Unsure about applying series comparison test Does this converge or diverge?
$$
\sum\limits_{n=1}^\infty (a_{n} = \frac{1}{2\sqrt{n} + \sqrt[3]{n}})
$$
The answer is: diverges by limit comparison to $\sum (b_{n} = \frac{1}{\sqrt{n}})$
If I look at $\lim_{n \to \infty}\frac{a_{n}}{b_{n}}$ I get
$$
\frac{\sqrt{n}}{2\sqrt{n} + \sqrt[3]{n}} = \frac{n^\frac{1}{2}} {2n^{\frac{1}{2}} + n^{\frac{1}{2}}n^\frac{2}{3}} = \frac{1}{2 + n^\frac{2}{3}} = \frac{0}{0 + 1} = 0
$$ as $n\to\infty$ 
I would expect it to be some $c > 0$ or $\infty$ 
Also, because $a_{n} < b_{n}$ I don't think I can use the comparison test. I'm pretty sure I am missing something but not sure what. 
Thanks
Update:
My algebra was wrong, how is this for finding the limit?
$$
\frac{\sqrt{n}}{2\sqrt{n} + \sqrt[3]{n}} = \frac{1} {2 + \frac{1}{n^\frac{1}{6}}} = \frac{1} {\frac{2\sqrt[6]{n} + 1}{\sqrt[6]{n}}} = \frac{\sqrt[6]{n}}{2\sqrt[6]{n} + 1} = \frac{1}{2 + 0} = \frac{1}{2}
$$ as $n \to\infty$
And since $\frac{1}{2} > 0$ and $b_{n}$ diverges, $a_{n}$ diverges

There seems to be a mistake in the algebra -- namely, you took $\sqrt[3]{n} = n^{1/2} n^{2/3}$. But the right-hand side of this is equal to $n^{1/2 + 2/3} = n^{3/6 + 4/6} = n^{7/6} \ne \sqrt[3]{n}$.
