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The power series:

$$\sum_{n=0}^{\infty}\frac{(x-3)^n}{3^{2n}}$$

needs to be examined for its radius of convergence. However, I do have some stuggles to understand the problem. Is there anybody who might provide an explanation?

Thank you in advance.....

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3 Answers 3

Hint: Group everything in terms of one exponent, like so:

$$\sum\limits_{n = 0}^{\infty} \left(\frac{x - 3}{9}\right)^n$$

What kind of series is this?

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For simplicity (up to a translation from $3$ to $0$), let's look at the same series centered at $0$: the radius of convergence will be the same.

For which range of $x$ does the series $\sum_{n=0}^\infty \frac{x^n}{3^{2n}}$ converge? If $|x|<9$, what can you say? If $|x| > 9$?

You might want to use the equivalent definition of radius of convergence $R$ for a power series $\sum a_n x^n$: $$ R = \sup\{r \geq 0 : \sum_{n=0}^\infty |a_n| r^n < \infty \} = \sup\{r \geq 0 : a_n r^n \xrightarrow[n\to\infty]{} 0 \} $$

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Here $$ \sum_{n=0}^\infty\frac{(x-3)^n}{3^{2n}}=\sum_{n=0}^\infty a_n(x-3)^n, $$ with $$ a_n=\frac{1}{3^{2n}}. $$ Then, using the ratio test (i.e., if $\lim_{n\to\infty}{|a_{n+1}|/|a_n|}=\ell$, then radius$=1/\ell$), let $$ \frac{a_{n+1}}{a_n}=\frac{\frac{1}{3^{2n+2}}}{\frac{1}{3^{2n}}}=\frac{1}{9}, $$ and hence, the ratio test we obtain that the radius of convergence is $$ r=9. $$

Note. The radius of convergence can be also found by the root test ((i.e., if $\lim_{n\to\infty}\sqrt[n]{|a_n|}=\ell$, then radius$=1/\ell$). Here $$ \sqrt[n]{\frac{1}{3^{2n}}}=\frac{1}{9}, $$ and hence $r=9$, as with the ratio test.

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2  
This is not how the ratio test gives the radius of convergence. –  John Habert Feb 6 at 15:32
1  
Those two power series are not identical. –  Cameron Buie Feb 6 at 15:34
    
What is wrong with that answer? I still dont get it? –  Mamba Feb 6 at 16:13

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