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Find the interval of convergence and radius of convergence for the series:

$$ \sum_{n= 0}^\infty \frac{x^n}{3^n} $$

I'm not sure if I'm correct, but would the interval of convergence be $(-3,3)$ (not inclusive)
and would R = 3?

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  • $\begingroup$ Make it $$\sum_{n= 0}^\infty (\frac{x}{3})^n$$ and notice that it's just a geometric series. $\endgroup$ – Shahar Mar 25 '14 at 22:32
  • $\begingroup$ You are correct, since the series does not converge for $|x| = 3$. $\endgroup$ – Christopher K Mar 25 '14 at 22:46
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This is the geometric series with parameter $$p=\frac{x}{3}$$ and converges for values of $|p|<1$. So your answer is correct.

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$$\lim_{n\to\infty}\left|\dfrac{x^{n+1}}{3^{n+1}}\dfrac{3^n}{x^n}\right|<1\\ \implies \lim_{n\to\infty}\left|\dfrac{x}{3}\right|<1\\ \implies |x|<3,\text{ so the sum converges when }x\in(-3,3),\text{ and the radius of convergence is therefore }3.$$

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  • $\begingroup$ Did you mean xE(-3,3)? $\endgroup$ – user136088 Mar 25 '14 at 22:35
  • $\begingroup$ Ah, yes.${}{}{}$ $\endgroup$ – user122283 Mar 25 '14 at 22:38
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$$r=\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|$$ if the limit exists. In your case you immediately get $r=3$.

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