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I can't figure out where I'm wrong, I just want to show that the series definitely converges. $\sum_{n=1}^{\infty}(\frac{\cos(n)}{n\sqrt{n}})$
so I started by checking the absolute value.
$\sum_{n=1}^{\infty}(\frac{|\cos(n)|}{n\sqrt{n}})$ and $0\leq|\cos(n)|\leq1$
$\frac{|\cos(n)|}{n\sqrt{n}}\leq \frac{1}{n\sqrt n}$
Therefore, the comparison test can be used.
and $\sum_{n=1}^{\infty}(\frac{1}{n\sqrt{n}})$ is a convergent series, and therefore also the original series

Infinity_hunter
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complex_stock
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    This is how I would do it, why do you think you're wrong? – Leonidas Lanier Oct 03 '22 at 12:13
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    why do you think that you are wrong ? – Surb Oct 03 '22 at 12:13
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    Because I answered this answer in the test and got 0 points, it seems perfectly fine to me too – complex_stock Oct 03 '22 at 12:16
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    @complex_stock Well then you need to consult your professor, because you are completely correct. – K.defaoite Oct 03 '22 at 12:18
  • @K.defaoitr Thanks for checking, I will – complex_stock Oct 03 '22 at 12:22
  • For an alternative, this converges by Dirchelet's test such as in the answer here, Dirchelet's test just being the more generic form of the "alternating series" test. That being said, I do think your answer is more common and elementary. – JMoravitz Oct 03 '22 at 14:40
  • It may depend exactly what you have as the Comparison Test. I would have started like you, and said that by [my version of] the Comparison Test that $\sum_{n=1}^{\infty}|\frac{\cos(n)}{n\sqrt{n}}|$ is convergent, and then, since absolute convergence implies convergence got that $\sum_{n=1}^{\infty}(\frac{\cos(n)}{n\sqrt{n}})$ converges. But nul points seems harsh. – ancient mathematician Oct 03 '22 at 15:01

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