# General case of radius of convergence of a power series

Show that if the series $\sum_{n=1}^{\infty} a_nx^n$ has a radius of convergence $L = R$ so the series $\sum_{n=1}^{\infty} a_nx^{kn}$ has radius of convergence $L = R^{\frac{1}{k}}$. Anyone could help me?

## 2 Answers

Let $y=x^k$ so the series $$\sum_{n=1}^\infty a_n y^n$$ is convergent for $|y|<R$ and divergent for $|y|>R$ i.e. it's convergent for $|x|<R^{1/k}$ and divergent for $|x|>R^{1/k}$ so the radius of the series $$\sum_{n=1}^\infty a_n x^{kn}$$ is $R^{1/k}$.

Hint: Let $y=x^k$. ${}{}{}{}{}$