<s>[INST] How to prove $\sum_{n=1}^{\infty} \frac{3}{\sqrt[3]{n^2+2}}$ diverges? $$\sum_{n=1}^{\infty} \frac{3}{\sqrt[3]{n^2+2}}$$
It seems clear to me that this series diverges because the dominant temr is $1/n^{2/3}$, a p-series with $p < 1$
However I need to prove divergence using something like the comparison test, integral test, or similar.
I can't work out a suitable comparison to make to prove divergence. Suggestions?
 [/INST] Note that for all positive integers $n$,
$$n^2 + 2 < n^2 + 4n + 4 = (n+2)^2.$$  Therefore, $$\frac{3}{\sqrt[3]{n^2 + 2}} > \frac{3}{\sqrt[3]{(n+2)^2}} > \frac{1}{(n+2)^{2/3}},$$ hence $$\sum_{n=1}^\infty \frac{3}{\sqrt[3]{n^2+2}} > \sum_{n=1}^\infty \frac{1}{(n+2)^{2/3}} = - 1^{2/3} - \frac{1}{2^{2/}} + \sum_{n=1}^\infty \frac{1}{n^{2/3}},$$ the last sum of which is a divergent $p$-series, hence the given sum also diverges.
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