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?
 A: $$
\frac{\sqrt n}{n\sqrt[3]n+2}
\ge \frac{\sqrt n}{n\sqrt[3]n+n\sqrt[3]n}
=\frac12\frac{\sqrt n}{n\sqrt[3]n}
=\frac12n^{-\frac56}
$$
hence your series diverges.
A: Intuitively, since the sequence is very similar to $n^{-5/6}$ it should diverge (as $\sum n^{-1}$ diverges)
For sufficiently large $n$ (say $n > N$), we have $n\sqrt[3]n +2 < n \sqrt n$.
Then $$\sum_{n\ge1}\frac {\sqrt n}{n\sqrt[3]n+2} \ge\sum_{n> N}\frac {\sqrt n}{n\sqrt[3]n+2}\ge\sum_{n> N}\frac {\sqrt n}{n\sqrt n} = \sum_{n> N}\frac {1}{n} \to \infty$$
and hence the sum diverges by comparison test.
A: It is well-known that:
$$\sum_{n \geq 1} \frac{1}{n^a}$$
diverges when $a \leq 1.$
In your case, since:
$$\sum_{n \ge 1} \frac{\sqrt{n}}{n \sqrt[3]{n} + 2} \sim \sum_{n \ge 1} \frac{1}{n^{\frac{5}{6}}},$$
then your series diverges.

The symbol $\sim$ stands for:
$$\lim_{n \to +\infty} \left(\frac{\sqrt{n}}{n \sqrt[3]{n} + 2} - \frac{1}{n^\frac{5}{6}}\right) = 0$$
A: 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
$$\frac{\sqrt{n}}{n \sqrt[3]{n} + 2} \ge \frac{\sqrt{n}}{n \sqrt[3]{n}+\sqrt n}  = \frac{1}{n^{5 / 6}+1} \ge \frac{1}{n^{5 / 6}+n^{5 / 6}}=\frac12\frac{1}{n^{5 / 6}}$$
