Limit of Sum to Infinity

Question

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.

Answer 1

$$\sqrt[3]{n^3+1}-n=\\\sqrt[3]{n^3+1}-n \cdot \frac{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)^1}+n^2}{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)^1}+n^2}\\=\\\frac{n^3+1-n^3}{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)^1}+n^2}\\=\frac{1}{\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)^1}+n^2}\\<\frac{1}{\sqrt[3]{(n^3)^2}+n\sqrt[3]{(n^3)^1}+n^2}\\\\\frac{1}{3n^2} $$ 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.

Answer 2

Yes, the idea is to take advantage of the identity $(a-b)(a^2+ab+b^2)=a^3-b^3$.

In your case $a=\sqrt[3]{n^3+1}$ and $b=n$. So, we must multiply and divide by $a^2+ab+b^2$.

Answer 3

Another way is to realize that $$n^3+1< (n+n^{-2})^3=n^3+3+3n^{-3}+n^{-6}$$ Therefore $$\sum_{n=1}^\infty\left(\sqrt[3]{n^3+1}-n\right)<\sum_{n=1}^\infty n^{-2}<\infty$$