# Determine whether the series converges or diverges: $\sum_{n=1}^\infty \frac{4\sqrt{n}-1}{n^2+2\sqrt{n}}$

### Determine whether the series converges or diverges.

$$\sum_{n=1}^\infty \frac{4\sqrt{n}-1}{n^2+2\sqrt{n}}$$

I am very confused by this. I believe I'm meant to use the comparison test (as I've read here) but it has me a little lost.

This page says to use the series $\sum_{}b_n$ because if it diverges then $\sum_{}a_n$ converges as well. I'm lost on how to build the series $\sum_{}b_n$ or where to start. Any direction would be very appreciated.

Thanks.

$$\frac{4\sqrt n-1}{n^2+2\sqrt n}<\frac{4\sqrt n}{n^2}=\frac4{n^{3/2}}$$
where we used $\frac ab<\frac cd$ when $a<c$ and $b>d$ with $a,b,c,d>0$.
• So I'm meant to check if $\sum \frac{4}{n^{3/2}}$ converges? This should prove the original series converges or diverges by rule of the comparison test? – Jacob Johnson May 8 '17 at 1:12
• @JacobJohnson Yes. Particularly, if $b_n>a_n>0$ and $\sum b_n$ converges, then $\sum a_n$ converges. If $\sum a_n$ diverges, then $\sum b_n$ diverges. – Simply Beautiful Art May 8 '17 at 1:13
• So I can confirm using the p-series test that because $\frac{3}{2}$ (p) is greater than 0 (?) then it converges? Thank you for your help. – Jacob Johnson May 8 '17 at 1:16
• No, you check that $\frac32>1$, so it converges. – Simply Beautiful Art May 8 '17 at 13:21