I've tried the Ratio and Root test, but both are inconclusive, and I can't find a comparison to prove that the following serie is converge or diverge.
$$\sum_{n=0}^\infty \frac{\sqrt[3]{n}}{\sqrt[3]{n^4}-1}$$
How to prove the divergence/convergence of the following series: $\sum_{n=0}^\infty \frac{\sqrt[3]{n}}{\sqrt[3]{n^4}-1}$?
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$\begingroup$
calculus
sequences-and-series
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1Please, try to make the titles of your questions more informative. E.g., *Why does $a
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0@MartinSleziak Ok! I'll take note, thank you – 2017-01-04
2 Answers
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Hint: for large enough $n$, $$ \frac{\sqrt[3]{n}}{\sqrt[3]{n^4}-1}\ge\frac{1}{n} $$
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$$\frac{\sqrt[3]{n}}{\sqrt[3]{n^4}-1}> \frac{\sqrt[3]n}{\sqrt[3]{n^4}}=\frac{1}{n^{4/3-1/3}}=\frac{1}{n}$$