The radius of convergence has to be determined 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.....
 A: 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?
A: 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 $\sum a_n x^n$:
$$
R = \sup\{r \geq 0 : \sum_{n=0}^\infty |a_n| r^n < \infty \} = \sup\{r \geq 0 : a_n r^n \xrightarrow[n\to\infty]{} 0 \}
$$
A: 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.
