# Find the limit of a sequence to the power of something

First of all, I'm sorry if my English is not very correct, I never used mathematical words in English before, but I hope it's readable.

I have an exercise in which they give me a sequence, called $\{a_n\}$, and there is this statement: If $\left\{\sqrt{a_n}\right\}$ is regular, then $\left\{\sqrt[3]{a_n}\right\}$ is regular too. I have to say if the statement is true or false, but I simply don't know where to start.

I searched online but I didn't find anything, so I'll be grateful to everyone who can help me... What should I use to solve exercises in which there is a sequence to the power of a number, in this case $1/2$ and $1/3$? Maybe there is a special rule and I don't know that...

Thank you and sorry if the question seems to be stupid (indeed it is) but I don't know another place to ask...

• Welcome to Math.SE! It might be useful here to explain what you mean by regular, since you say yourself that you are not sure about mathematical words in English: a mathematical notion of regularity will overcome this possible language barrier. – Hrodelbert Dec 18 '15 at 10:32
• By regular I mean that the limit of the sequence exists (and it can be either a real number or positive/negative infinity), for example $1/n$ is regular, $n$ is regular, $(-1)^n$ is not – RaffoSorr Dec 18 '15 at 10:37
• @Raffolox perhaps it is better to use convergent than regular. – Nizar Dec 18 '15 at 10:39
• Nope, because a divergent sequence is also regular... I don't know if everybody uses the expression "regular" for a sequence, but a regular sequence is a convergent or divergent sequence (while there are other sequences that are not convergent and are not divergent too) – RaffoSorr Dec 18 '15 at 10:45

First, suppose that $\{ \sqrt{a_n} \}_n$ is convergent. So, consider the function $f(x)$ defined by: $$f(x)=x^{\frac{2}{3}}$$ This function is continuous on $[0, + \infty)$ and since the sequence $\{ \sqrt{a_n}\}_n$ is convergennt, then the sequence defined by $\sqrt[3]{a_n}$ which is equal to $f(\sqrt{a_n})$ is convergent.
Now, if $\{ \sqrt{a_n} \}_n$ is not convergent, and since it is regular, then its limit is either $+\infty$ ( it cannot be $-\infty$ as its terms are all positive). Then the limit of the sequece $\{ \sqrt[3]{a_n}\}$ is also $+\infty$.