Q:

Is this claim true? (About the sequence $\{a_n^{\frac{1}{n}}\}$)

Suppose that $\{(a_n)^{\frac{1}{n}}\}\subseteq \mathbb{R}$ converges and denote with $\alpha$ its limit ($a_n>0$ for all $n$). Is it true that $\{(a_{n+1})^{\frac{1}{n}}\}$ converges to the same limit than the first sequence? Does it even converges necessarely?

I'm trying to solve a problem and, if this result happens to be true, the problem simplifies a lot. Can you give me hand with this please? Counterexamples or hints of a possible proof would be extremely useful (I've tried a lot, but I have not reached any of those).

A:

HINT: $$a_{n+1}^{1/n} = {\left(a_{n+1}^{1/(n+1)}\right)}^{(n+1)/n}$$

