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Can someone please offer any tips on how to manipulate the series below, or what convergence test I should use to determine its convergence based on $\alpha$?

$$\sum_{n\ge1} \frac{\sqrt[3]{n+1}+\sqrt[3]{n}}{n^\alpha},\ \alpha \in \mathbb{R}$$

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up vote 2 down vote accepted

Hint: It is easy to see that the numerator is $\gt 2n^{1/3}$. And since $n+1\le 2n$, the numerator is $\le (1+2^{1/3})n^{1/3}$. These inequalities will be sufficient for comparison tests.

We really don't need such tight bounds. Any positive constants $a$ and $b$ such that $an^{1/3}\le (n+1)^{1/3}+n^{1/3}\le bn^{1/3}$ are good enough for the comparisons.

Alternately, you could divide top and bottom by $n^{1/3}$. On top you get $\sqrt[3]{1+\frac{1}{n}}+1$, and at the bottom you get $n^{\alpha-1/3}$.

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