Convergence of $\sum _{n=1}^{\infty \:}\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}}$ $$\sum _{n=1}^{\infty \:}\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}}$$

Upon research I know this series diverges because of the comparison test but I do not know what to compare the series to as I am having trouble finding a series smaller that diverges. And I know when I use the comparison test, my comparison series will diverge because of the $p$-series test.
Try multiplying the top and the bottom by $\sqrt{n+1}+\sqrt{n}$. After simplifying we get
$$\frac{1}{\sqrt{n}(\sqrt{n+1}+\sqrt{n})}\geq \frac{1}{\sqrt{n+1}(\sqrt{n+1}+\sqrt{n+1})}=\frac{1}{2(n+1)}$$
If you are stuck, I'd try doing a little algebra first. Note that $$\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}} = \frac{\sqrt{n+1}}{\sqrt{n}}-\frac{\sqrt{n}}{\sqrt{n}} \\ = \sqrt{\frac{n+1}{n}}-1 $$ Now you can compare $$\sqrt{\frac{n+1}{n}}-1 \geq \sqrt{\frac{0+1}{n}}-1 \\ = \frac{1}{\sqrt{n}}-1$$ and hence $$-1+\left(\sum_{n=1}^\infty \frac{1}{\sqrt{n}}\right) \leq \sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}}$$