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

$$\sqrt[3]{n^3+n}=n\left(1+\frac1{n^2}\right)^\frac13=n\left(1+\frac1{3n^2}+o(n^{-2})\right)$$

$$\sqrt[3]{n^3-n}=n\left(1-\frac1{n^2}\right)^\frac13=n\left(1-\frac1{3n^2}+o(n^{-2})\right)$$

thus

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

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

gimusi
gimusi
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.

user8734617
user8734617
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

Related Questions



Limit approach to finding $1+2+3+4+\ldots$

Updated June 18, 2015 23:08 PM

Power Series involving Double Factorials

Updated June 02, 2017 02:20 AM