If $\sum_{n=0}^{\infty}a_nz^n$ $(z \in \mathbb C)$ converges then it converges absolutely for every $z$ satisfying $|z|<|z_0|$.

Question

Show that if the power series $\sum_{n=0}^{\infty}a_nz^n$ $(z \in \mathbb C)$ converges for some fixed $z=z_0$, then it converges absolutely for every $z$ satisfying $|z|<|z_0|$.

My attempt:

If $\sum_{n=0}^{\infty}a_nz^n$ converges then we must have $\left|\frac{a_{n+1}}{a_n} z\right| \le 1$, else the series would diverge by the ratio test. So if $|z|<|z_0|$ then $\left|\frac{a_{n+1}}{a_n}z\right|<1$, so by the ratio test $\sum_{n=0}^{\infty}a_n z^n$ converges absolutely.

Can someone please verify if my proof is correct? The examiner says the ratio test does not help here but I can't see where I have gone wrong.

Answer 1

You can't compute a ratio if you can't verify that the denominator doesn't cancel.

Simply do this : if $\sum a_nz_0^n$ converges, then $\left|a_nz_0^n\right|$ is bounded, say by $M$.

Now let $z$ be a complex such that $|z|<|z_0|$. You have : $$\left|a_nz^n\right| = \left|a_nz_0^n\right|\left|\frac{z}{z_0}\right|^n \le M\left|\frac{z}{z_0}\right|^n$$ so the term $\left|a_nz^n\right|$ is bounded by the term of a convergent geometric series, therefore the series $\sum a_nz^n$ is absolutely convergent.

Remember : the ratio test works only if you can assure $u_n\ne0$ for every $n$.

Answer 2

If $\sum a_nz_0^n$ converges then $\lim_{n\to+\infty}a_nz_0^n=0$ and for great enough $n$,

$$|a_nz_0^n|<1$$ thus

$$|a_nz^n|=|a_nz_0^n||\frac{z}{z_0}|^n$$$$<|\frac{z}{z_0}|^n$$

use grometric series and comparison test.