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

I have to prove that : Given a power series $\sum_n a_n x^n$ diverges for $x = x'$ then it diverges for every $x$ such that $| x | > |x'|$ . Please suggest how to proceed.

I am trying to use the First comparison test, knowing the series $\sum_n a_n x'^n$ is not convergent which implies that it is not absolutely convergent.

Then comparing the $n$'th terms of the series we have

$$ | a_n x'^n | < | a_n x ^n | $$

Since the series on the left is not convergent, we have that the series on the right is not convergent either.

But I am getting the not absolute convergence of the series by this approach...Please help !

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    Have you tried to use the contrapositive assumption?2012-04-06
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    @JyrkiLahtonen: thanks for your reply. i guess you mean by assuming on the contrary, the series to be convergent. but again getting stuck on the fact convergence does not yield absolute convergence.2012-04-06
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    @preeti: There may be a theorem in your textbook stating that if $\sum_n a_n x^n$ converges at $x$, then it converges at $x'$ whenever $|x'|<|x|$?2012-04-06
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    @JyrkiLahtonen:Thanks for the help. You are right !2012-04-06
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    @preeti: If you figured it out, do consider writing it up here as an answer. That way you get experience writing down an argument, you get feedback on it, the question won't get stuck in the unanswered files etc. You may have to wait a few hours though before the system lets you do that.2012-04-06
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    @JyrkiLahtonen: Thanks, will do so.2012-04-06

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Try to prove it by contradiction. Have you proven that

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

Suppose that a series diverges for $z=z_0$. Then assume it converges for $z=z_1$ such taht $|z_1|>|z_0|$. Then what would happen to the series for $z_0$ given the above theorem is true?