb N$ so that 
$$ a_n < \rho +\epsilon$$
whenever $n\ge N_1$. Then 
$$ a_n^{\frac{n-m}{n}} <\left(\rho +\epsilon \right)^{\frac{n-m}{n}}. $$
when $n\ge N_1$. This implies 
$$\limsup_{n\to \infty} a_n^{\frac{n-m}{n}}  \le \limsup_{n\to \infty}\left(\rho +\epsilon \right)^{\frac{n-m}{n}} = \rho +\epsilon$$
(of course we used that $x\mapsto (\rho+\epsilon)^x$ is continuous when $\rho+ \epsilon >0$). As $\epsilon >0$ is arbitrary, we have 
$$\limsup_{n\to \infty} a_n^{\frac{n-m}{n}}  \le  \rho. $$