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?

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!
