# Radius of Convergence for two power series is equal.

Given any two power series $$\sum_{n=0}^{\infty}{a_n{(z\ -\ a)}^n}$$ and $$\sum_{n=0}^{\infty}{b_n{(z\ -\ b)}^n}$$ , if there is some m ∈ N such that $$a_n$$= $$b_n$$ for all n ≥ m, then $$\sum_{n=0}^{\infty}{a_n{(z\ -\ a)}^n}$$ and $$\sum_{n=0}^{\infty}{b_n{(z\ -\ b)}^n}$$ have the same radius of convergence.

I have found this stated in my book, but it is given without proof and I am unsure as to how this is simply proved.

It's because the radius of convergence is$$\frac1{\limsup_{n\in\mathbb N}\sqrt[n]{\lvert a_n\rvert}}=\frac1{\limsup_{n\in\mathbb N}\sqrt[n]{\lvert b_n\rvert}}$$(since $$a_n=b_n$$ is $$n$$ is large enough).
A more basic approach consists in using that fact that, if you have two series $$\sum_{n=0}^\infty z_n$$ and $$\sum_{n=0}^\infty w_n$$ and if $$z_n=w_n$$ if $$n$$ is large enough, then either both series converge or both series diverge.