<s>[INST] Comparison test Use the comparison test to determine whether or not 
$$\sum\limits_{n=1}^\infty \dfrac{\sqrt{n}}{1+n+n^2}$$
converges. 
Would I use $\frac{1}{n^\frac{3}{2}}$ for the test and then show,
$$\lim_{x \to \infty} \dfrac{\sqrt{n}}{1+n+n^2}\frac{n^\frac{3}{2}}{1}=\lim_{x \to \infty} \dfrac{n^2}{1+n+n^2} = 1$$ 
$$\sum\limits_{n=1}^\infty \frac{1}{n^\frac{3}{2}}\Rightarrow \sum\limits_{n=1}^\infty \dfrac{\sqrt{n}}{1+n+n^2}\Rightarrow converges $$
 [/INST] $$\frac{\sqrt n}{1+n+n^2}\leq \frac{\sqrt n}{n^2}=\frac{1}{n^{3/2}}$$
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