Relations between radius of convergence Let $R$ denote the radius of convergence. Then $\sum_{n=0} a_{2n} x^{2n}$ has $R = 2$, and $\sum_{n=0} a_{3n} x^{3n}$ has $R = 3$. How to prove that $\sum_{n=0} a_{n} x^{n}$ has $R \leq 2$?
For simplification suppose that $a_n > 0\quad \forall n$. Then the following could be written:$$\frac {a_{2n} }{a_{2n+2}} = 2; \frac {a_{3n} }{a_{3n+3}} = 3 \Rightarrow a_{2n+2} = \frac 12 a_{2n}; a_{3n+3} = \frac 13 a_{3n}$$All three series share common $a_0$, thus $a_{2} = \frac 12 a_{0}; a_{3} = \frac 13 a_{0}$.

We have the radius of convergence $1/R=\limsup_{n\ge 1}|b_n|^{1/n}$ for a power series $\sum_{n\ge 0}b_nz^n$.
Clearly, $ \limsup_{n\ge 1}|a_n|^{1/n}\ge \max (\limsup_{n\ge 1}|a_{2n}|^{1/2n},\limsup_{n\ge 1}|a_{3n}|^{1/3n})$, so we conclude that $R\le \min(2,3)=2$.
