Convergence of $\sum_{n=1}^{\infty}\frac{1}{n}\Big(\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}\Big)$

Question

$\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}}{n}$

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?

Answer 1

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.

Answer 2

$$ 0<\frac{\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}}{n}= \frac{\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}}{n}\cdot \frac{\sqrt[3]{(n+1)^{4}} +\sqrt[3]{(n+1)^{2}}\sqrt[3]{n^{2}}+ \sqrt[3]{n^{4}}} {\sqrt[3]{(n+1)^{4}} +\sqrt[3]{(n+1)^{2}}\sqrt[3]{n^{2}}+ \sqrt[3]{n^{4}}} \\ =\frac{(n+1)^2-n^2} { n(\sqrt[3]{(n+1)^{4}} +\sqrt[3]{(n+1)^{2}}\sqrt[3]{n^{2}}+\sqrt[3]{n^4}) }\le \frac{2n+1}{n\cdot 3n^{4/3}}\le\frac{3n}{3n^{7/3}}=\frac{3}{n^{4/3}}. $$ Hence, due to the Comparison Test, the series converge, as $\sum_{n=1}^\infty \frac{1}{n^a}$, converges whenever $a>1$.

Answer 3

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 $\displaystyle\lim_{h\to 0} \frac{e^h-1}{h}$.