does this series converge? $\sum_{n=1}^\infty{\left( \sqrt[3]{n+1} - \sqrt[3]{n-1} \right)^\alpha}$

show the the following series converge\diverge

$\sum_{n=1}^\infty{\left( \sqrt[3]{n+1} - \sqrt[3]{n-1} \right)^\alpha}$

all the test i tried failed (root test, ratio test,direct comparison)

please dont use integrals as this is out of the scope for me right now

Hint rationalize:

$$\left( \sqrt[3]{n+1} - \sqrt[3]{n-1} \right)^\alpha = \left( \frac{2}{(\sqrt[3]{n+1})^2+ \sqrt[3]{n-1}\sqrt[3]{n+1}+ (\sqrt[3]{n-1})^2} \right)^\alpha$$

Compare with $$\sum\frac{1}{n^{\frac{2\alpha}{3}}}$$

• could you explain the algebra you did in the first step? – guynaa Nov 30 '13 at 19:01
• @user1333057 It is rationalization for cubic roots. $$\sqrt[3]{a}-\sqrt[3]{b}=\left( \sqrt[3]{a}-\sqrt[3]{b} \right) \frac{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}\\=\frac{a-b}{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}$$ The top is just the formula $$x^3-y^3=(x-y)(x^2+xy+y^2)$$ with $x=\sqrt[3]{a}, y=\sqrt[3]{b}$ – N. S. Nov 30 '13 at 19:05
• wow, i would have never have guessed that, what made you try thas route? – guynaa Nov 30 '13 at 19:07
• @guynaa It is a standard tool, the problem is that in Calculus we teach square root rationalization, but the same ideas work for higher roots. Is just that the expression by which you rationalize is more complicated. – N. S. Nov 30 '13 at 19:12
• @GinKin True, don't forget about the equality $\alpha \leq 0, 0 < \alpha \leq \frac{3}{2}$. Also when one covers the $p$ series, often the cases $p \leq 0$ and $0 < p \leq 1$ are studied as a single case $p \leq 1$. – N. S. Dec 1 '13 at 16:57

Ratio test is inconclusive, but using Raabe's test we can see that the series converges when $\alpha>\frac{3}{2}$.