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?

Recall that $(1+\varepsilon)^p\approx 1+p\varepsilon$; thus
$$ \frac{\sqrt[3]{(n+1)^2} - \sqrt[3]{n^2}}{n}
= \frac{n^{2/3} \sqrt[3]{(1+\frac1n)^2} - \sqrt[3]{n^2}}{n}
\approx \frac{n^{2/3}\left(1+\frac2{3n}\right) - n^{2/3}}{n}
= \frac{2}{3n^{4/3}}
$$
So it should converge.
To justify the conclusion more formally, we can replace $(1+\varepsilon)^p\approx 1+p\varepsilon$ with the more exact statement
$$ 1+p\varepsilon\le (1+\varepsilon)^p\le e^{p\varepsilon} $$
(The left-hand inequality is Bernoulli's inequality; the right-hand one is $1+x\le e^x$.)  The informal work above showed that the series converges, so we'll take the upper bound; using it in place of $\approx$ above gives
$$ \frac{\sqrt[3]{(n+1)^2} - \sqrt[3]{n^2}}{n}
\le n^{2/3} \cdot \frac{e^{2/(3n)} - 1}{n}
= \frac2{3n^{4/3}} \cdot \frac{e^{2/(3n)} - 1}{\bigl(\frac2{3n}\bigr)}
$$
and now apply the standard limit $\displaystyle\lim_{h\to 0} \frac{e^h-1}{h}$.
