Radius of convergence of power series where power increases by increments of 2

Question

I know that to determine the radius of convergence of the series $$ \sum_{n=0}^\infty a_nx^n $$ I need to find $$ \lim_{k\rightarrow \infty} \left| \frac{a_{k+1}}{a_k} \right| = c$$ Then the radius of convergence $R$ $$R = \frac{1}{c}$$ However how do I calculate the radius of a convergence for the series $$ \sum_{n=0}^\infty a_nx^{2n} $$ Or more generally $$ \sum_{n=0}^\infty a_nx^{Bn}, \quad B\in\mathbb{N} $$

Answer 1

No, that's not what you need. The Ratio Test is only one way to test convergence of a series. It doesn't always work. But in the case of a series where some of the terms are zero, what might work is using the Ratio Test on consecutive nonzero terms. Thus your series $\sum_{n} a_n x^{Bn}$ converges if $$ \lim_{n \to \infty} \frac{|a_{n+1} x^{B(n+1)}|}{|a_n x^{Bn}|} = |x|^B \lim_{n \to\infty} \frac{|a_{n+1}|}{|a_n|} < 1$$ and diverges if that limit $> 1$. If $\lim_{n \to \infty} |a_{n+1}|/|a_n| = c$ with $0 < c < \infty$, the radius of convergence is then $1/c^{1/B}$.

Answer 2

For your problem, let $y=x^B$, so the series becomes $a_ny^n$ and the ratio test gives the radius of convergence for $y$ as $\frac{1}{c}$, so the radius of convergence for $x$ is $(\frac{1}{c})^\frac{1}{B}$.

Answer 3

Use the root test:

$$\frac{1}{R} = \lim \sup |a_n^{1/n}|.$$

For the series:

$$S = \sum_{n=0}^\infty a_n x^{Bn}=\sum_{k=0}^\infty b_{k}x^{k}$$

with $$b_k = \left\{ \begin{aligned} &a_{k/B}, &k \textrm{ mod }B=0 \\ &0, &\textrm{ otherwise } \end{aligned} \right.$$

$$\frac{1}{R_S} = \lim \sup |b_k^{1/k}| = \lim \sup|a_{k/B}^{1/k}|= \lim \sup|a_{n}^{\frac{1}{nB}}|=\lim_{n\rightarrow \infty} (|a_n|^{1/n})^\frac{1}{B}=c^{1/B}.$$

Answer 4

Consider the series:

$\begin{equation*} \sum_{n \ge 0} a_n x^{B n} \end{equation*}$

From the respective theory, you know that for the series:

$\begin{equation*} \sum_{n \ge 0} a_n y^n \end{equation*}$

there is a radius of convergence $R$ such that it converges if $\lvert y \rvert < R$ and diverges whenever $\lvert y \rvert > R$. Now you can use the comparison test (pick $y_0$ so it is $\lvert y_0 \rvert < R$ and compare with the original series at $x_0 = y_0^{1/B}$ to prove convergence; pick a larger one to prove divergence similarly) to show that your original series converges if $\lvert x \rvert < R^{1/B}$ and diverges whenever $\lvert x \rvert > R^{1/B}$.