![\displaystyle\sum_{k\ge1}\frac{\sqrt[3]{k^2+1}}{\sqrt{k^3+2}}\sim\sum_{k\ge1}\frac{k^{2/3}}{k^{3/2}}=\sum_{k\ge1}\frac1{k^{5/6}}](/tpl/images/0468/9395/3e05d.png)
diverges for 
. 5/6 falls in this interval, so this series must diverge, and so must the original one by comparison.
Determine if convergent or divergent using comparison tests
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Mathematics, 24.01.2020 14:31, annekesimonsen
Determine if convergent or divergent using comparison tests
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