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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?

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

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

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