<s>[INST] Proof of sum convergence? 
$$\sum_{n=1}^{\infty} \frac{2\sqrt{n} + 1}{n^2 + n + 1}$$

This seems like a problem that could be handled by the comparison test.
So we need an $f(n) > \displaystyle \frac{2\sqrt{n} + 1}{n^2 + n + 1}$
So we need $\text{num}f(n) > 2\sqrt{n} + 1$ and $\text{denom}f(n) < n^2 + n + 1$. 
So obviously, we can let $\text{denom}f(n) = n^2$ and we can try $\text{num}f(n) = 3n$
So we have $f(n) = \frac{3}{n} >  \frac{2\sqrt{n} + 1}{n^2 + n + 1}$
but this fails?
 [/INST] You're on the right track, but you gave up too much in the numerator. Instead, observe that
$$ \frac{2\sqrt{n}+1}{n^2+n+1}\leq \frac{2\sqrt{n}+1}{n^2}\leq \frac{3\sqrt{n}}{n^2}=\frac{3}{n^{\frac{3}{2}}}$$ 
and $\sum_{n=1}^{\infty}\frac{3}{n^{\frac{3}{2}}}$ converges.
</s>