Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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. –  Hardy Mar 4 '12 at 1:55
1  
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
1  
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
1  
(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
show 1 more comment

1 Answer

up vote 4 down vote accepted

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 $|z|>|z_0|$.

share|improve this answer
    
Thanks so much mate that was very helpful. –  Hardy Mar 4 '12 at 1:42
    
@Hardy Anytime! Glad to help. –  Pedro Tamaroff Mar 4 '12 at 1:47
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.