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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?

Meecolm
  • 683

1 Answers1

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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

$$\limsup_{n\to \infty} a_n^{\frac{n-m}{n}} \le \rho. $$