<s>[INST] Convergence of $\sum_{n=1}^{\infty}\frac{1}{n}\Big(\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}\Big)$ $\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}}{n}$
Converging or Diverging? I guess I have to lower the fraction so that the roots will get away and I will have $\frac{1} {n}$ that diverges. But I have no idea how to do that.
Any ideas?
 [/INST] $$
0<\frac{\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}}{n}=
\frac{\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}}{n}\cdot
\frac{\sqrt[3]{(n+1)^{4}} +\sqrt[3]{(n+1)^{2}}\sqrt[3]{n^{2}}+ \sqrt[3]{n^{4}}}
{\sqrt[3]{(n+1)^{4}} +\sqrt[3]{(n+1)^{2}}\sqrt[3]{n^{2}}+ \sqrt[3]{n^{4}}} \\
=\frac{(n+1)^2-n^2}
{
n(\sqrt[3]{(n+1)^{4}} +\sqrt[3]{(n+1)^{2}}\sqrt[3]{n^{2}}+\sqrt[3]{n^4})
}\le \frac{2n+1}{n\cdot 3n^{4/3}}\le\frac{3n}{3n^{7/3}}=\frac{3}{n^{4/3}}.
$$
Hence, due to the Comparison Test,
the series converge, as $\sum_{n=1}^\infty \frac{1}{n^a}$, converges whenever $a>1$. 
</s>