Determining whether the series $\sum_{n=1}^{\infty} \frac{\sqrt{n}+\sin(n)}{n^2+5}$ is convergent or divergent by comparison test

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?

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.

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

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!

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