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Wolfram Alpha says the following series converges, but I can't figure out how to prove it.

$$\sum_{n=1}^{\infty}\frac{\sin(n)+\sqrt{n}}{n^2+5}$$

Can I use a comparison test with a simple harmonic p series, or is there a better way?

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marked as duplicate by Mark Viola sequences-and-series Mar 19 '17 at 17:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ That is quite trivial, my friend: given that $\sum_{n\geq 1}\frac{2\sqrt{n}}{n^2}$ is absolutely convergent, your series is absolutely convergent too. $\endgroup$ – Jack D'Aurizio Mar 19 '17 at 17:50
  • $\begingroup$ Thanks for whoever found the duplicate. I must be in the same class as this guy... $\endgroup$ – user427232 Mar 19 '17 at 17:54
  • $\begingroup$ You're welcome! Just FYI ... You can see the person who marked this as a duplicate. $\endgroup$ – Mark Viola Mar 19 '17 at 18:05
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Comparison: $$ \left|\frac{\sin n + \sqrt n}{n^2+5}\right| \le \frac{2\sqrt n}{n^2} = \frac 2 {n^{3/2}} $$

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Hint. If a series is absolutely convergent then it is convergent, observe that, for $n\ge1$, $$ \left|\frac{\sin(n)+\sqrt{n}}{n^2+5} \right|\le\frac{\left|\sin(n)+\sqrt{n}\right|}{n^2+5} \le \frac{\left|2\sqrt{n}\right|}{n^2} =2\frac1{n^{3/2}}. $$

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