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Given the series $\sum a_n, $ a necessary (but not suficient) condition of convergence is: lim $a_n = 0.$ In the case of power series, which is a specific form of a series, $\sum a_nz^n,$ a necessary and sufficient condition of convergence for $|z|<|z_0|$ is boundedness of the general term $a_n|z_0|^n.$ Here let $z$ be complex.

I cant see how the necessary condition for a series is satisfied, lim $a_n|z_0|^n=0.$ Can somebody help me out ? I dont know what I am missing here, since being bounded must not mean $0.$

Manw thanks.

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    $\begingroup$ The condition $\lim a_n|z_0|^n$ is not satisified. That's not necessary, because we're not saying that $\sum a_n z_0^n$ converges. $\endgroup$ – David C. Ullrich Jun 22 '18 at 13:17
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Assume $a_n|z_0|^n$ to be bounded hence $|a_n|\cdot|z_0|^n \le c$ for a constant $c>0$ then $$|a_nz^n| = |a_n|\cdot|z_0|^n\left|\frac{z}{z_0}\right|^n \le cq^n$$ for $$q = \left|\frac{z}{z_0}\right| < 1$$

And we have absolute convergence by a majorizing series.

And can additionally conclude $$0 \le \lim_{n\to\infty} |a_nz^n| \le \lim_{n\to\infty}cq^n = 0$$

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  • $\begingroup$ What is $z_n$ in this? $\endgroup$ – Kavi Rama Murthy Jun 22 '18 at 10:00
  • $\begingroup$ fixed, thx for the hint $\endgroup$ – Gono Jun 22 '18 at 10:01

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