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.

  • $\begingroup$ Thank you, I think I my troubles were based on not completely understanding what was meant by the limsup, this makes it clear to me :) $\endgroup$ – Suiri Dec 1 '18 at 16:31

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