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...

  • $\begingroup$ 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. $\endgroup$ – Hrodelbert Dec 18 '15 at 10:32
  • $\begingroup$ 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 $\endgroup$ – RaffoSorr Dec 18 '15 at 10:37
  • $\begingroup$ @Raffolox perhaps it is better to use convergent than regular. $\endgroup$ – Nizar Dec 18 '15 at 10:39
  • $\begingroup$ 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) $\endgroup$ – 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 $+\infty$.

  • $\begingroup$ It's perfect, thank you a lot! $\endgroup$ – RaffoSorr Dec 18 '15 at 14:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.