I know the definition of finite limit that $\lim a_n =L$ if $\forall \epsilon > 0 $, $\exists n_{\epsilon} \in \mathbb{N} $ such that $\forall n> n_{\epsilon} $, $|a_n - L| < \epsilon$. So I can start with

$$ \left| 3^{1/n} - 1 \right| < \epsilon $$

and then I'm pretty stuck. I can try to exponentiate the first term $$ \left| e^\frac{\ln 3}{n} - 1 \right| < \epsilon, $$ but I don't see how it helps...

  • $\begingroup$ For each epsilon you can find N such that the nth (n> N) power of numbers larger than $1+\epsilon $ is larger than three and the nth (n> N) power of numbers smaller than $1-\epsilon $ is smaller than 3. Prove that and then finish $\endgroup$ – Bananach Aug 13 '16 at 17:18


Recall ( SEE THIS ANSWER )that for $x<1$, the exponential function satisfies the inequalities

$$1+x\le e^x\le \frac{1}{1-x}\tag 1$$

Can you finish now?

SPOILER ALERT: Scroll over the highlighted area to reveal the solution.

Using $(1)$ we can write for $n\ge 2$ $$\begin{align}\left|3^{1/n}-1\right|&=\left|e^{\frac1n \log(3)}-1\right|\\\\ & \le \frac{\frac1n \log(3)}{1-\frac1n \log(3)}\\\\&=\frac{\log(3)}{n-\log(3)}\\\\ &<\epsilon \end{align}$$ whenever $n>\max\left(2,\log(3)+\frac{\log(3)}{\epsilon}\right)$. And we are done!

  • $\begingroup$ Thank you! The assertion $e^x \le 1/(1-x)$ is trivial? I can see that $ 1+x \le e^x$ because $1+x$ is just first order approx. of $e^x$ at $0$, but the second inequality isn't that obvious for me. $\endgroup$ – J.Doe Aug 13 '16 at 17:24
  • $\begingroup$ You're welcome. My pleasure. I've added a link to an answer in which I showed both inequalities using only the limit definition of the exponential function and Bernoulli's Inequality. $\endgroup$ – Mark Viola Aug 13 '16 at 17:26

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