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Prove $\displaystyle\lim_{n\to\infty} \left(\frac {2}{3}\right)^n=0$ with the definition of limit.

From the definition and since $n\in\mathbb N$ I get that ${\Large\mid} \left(\frac {2}{3}\right)^n{\Large\mid}=\left(\frac {2}{3}\right)^n$ but now I'm not sure what to do, I don't see how taking a $\log$ here would help.

I thought of different approach, proving by induction that $\left(\frac {2}{3}\right)^n < \frac 1 n$: The first steps are trivial, then for $n+1$: $\frac {1}{n+1}>\frac {2}{3}\left(\frac {2}{3}\right)^n$, then from the induction hypothesis: $\frac {1}{n+1}>\frac {2}{3}\frac 1 n$ and we get that this true for $n>2$ so we can choose $N_{\epsilon}=\frac 1 {\epsilon}$.

Is there another way that does not involve induction?

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HINT. By limit definition: $$|a_n-l|<\epsilon$$ i.e. $$\left|\left(\frac{2}{3}\right)^n\right|=\left(\frac{2}{3}\right)^n<\epsilon$$ whence $$-n\ln(3/2)<\ln\epsilon$$ and $$n>-\frac{\ln\epsilon}{\ln \frac{3}{2}}$$

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    $\begingroup$ Oh right I forgot it becomes negative because it's a fraction... Thanks. $\endgroup$ – shinzou Nov 23 '14 at 20:04
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Clearly the sequences $a_n = \frac23^n$ and $b_n = \frac23^{n+1}$ have the same limit point $L$ since $b_n = a_{n+1}$, and the limit exists since it is monotonically decreasing and positive. Yet we also know that $b_n = \frac23a_n$, so we have $$\lim_{n\to\infty} b_n = \frac23\cdot\lim_{n\to\infty} a_n,$$ or $$L = \frac23L,$$ thus $L = 0$.

This doesn't use induction, but I suppose it doesn't really use the definition of limit either...

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  • $\begingroup$ How do you know $\lim b_n=\frac{2}{3}\lim a_n$ when we do not know whether $\lim a_n$ exists? There seems to have some circular reasoning. $\endgroup$ – Eric Sep 23 '16 at 19:37
  • $\begingroup$ @Eric This answer wasn't helpful because I cited a fact that monotonically decreasing bounded below sequences have limit points instead of using the definition of a limit. $\endgroup$ – Ben Sep 23 '16 at 20:18
  • $\begingroup$ Sorry! I overlooked that sentence. My mistake. :-) This proof is correct, and I think it is helpful! The monotone convergence theorem is pretty basic so that accepting this theorem to proof this is OK for me! $\endgroup$ – Eric Sep 24 '16 at 5:05

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