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I'm asked to prove the convergence or divergence of the below series using the comparison test.

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

I've reduced the above series to $\sqrt[3]{n}(\sqrt[3]{(1-\frac{1}{n}})-1)=U_n$, but after taking $V_n$ as $\sqrt[3]{n}$ and applying the limit, $\lim_{n\to {\infty}}\frac{U_n}{V_n}$ yields me zero, but the value has to be greater than zero for the comparison test to work. So I'm in a dilemma, am I doing something wrong?

Feel free to correct me if I'm wrong in my calculations.

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    $\begingroup$ What's the meaning of the first and the second $=$ signs from your question? $\endgroup$ Sep 9 '18 at 16:18
  • $\begingroup$ I've assigned the series to $U_n$ $\endgroup$ Sep 9 '18 at 16:21
  • $\begingroup$ That doesn't answer my question. I want to know the meaning of the $=$ sign in $$\sum_{n=1}^{\infty}=\sqrt[3]{(n-1)}-\sqrt[3]{n},$$for instance. $\endgroup$ Sep 9 '18 at 16:22
  • $\begingroup$ Sorry, I'm in a hurry I didn't notice that. $\endgroup$ Sep 9 '18 at 16:28
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HINT

By binomial expansion prove that

$$\sqrt[3]{n}-\sqrt[3]{(n-1)}\ge \frac13 \frac1{n^{2/3}}$$

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HINT

Firstly, remember that $x^{3} - y^{3} = (x-y)(x^{2} + xy +y^{2})$. Hence we get: \begin{align*} \sqrt[3]{n} - \sqrt[3]{n-1} & = \sqrt[3]{n} - \sqrt[3]{n-1}\times\frac{\sqrt[3]{n^{2}} + \sqrt[3]{n(n-1)} + \sqrt[3]{(n-1)^{2}}}{\sqrt[3]{n^{2}} + \sqrt[3]{n(n-1)} + \sqrt[3]{(n-1)^{2}}}\\ & = \frac{1}{\sqrt[3]{n^{2}} + \sqrt[3]{n(n-1)} + \sqrt[3]{(n-1)^{2}}} \geq \frac{1}{3\sqrt{n^{3}}} \end{align*}

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  • $\begingroup$ Thanks, but that's not the question I posted. $\endgroup$ Sep 16 '18 at 13:36

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