1. ## Radius of Convergence for Complex Power Series

I'm trying to find the Radius of Convergence of a complex power series, and I can see that I can use the Ratio test (i.e. the limit of a_{n}/a_{n+1} exists). However, the first coefficient, a_{0} = 0 (but a_{n} is non zero for all other n).

Is it true that the radius of convergence of a power series from 0 to infinity is the same as the the radius of convergence for the same power series but changing to a sum from m to infinity. (for some arbitrary natural number m). That is, the radius of convergence is independent of where you chose to start your summation from?

This seems logical to me but I'm not 100% certain.

Thanks for any help!

3. ## Re: Radius of Convergence for Complex Power Series

Originally Posted by Ant
Is it true that the radius of convergence of a power series from 0 to infinity is the same as the the radius of convergence for the same power series but changing to a sum from m to infinity. (for some arbitrary natural number m). That is, the radius of convergence is independent of where you chose to start your summation from?

This seems logical to me but I'm not 100% certain.
Be certain. It is true.

It's completely analogous to the fact that, for any positive integer m, the first m terms of a series of real numbers has no bearing on whether or not the infinite series of those real numbers converges (because it doesn't impact the convergence/non-convergence of the limit of the partial sums).

There are two ways to see this, one of which you stated:

1) The ratio test will be $\lim_{n \to \infty} \left\lvert \frac{a_{n+1}}{a_n} \right\rvert \lVert z - z_0 \rVert ,$ so the first m terms of $\{a_n(z-z_0)^n \}_{n \in \mathbb{N}}$ are irrelevant.

Also, remember that the ratio test isn't the definition of convergence (hence nor of the radius of convergence). It's just a test, and so when it doesn't apply, you'll need some other argument to determine the radius of convergence.

Would you dispute that $\cos(z) = \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n)!}z^{2n}$ converges everywhere,

simply because every other term there is 0, and so a naively direct application of the ratio test can't apply?

2) Consider it as $f(z) = p_m(z-z_0) + \sum_{n = m+1}^{\infty} a_n(z-z_0)^n$, where $p_m$ is a polynomial of degree m.

Clearly the radius of convergence of $f$ should be the same as the radius of convergence of $f - p_m$, since

$p_m$ has nothing to do with the convergence of the power series. Yet an additional persepctive is that, since $p_m$ is entire,

and since the radius of convergence is the miminal distance from $z_0$ to a singularity of $f$, that must be the same

distance as the it is for $f - p_m$.