I was working on finding the radius of convergence of $$\sum _{n\ge1} \frac{(-1)^n}{\sqrt n 2^n}x^{n^2} \,\,\,\,\,(*)$$ The radius for this example happens to be $R=1$ since $$R=\frac{1}{\limsup _{n\to \infty}\vert a_n \vert^{\frac{1}{n^2}}}$$, where $a_n=\frac{(-1)^n}{\sqrt n 2^n}.$
- Is there a way to find $R$ using the standard formula for the radius of covergence, which is $$R=\frac{1}{\limsup \vert a_n\vert ^{1/n}},$$ instead of introducing an exponent of $1/n^2?$
- What would the formula for $R$ be if we replace $n^2$ with a general positive integer valued polynomial $f(n)$ in $(*)$? Would it simply be $$R=\frac{1}{\limsup _{n\to \infty}\vert a_n \vert^{\frac{1}{f(n)}}}$$
Let me suggest to go back to elementary properties of the radius of convergence $R$, characterized by the following pair of implications. Consider some series $\sum\limits_n\alpha_nx^n$ with radius of convergence $R$.
In your first example, $$\alpha_n=\left\{\begin{array}{cc}a_k&\text{if}\ n=k^2\\ 0&\text{otherwise}\end{array}\right.$$ hence $|\alpha_nx^n|\to0$ if and only if $|a_nx^{n^2}|\to0$ if and only if $|x|\leqslant1$. Likewise, if $|x|\gt1$, $|a_nx^{n^2}|$ is unbounded. This proves that $R=1$. Note that $$ 1/R=\limsup|\alpha_n|^{1/n}=\limsup|a_k|^{1/k^2}, $$ hence there is no need to adapt the general formula. In particular, the formula you suggest in the last part of your question is correct.