<s>[INST] Limit of Sum to Infinity Is $ \Sigma_{n=0}^\infty (\sqrt[3]{n^3+1} - n)$ convergent or divergent?
For expressions of the form $\sqrt{n^2+1} - n$, I believe the common trick is to multiply by the "conjugate" $\frac{\sqrt{n^2+1} + n}{\sqrt{n^2+1} + n}$. 
Is there a similar trick for other roots (i.e. not square roots) as well?
Thanks.
 [/INST] $$\sqrt[3]{n^3+1}-n=\\\sqrt[3]{n^3+1}-n \cdot \frac{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)^1}+n^2}{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)^1}+n^2}\\=\\\frac{n^3+1-n^3}{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)^1}+n^2}\\=\frac{1}{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)^1}+n^2}\\<\frac{1}{\sqrt[3]{(n^3)^2}+n\sqrt[3]{(n^3)^1}+n^2}\\\\\frac{1}{3n^2} $$
Also 
$$ \sum_{n=1}^{\infty}\frac{1}{3n^2}=\\\frac{1}{3}\sum_{n=1}^{\infty}\frac{1}{n^2}=\\\frac{1}{3}\frac{\pi^2}{6}$$
Thus the original series is convergent.
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