Convergence of some nasty series

by Sebastian Kupisiewicz   Last Updated February 13, 2018 13:20 PM

i've got some problem with this series: $$ \sum_{n=1}^{\infty} \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)} $$ I know that the series diverges by a comparison test, but I've got a lot of trouble to prove it. I've tried some algebraic tricks like: $$ a - b = \frac{a^{3} - b^{3}}{a^{2} + ab + b^{2}},$$ where $ a = \sqrt[3]{(n^{3}+n)} $ and $ b = \sqrt[3]{(n^3-n)} $.

It gave me $\sum_{n=1}^{\infty} \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)} = \sum_{n=1}^{\infty} \frac{2n}{\sqrt[3]{(n^{3}+n)^{2}} + \sqrt[3]{n^{6} - n^{2}} + \sqrt[3]{(n^3-n)^{2}}}$ but for me it looks like a blind valley. I would appreciate every help, thank you/

Answers 3

Note that since by binomial expansion for $x\to 0$

  • $(1+x)^a=1+ax +o(x)$

we have that




$$\sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)}=\frac2{3n}+o(n^{-1})$$

therefore the given series diverges by limit comparison test with $\sum \frac1n$.

February 13, 2018 13:11 PM

Using asymptotic calculus:

$$\sqrt[3]{n^3+n}-\sqrt[3]{n^3-n}=n\left(\sqrt[3]{1+\frac{1}{n^2}}-\sqrt[3]{1-\frac{1}{n^2}}\right)=n\left(\left(1+\frac{1}{n^2}\right)^{1/3}-\left(1-\frac{1}{n^2}\right)^{1/3}\right)=n\left(\left(1+\frac{1}{3n^2}+o\left(\frac{1}{n^2}\right)\right)-\left(1-\frac{1}{3n^2}+o\left(\frac{1}{n^2}\right)\right)\right)=n\left(\frac{2}{3n^2}+o\left(\frac{1}{n^2}\right)\right)=\frac{2}{3n}+o\left(\frac{1}{n}\right)\sim\frac{2}{3n} (n\to\infty)$$

thus $\sum_{n=1}^{\infty}\sqrt[3]{n^3+n}-\sqrt[3]{n^3-n}$ diverges because $\sum_{n=1}^{\infty}\frac{2}{3n}$ diverges.

February 13, 2018 13:14 PM

Alternative approach: $\sqrt[3]{x}$ is a concave function on $\mathbb{R}^+$, hence for any $n\geq 1$ $$ \sqrt[3]{n^3+n}-\sqrt[3]{n^3-n}=n\left(\sqrt[3]{1+\frac{1}{n^2}}-\sqrt[3]{1-\frac{1}{n^2}}\right)\geq \frac{2}{3n} $$ and the conclusion is straightforward.

Jack D'Aurizio
Jack D'Aurizio
February 13, 2018 13:19 PM

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