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I am given the series:

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

and I am asked to determine whether it is convergent or not. I know I need to use the comparison test to determine this. I can make a comparison with a harmonic p series ($a_n=\frac{1}{n^p}$ where p > 1, series converges). I argue that as the denominator grows more rapidly than the numerator, I need only look at the denominators:

$$\frac{1}{n^2+5}\le\frac{1}{n^2}$$

$\frac{1}{n^2}$ is a harmonic p series where $p>1$ which converges. As $\frac{\sqrt{n}+\sin(n)}{n^2+5}$ is less than that, by the comparison test, $\sum_{n=1}^{\infty} \frac{\sqrt{n}+\sin(n)}{n^2+5}$ is convergent.

Is this a valid argument for this question?

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Note quite. The numerator contains the term $\sqrt n+\sin(n)\ge \sqrt n-1$. However, we have

$$\frac{\sqrt n+\sin(n)}{n^2+5}\le \frac{2\sqrt n}{n^2}=\frac{2}{n^{3/2}}$$

Inasmuch as the series $\sum_{n=1}\frac{1}{n^{3/2}}$ converges, the series of interest does likewise by comparison.

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    $\begingroup$ $\sin(n) < \sqrt{n}$ right! , nice but what is the motivation behind the first line?,also is my solution correct ? $\endgroup$ – BAYMAX Mar 19 '17 at 6:02
  • $\begingroup$ What do you mean? We are trying to find a convergence series that dominates the series of interest. We found one here. $\endgroup$ – Mark Viola Mar 19 '17 at 6:03
  • $\begingroup$ I couldnot understand this $\sqrt n+\sin(n)\ge n-1$ ? $\endgroup$ – BAYMAX Mar 19 '17 at 6:07
  • $\begingroup$ Ok,can you please check whether my answer is correct ? $\endgroup$ – BAYMAX Mar 19 '17 at 6:10
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    $\begingroup$ Yes, it certainly appears to be solid! $\endgroup$ – Mark Viola Mar 19 '17 at 6:14
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Since $\sin(n) \leq 1$,so $\sum_{n=1}^{\infty} \frac{\sqrt{n}+\sin(n)}{n^2+5}\leq \sum_{n=1}^{\infty} \frac{\sqrt{n}+1}{n^2+5}$,

Now for large $n$, $(n^{2} + 5)$ can be taken to be $n^{2}$ ,

so the series becomes $\sum_{n=1}^{\infty} \frac{\sqrt{n}+1}{n^2} = \sum_{n=1}^{\infty} \frac{1}{n^{1.5}}+\sum_{n=1}^{\infty} \frac{1}{n^2}$ and both the series are converging so,$\sum_{n=1}^{\infty} \frac{\sqrt{n}+1}{n^2+5}$converges!

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  • $\begingroup$ (+) for this solid solution $\endgroup$ – Mark Viola Mar 19 '17 at 6:23
  • $\begingroup$ Thanks for verifying the solution! @Dr.MV $\endgroup$ – BAYMAX Mar 19 '17 at 7:08

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