# 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 $$0 and let $$\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:

$$a_k \;=\; \begin{cases} a_{n_k}, & \text{k = n_k} \\ 0, & \text{else} \end{cases}$$

$$\sqrt[k]{|a_k|} \;=\; \begin{cases} |a_{n_k}|^\frac{1}{k}, & \text{k = n_k} \\ 0, & \text{else} \end{cases}$$

$$\lim_{k \to \infty}{|a_{n_k}|^{\frac{1}{k}}} \;=\; \lim_{k \to \infty}{|a_{n_k}|^{\frac{1}{Ln_k}}} \;=\;\; ?$$

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 $n$.