On the convergence of the series $ \sum_n \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)} $

Question

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.

Answer 1

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 $\sum \frac1n$.

Answer 2

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.

Answer 3

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.

Answer 4

Let $$a_n=\sqrt[3]{n^3+n}-\sqrt[3]{n^3-n}=\sqrt[3]n\left(\sqrt[3]{n^2+1}-\sqrt[3]{n^2-1}\right)$$

Defining $f(x)=\sqrt[3]x$ we can apply Mean Value Theorem in $[n^2-1,n^2+1]$.

$$f'(x)=\frac1{3\sqrt[3]{x^2}}$$

For some $\xi\in[n^2-1,n^2+1]$, we have

$$a_n=\frac23\sqrt[3]{\frac n{\xi^2}}\ge\frac23\sqrt[3]{\frac n{(n^2+1)^2}}\ge\frac23\sqrt[3]{\frac n{4n^4}}\ge\frac1{3n}$$

Answer 5

Your idea was good: $$\sqrt[3]{(n^3 + n)^2} = n^2\sqrt[3]{(1 + 1/n^2)^2},$$ $$\sqrt[3]{n^6 - n^2} = n^2\sqrt[3]{1 - 1/n^4},$$ $$\sqrt[3]{(n^3 - n)^2} = n^2\sqrt[3]{(1 - 1/n^2)^2}.$$ So $$ \sqrt[3]{(n^3 + n)^2} + \sqrt[3]{n^6 - n^2} + \sqrt[3]{(n^3 - n)^2}\sim 3n^2 $$ and $$ \frac{2n}{\sqrt[3]{(n^3 + n)^2} + \sqrt[3]{n^6 - n^2} + \sqrt[3]{(n^3 - n)^2}}\sim\frac{2}{3n}. $$