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$$\sum\limits_{n = 1}^\infty {\frac{{\sin (n)}}{{\sqrt {{n^3} + {{\cos }^3}(n)} }}} $$ I tried to check with Maplesoft and Microsft Excel and seems this series is divergent.

Is my conjecture true? How can I prove it?

Thanks in advance.

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  • $\begingroup$ Can you find a simpler series to compare with? $\endgroup$ – Daniel Fischer Nov 11 '15 at 13:18
  • $\begingroup$ Try to show that this series is absolutely convergent by considering $|\sin n | < 1$ $\endgroup$ – Crostul Nov 11 '15 at 13:20
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For $n \geq 2$ we have $$ \frac{|\sin n|}{\sqrt{n^{3}+\cos^{3}(n)}} \leq \frac{1}{\sqrt{n^{3} + \cos^{3}(n)}} \leq \frac{1}{\sqrt{n^{3}-1}}. $$ We have $$ \frac{1}{\sqrt{n^{3}-1}} \sim \frac{1}{n^{3/2}} $$ as $n \to \infty$, so the series $\sum_{n \geq 2}\frac{1}{\sqrt{n^{3}-1}}$ converges by limit comparison test; hence by comparison test we conclude that the series $$ \sum_{n \geq 1}\frac{\sin n}{\sqrt{n^{3}+\cos^{3}(n)}} $$ converges absolutely, and the convergence follows.

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    $\begingroup$ @JohnMa Thank you. :) $\endgroup$ – Megadeth Nov 11 '15 at 14:57
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Hint $$\left|\frac{\sin(n)}{\sqrt{n^3+\cos^3(n)}}\right|\leq \frac{1}{\sqrt{n^3-1}}$$

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Observe that $\frac{\sin(n)}{\sqrt{n^3+\cos^3(n)}}\approx\frac{1}{n^{3/2}}$. Then apply limit comparison test.

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  • $\begingroup$ It's not quite true. That sine will have no limit. $\endgroup$ – user99914 Nov 11 '15 at 13:24

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