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Show that if a power-series converges for any value of $z_{0}$ of $z$, it will be absolutely convergent for all values of $z$ whose representation points are within a circle which passes through $z_{0}$ and has its center at the origin.

Proof Attempt Let z be such a point, so we have $\left| z\right| < \left| z_{0}\right| $. Now, since $\sum _{n=0}^{\infty }a_{n}z_{0}^{n}$ converges, $a_{n}z_{0}^{n}\rightarrow 0$ as $n\rightarrow \infty $, so we can find M(independent of n) such that $\left| a_{n}z_{0}^{n}\right| < M$ and we observe that $\left| a_{n}z^{n}\right| < M\left| \dfrac {z} {z_{0}}\right| ^{n}$. So every term in the series $\sum _{n=0}^{\infty }\left| a_{n}z^{n}\right| $ is less than the corresponding term in the convergent geometric series $\sum _{n=0}^{\infty }M\left| \dfrac {Z} {Z_{0}}\right| ^{n}$ the series is therefore convergent and so the power-series is absolutely convergent , as the series of moduli of its terms is a convergent series. I am unsure how to tackle the second part(converse) of the problem. Any help would be much appreciated.

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    (Don't worry, it becomes second nature to remember how these things fit together with a bit of practice).2012-03-04

1 Answers 1

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What you're saying is the following.

If a power series converges for a number $z=z_0\neq0$ then it will converge absolutely for any $z$ such that $|z|<|z_0|$.

The idea is that since $\sum a_k z_0^k$ converges then $a_k z_0^k\to0$. Then, for some $n\geq N$ we have that $|a_n z_0^n| < 1$. Let $C$ be a circle with radius $0< R < |z_1|$. Then if $z\in C$ and $n\geq N$ we have that $|z| \leq R$ and

$|a_n z^n|=|a_n z_0^n|\left|\frac{z^n}{z_0^n}\right|<\left|\frac{z^n}{z_0^n}\right|\leq \left|\frac{R}{z_0 }\right|^n=q^n$

Since $0, $\sum a_k z^k$ is dominated by $\sum q^k$, so that by Weierstrass' $M$ criterion, the original series converges absolutely and uniformly in $C$.


ADD: You might also want to prove the following (by contraposition):

If a power series diverges for a number $z=z_0\neq0$ then it will diverge for any $z$ such that $|z|>|z_0|$.

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    @Hardy Anytime! Glad to help.2012-03-04