Convergence of a series if the coefficients are periodic. Suppose that $c_n$ is a periodic sequence -- although, for simplicity, I think we can just imagine that $c_{n+2}=c_n$ for all $n\ge 0$, since I doubt the period should matter.
Can we say anything about the radius of convergence of $\sum_{n\ge 0}c_nx^{2n}$?  I would first think of the root test and examine
$$ \lim_{n\to\infty}\sqrt[n]{|c_nx^{2n}|} = \lim_{n\to\infty}|c_n|^{1/n}|x|^2 $$
and with a little argument I think we can justify $\lim_{n\to\infty}|c_n|^{1/n} = 1$ so long as $c_n\ne 0$ for any $n\ge 0$.  This suggests that the interval of convergence is given by $|x|<1$.
On the other hand by the ratio test we need
$$ \limsup_{n\to\infty}\left| \frac{c_{n+1}x^{2n+2}}{c_nx^{2n}} \right| = \limsup_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right||x|^2<1$$
But here it seems like we don't get the same result.
It seems like the ratio test is just inconclusive in this case, and that could make sense -- there is no reason why the ratio and root tests have to say the same things in all cases.  But it does make me wonder if I've reasoned through all of this correctly.
So is it correct that the root test implies convergence for any nonzero sequence $c_n$ in the interval $|x|<1$ as above, whereas the root test is inconclusive?
 A: Here's another way of seeing that the radius of convergence is indeed $1$ (assuming that the sequence $\{c_n\}$ is not identically zero).
Suppose that $p$ is a positive integer such that $c_{n+p}=c_n$ for all $n\ge0$. Then
$$
\sum_{n=0}^\infty c_nx^n = \sum_{a=0}^{p-1} \sum_{k=0}^\infty c_{kp+a}x^{kp+a} = \sum_{a=0}^{p-1} c_a x^a \sum_{k=0}^\infty x^{kp},
$$
and the inner sum converges when $|x|<1$ (to $\dfrac1{1-x^p}$) and diverges otherwise.
A: The radius of convergence equals
$$R= \lim \inf_{n\to \infty}\  \frac{1}{\sqrt[n]{|c_n|}}$$
If $(c_n)_n$ is periodic,  this is $\infty$ or $1$, according to $c_n \equiv 0$ or not.
Note that in general we have
$$\lim \inf\left| \frac{c_n}{c_{n+1}} \right| \le \lim \inf \frac{1}{\sqrt[n]{|c_n|}} \le \lim \sup \frac{1}{\sqrt[n]{|c_n|}}\le \lim \sup \left| \frac{c_n}{c_{n+1}} \right|$$
so we should only use $\left| \frac{c_n}{c_{n+1}} \right|$ when this sequence is  convergent.
$\bf{Added:}$ Let's see how this works with the sequence
$$c_n = \begin{cases} 2^n \text{ if } n \text{ even} \\
                     3^n \text{ if } n \text { odd} 
         \end{cases} $$
We have
$$\lim \inf_{n\to \infty}\frac{1}{\sqrt[n]{|c_n|}} = \frac{1}{3}\\
\lim \sup_{n\to \infty}\frac{1}{\sqrt[n]{|c_n|}} = \frac{1}{2}$$
$$\lim \inf_{n\to \infty} \frac{|c_n|}{|c_{n+1}|} = 0\\
  \lim \sup_{n\to \infty} \frac{|c_n|}{|c_{n+1}|} = \infty$$
So the radius of convergence of the power series $\sum_{n\ge 0} c_n x^n$ is $\frac{1}{3}$. Moreover, the ratio test does not work.
We can also find the radius of convergence by computing the function and see the largest disk centered at $0$ on which the function is defined. So we compute
$$\sum c_n x^n = \sum_{n\ge 0} 2^{2n} x^{2n} + \sum_{n\ge 0} 3^{2n+1} x^{2n+1} = \frac{1}{1- (2 x)^2} + \frac{3 x}{1- (3 x)^2}$$
Examining the denominators, we see that the largest disk has around $0$ on which the function is defined has radius $\frac{1}{3}$, which also equals the radius of convergence of the series ( a general fact,  due to Cauchy).
