Prove: the radius of convergence of the series $\sum_{k=1}^{\infty}a_kx^{n_k}$ is $R^L$. 
Let $\sum_{n=1}^{\infty}a_nx^n$ be a power series with a radius of
convergence $0<R<\infty$ and let $\left \{ n_k \right \}_{k=1}^\infty$
be a strictly monotonically increasing sequence of natural numbers,
such that the limit $L=\lim_{k\to\infty}\frac{k}{n_k}$ exists.
Prove: the radius of convergence of the series $\sum_{k=1}^{\infty}a_kx^{n_k}$ is $R^L$.

My attempt:
$$  a_k \;=\;
\begin{cases}
a_{n_k},  & \text{$k = n_k$} \\
0, & \text{else}
\end{cases}$$
$$  \sqrt[k]{|a_k|} \;=\;
\begin{cases}
|a_{n_k}|^\frac{1}{k},  & \text{$k = n_k$} \\
0, & \text{else}
\end{cases}$$
$$\lim_{k \to \infty}{|a_{n_k}|^{\frac{1}{k}}} \;=\; \lim_{k \to \infty}{|a_{n_k}|^{\frac{1}{Ln_k}}} \;=\;\; ?$$
 A: We are given that
$$ \limsup_{n\to\infty}\sqrt[n]{|a_n|}=\frac1R$$
Letting $$b_n=\begin{cases}a_{k},&n=n_k\text{ for some }k\\0,&\text{otherwise}\end{cases} $$
(I'm avoiding confusion between the coefficients here)
we want to show that
$$ \limsup_{n\to\infty}\sqrt[n]{|b_n|}=\frac1{R^L}.$$
We do so by showing two things:


*

*If $c>\frac1{R^L}$, then there are at most finitely many $n$ with $\sqrt[n]{|b_n|}>c$.

*And if $c<\frac1{R^L}$, then there are infinitely many $n$ with $\sqrt[n]{|b_n|}>c$


Note that $\sqrt[n]{|b_n|}>c>0$ means $|b_n|>c^n$. This can happen only if $n=n_k$ for some $k$, and then is equivalent to $|a_k|>c^n$.
Now if $c=\frac q{R^L}$ with $q>1$, then 
$$c^n=\frac {q^n}{R^{Ln}} $$
Given $\epsilon>0$, we know that $|L-\frac kn|<\epsilon$ for almost all $n$ and matching $k$, thus $|Ln-k|<k\epsilon$ and
$$\sqrt[k]{|a_k|}>\sqrt[k]{c^n}=\sqrt[k]{\frac{(q^{1/L})^{nL}}{R^{nL}}}>\sqrt[k]{\frac{(q^{1/L})^k(q^{1/L})^{-k\epsilon}}{R^kR^{k\epsilon}}} =\frac1R\cdot \frac{q^{1/L}}{R^\epsilon q^{\epsilon/L}}$$
If we pick $\epsilon$ small enough, the last factor will be $>1$, hence this inequality can hold at most for finitely many $k$, and so $\sqrt[n]{|b_n|}>c$ only for finitely many $n$.
On the other hand, if $c=\frac q{R^L}$ with $0<q<1$, we find by a similar argumentation that, given $\epsilon>0$,
$$\sqrt[k]{c^n}<\frac 1R\cdot q^{1/L}R^\epsilon q^{\epsilon/L} $$
for almost all $k$ and $n=n_k$.
If $\epsilon$ is small enough, the right hand side is $<\frac 1R$, hence
 $\sqrt[k]{|a_k|}>\frac 1R\cdot q^{1/L}R^\epsilon q^{\epsilon/L}$ for infinitely many $k$, hence $\sqrt[n]{|b_n|}>c$ for infinitely many $n$.
