Find the radius of convergence of $\sum_{n=0}^{\infty}{\ln\left(\cos{\frac{1}{3^n}}\right)x^n}$ I need to find the radius of convergence of
$$\sum\limits_{n=0}^{\infty}{\ln\left(\cos{\frac{1}{3^n}}\right)x^n}$$.
I came up with the following solution:
Since $\cos{\frac{1}{3^n}\sim1-\frac{1}{2(3^n)^2}}$, then
$$\ln\left(\cos{\frac{1}{3^n}}\right)\sim\ln\left(1-\frac{1}{2(3^n)^2}\right)\sim-\frac{1}{2(3^n)^2}$$
Therefore,
$$\sum_{n=0}^{\infty}{\ln\left(\cos{\frac{1}{3^n}}\right)x^n}\sim\sum_{n=0}^{\infty}{-\frac{x^n}{2(3^n)^2}}=-\sum_{n=0}^{\infty}{\frac{x^n}{2(3^n)^2}},$$ where we can assess easily what the radius is using the ratio test:
$$R = \lim\limits_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|.$$
Is the solution correct? If not, could you suggest an alternative solution, please?

As a few commentors have pointed out: Yes, this looks good. Ratio test is always going to be your friend on questions such as these. Let's go ahead and work it out so you're sure you got the right final answer:
\begin{align}
\frac{a_{n+1}}{a_n} &= \frac{{\frac{x^{n+1}}{2(3^{n+1})^2}}}{{\frac{x^n}{2(3^n)^2}}}\\
&= {\frac{x^{n+1}}{2(3^{n+1})^2}}\cdot\frac{2(3^n)^2}{x^n}\\
&= {\frac{x}{(3^{n+1})^2}}\cdot\frac{(3^n)^2}{1}\\
&= {\frac{x}{3^2}}\cdot\frac{1}{1}\\
&= {\frac{x}{9}}\\
\end{align}
so we have that radius of convergence is $-9 \leqslant x \leqslant 9$.
