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,
$$\lim_{x \to \infty} \dfrac{\sqrt{n}}{1+n+n^2}\frac{n^\frac{3}{2}}{1}=\lim_{x \to \infty} \dfrac{n^2}{1+n+n^2} = 1$$ 
$$\sum\limits_{n=1}^\infty \frac{1}{n^\frac{3}{2}}\Rightarrow \sum\limits_{n=1}^\infty \dfrac{\sqrt{n}}{1+n+n^2}\Rightarrow converges $$
 A: $$\frac{\sqrt n}{1+n+n^2}\leq \frac{\sqrt n}{n^2}=\frac{1}{n^{3/2}}$$
A: I want to clear up something and sort of add a note on DonAntonio's answer. It seems you're saying your original sum is equal to $\sum_{n=0}^{\infty}\dfrac{1}{n^{3/2}}$. Instead, it really is similar to, but not equal to. Thus, we should use the Comparison Test and watch our inequalities.
Looking at behavior near $\infty$ (that is, looking at the terms that dominate), we have that:
$$\frac{\sqrt{n}}{1+n+n^2} \thicksim \frac{\sqrt{n}}{n^2} = \dfrac{1}{n^{3/2}}$$
Note that 
$$\frac{\sqrt{n}}{1+n+n^2} \le \dfrac{1}{n^{3/2}} \forall n \in \mathbb N$$
Thus we can say our original sum is less than $\sum_{n=0}^{\infty}\dfrac{1}{n^{3/2}}$ which is a convergent p-series ($p>1$) which means our original sum converges by comparison.
Addendum
It appears you used the Limit Comparison Test. For the LCT, we have an assumption that
$$\lim_{n\to\infty} \frac{a_n}{b_n}$$ exists and equals $L$. Then we have two cases:
$$L>0 \implies \sum{a_n} \ \ \ \text{converges} \iff \sum{b_n} \ \ \ \text{converges}$$
and
for the case $L=0$ : if $\sum{b_n}$ converges, then $\sum{a_n}$ converges. 
Note that the second case only works one way. Your reasoning using the LCT seems fine though by stating $\sum{a_n}$ converges because $L>0$ and $\sum{b_n}$ is a convergent p-series.
A: Your answer is correct if you can assume the knowledge that $\sum_n n^{-3/2}<\infty$.
