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April 6th, 2014, 07:34 PM   #1
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Verifying analytic part in example converges

$\displaystyle f(z)= \frac{sin(z)}{z^4}=\frac{1}{z^3} - \frac{1}{3!z} + \frac{z}{5!}- \frac{z^3}{7!} + \frac{z^5}{9!}.....$

Would like to know how to verify that the analytic part...
$\displaystyle \frac{z}{5!}- \frac{z^3}{7!} + \frac{z^5}{9!}.....$ converges.

Thanks!
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April 7th, 2014, 04:53 PM   #2
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The proof is a small variation on the proof that the series for sin(z) converges.
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