<s>[INST] Study the convergence of $ \sum_{n \ge 1} \frac{\sqrt{n}}{n \sqrt[3]{n} + 2}$ I have to study the convergence of the following series
$$\sum_{n \ge 1} \frac{\sqrt{n}}{n \sqrt[3]{n} + 2}$$
What I tried was to compare the following terms:
$$\frac{\sqrt{n}}{n \sqrt[3]{n} + 2} \le \frac{\sqrt{n}}{n \sqrt[3]{n}}  = \frac{1}{n^{5 / 6}}$$
We know that
$$\frac{1}{n^{5/6}} \rightarrow 0 \text{, as } n \rightarrow \infty$$
so using the First Comparison Test I concluded that the series
$$\sum_{n \ge 1} \frac{\sqrt{n}}{n \sqrt[3]{n} + 2}$$
is convergent. However, I think I got something wrong. The First Comparison Test tells us that if the series is convergent, then the limit of the term that makes the series is $0$, not the other way around. So I think my argument is invalid. How should I approach this?
 [/INST] Your inequality
$$\frac{\sqrt{n}}{n \sqrt[3]{n} + 2} \le \frac{\sqrt{n}}{n \sqrt[3]{n}}  = \frac{1}{n^{5 / 6}}$$
is fine but since $\sum  \frac{1}{n^{5 / 6}}$ diverges it is inconclusive.
As an alternative we can use limit comparison test with $\sum  \frac{1}{n^{5 / 6}}$ to conclude for divergence or by direct comparison we have
$$\frac{\sqrt{n}}{n \sqrt[3]{n} + 2} \ge \frac{\sqrt{n}}{n \sqrt[3]{n}+\sqrt n}  = \frac{1}{n^{5 / 6}+1} \ge \frac{1}{n^{5 / 6}+n^{5 / 6}}=\frac12\frac{1}{n^{5 / 6}}$$
</s>