If $\sum_{n=3}^{\infty}a_n x^{n}$ has radius of convergence equal $R$, find radius of convergence $\sum_{n=3}^{\infty}a_n x^{n^2}$.
What I have done
If $\sum_{n=3}^{\infty}a_n x^{n}$ has radius of convergence equal $R$, that means there is subsequence $a_{n_k}$ such that $$ \sqrt[n]{a_{n_k}} \rightarrow^{k\rightarrow \infty} \frac{1}{R}$$ Now, if $$ \exists_{n_0}\forall_{k>n_0} \exists_{m} n_k = m^2 $$ then my result is also $\frac{1}{R}$. In another case I have finite numbers of elements which are equal to my oryginal subsequence so I will get $$\frac{1}{R'} < \frac{1}{R} \text{ so } R' > R$$
is it ok?
You know that $R^{-1} = \limsup|a_n|^{1/n}$. Now, your new series is $\sum_kb_kx^k$ with $b_k = 0$ if $k\notin Q$ and $b_k = a_{\sqrt k}$ if $k\in Q$, where $Q = \{n^2 : n\in\mathbb N\}$. Thus, for the new radius of convergence $r$ you get $$ r^{-1} = \limsup_k|b_k|^{1/k} = \limsup_n|a_n|^{1/n^2} = \limsup_n(|a_n|^{1/n})^{1/n}. $$ Now, if you know that $c_n\to c$, then $c_n^{1/n}\to 1$ as $n\to\infty$. What can you thus say about $r$?