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.
$$\sum_{n=0}^\infty \frac{\sqrt[3]{n}}{\sqrt[3]{n^4}-1}$$
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1Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Martin Sleziak Jan 04 '17 at 14:13
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@MartinSleziak Ok! I'll take note, thank you – JohnAdams Jan 04 '17 at 14:15
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$$\frac{\sqrt[3]{n}}{\sqrt[3]{n^4}-1}> \frac{\sqrt[3]n}{\sqrt[3]{n^4}}=\frac{1}{n^{4/3-1/3}}=\frac{1}{n}$$
Jimmy R.
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Hint: for large enough $n$, $$ \frac{\sqrt[3]{n}}{\sqrt[3]{n^4}-1}\ge\frac{1}{n} $$
