This is a relatively simple problem. I'm just making sure I have the right idea here. I'd like to prove that the sequence $\displaystyle a_n = 1 + \frac{1}{n^{1/3}}$ converges. My proof is:

We conjecture that $a_n$ converges to 1. Thus, we must show that, for all $\epsilon \in \mathbb{R}$, there exists an $N(\epsilon) \in \mathbb{N}$ such that $|a_n - 1| < \epsilon$ for all $n > N(\epsilon)$.

From that inequality, I do some algebra and find that: $~~\displaystyle n > \frac{1}{\epsilon^3}$, so, if we choose $\displaystyle N(\epsilon) > \frac{1}{\epsilon^3}$, we've shown that the sequence converges to 1.

Is this correct?

  • $\begingroup$ I believe you are saying "series" when you mean to be saying "sequence". $\endgroup$ – David H Sep 22 '14 at 12:11
  • $\begingroup$ @DavidH I believe you're correct! $\endgroup$ – AmagicalFishy Sep 22 '14 at 12:11

Just so this won't go unanswered:
Yes, you're correct. If $n>1/\epsilon^{3}$ then $n^{-1/3}<\epsilon$, for all $\epsilon>0$.

  • $\begingroup$ Excellent! Thanks $\endgroup$ – AmagicalFishy Sep 23 '14 at 11:53

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.