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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

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2 Answers 2

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

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  • $\begingroup$ could you explain the algebra you did in the first step? $\endgroup$
    – guynaa
    Commented Nov 30, 2013 at 19:01
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    $\begingroup$ @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 $x=\sqrt[3]{a}, y=\sqrt[3]{b}$ $\endgroup$
    – N. S.
    Commented Nov 30, 2013 at 19:05
  • $\begingroup$ wow, i would have never have guessed that, what made you try thas route? $\endgroup$
    – guynaa
    Commented Nov 30, 2013 at 19:07
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    $\begingroup$ @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. $\endgroup$
    – N. S.
    Commented Nov 30, 2013 at 19:12
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    $\begingroup$ @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 $p \leq 1$. $\endgroup$
    – N. S.
    Commented Dec 1, 2013 at 16:57
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Ratio test is inconclusive, but using Raabe's test we can see that the series converges when $\alpha>\frac{3}{2}$.

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