# Prove that if $\sum a_k z_1^k$ converges, then $\sum a_k z^k$ also converges, for $|z|<|z_1|$

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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What is the "converse" you need to prove? As far as I can see, your proof is complete as it stands. – Henning Makholm Mar 4 '12 at 1:52
@HenningMakholm Sorry i am new to proofs so i was under the assumption that i would need to prove something to come to the conclusion that something forms a circle than just starting with the assumption as i had. – Comic Book Guy Mar 4 '12 at 1:55
If you want to prove "A if and only if B", then you need both directions: assume A and conclude B, then wipe the slate and assume B, conclude A (or one can do the two halves in the opposite order). However, in this case you just want to prove "if A then B", so there is only one direction to do: assume A, conclude B. – Henning Makholm Mar 4 '12 at 2:07
In both cases, you're allowed to replace any "assume this, conclude that" step with "assume not-that, conclude not-this" (that is, exchange the assumption and conclusion, and negate both) if the latter happens to be easier to do (which depends on that "this" and "that" are). That's the contrapositive, a term that is sometimes confused with the "converse". But you need only to prove either an implication or its contrapositive -- they are logically equivalent. – Henning Makholm Mar 4 '12 at 2:11
(Don't worry, it becomes second nature to remember how these things fit together with a bit of practice). – Henning Makholm Mar 4 '12 at 2:13

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 $0<q<1$, $\sum a_k z^k$ 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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Thanks so much mate that was very helpful. – Comic Book Guy Mar 4 '12 at 1:42
@Hardy Anytime! Glad to help. – Pedro Tamaroff Mar 4 '12 at 1:47