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?
 A: $$
0<\frac{\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}}{n}=
\frac{\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}}{n}\cdot
\frac{\sqrt[3]{(n+1)^{4}} +\sqrt[3]{(n+1)^{2}}\sqrt[3]{n^{2}}+ \sqrt[3]{n^{4}}}
{\sqrt[3]{(n+1)^{4}} +\sqrt[3]{(n+1)^{2}}\sqrt[3]{n^{2}}+ \sqrt[3]{n^{4}}} \\
=\frac{(n+1)^2-n^2}
{
n(\sqrt[3]{(n+1)^{4}} +\sqrt[3]{(n+1)^{2}}\sqrt[3]{n^{2}}+\sqrt[3]{n^4})
}\le \frac{2n+1}{n\cdot 3n^{4/3}}\le\frac{3n}{3n^{7/3}}=\frac{3}{n^{4/3}}.
$$
Hence, due to the Comparison Test,
the series converge, as $\sum_{n=1}^\infty \frac{1}{n^a}$, converges whenever $a>1$. 
A: HINT: try to show that
$$\frac{\sqrt[3]{(n+1)^2} -  \sqrt[3]{n^2}}{n} = O \left( \frac{1}{n^{4/3}}\right)$$
so the series is convergent.
A: 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 $\displaystyle\lim_{h\to 0} \frac{e^h-1}{h}$.
