How can I find the limit of this sequence $(n^2 + 1)^{1/3} - (n^2 + n)^{1/3}$?

Question

How can I find the limit of this sequence $(n^2 + 1)^{1/3} - (n^2 + n)^{1/3}$ ? I know that if it is a square root I will multiply by the conjugate. but if it is a cubic root what shall I do?

Thanks

Answer 1

Use the fact that $$ a^3-b^3 = (a-b)(a^2+ab+b^2) \quad\Leftrightarrow\quad a-b = \frac{a^3-b^3}{a^2+ab+b^2}. $$

Then \begin{align} (n^2+1)^{1/3}-(n^2+n)^{1/3} &= \frac{(n^2+1)-(n^2+n)}{(n^2+1)^{2/3}+(n^2+1)^{1/3}(n^2+n)^{1/3}+(n^2+n)^{2/3}}\\[0.2cm] &= \frac{1-n}{(n^2+1)^{2/3}+(n^2+1)^{1/3}(n^2+n)^{1/3}+(n^2+n)^{2/3}}\\[0.2cm] &= \frac{\frac{1}{n}-1}{\left(n^{1/2}+n^{-3/2}\right)^{2/3} + \left(n+1+\frac{1}{n}+\frac{1}{n^2}\right)^{1/3} + \left(n^{1/2}+n^{-1/2}\right)^{2/3}}\\[0.2cm] &\to \frac{0-1}{(\infty+0)^{2/3}+(\infty+1+0+0)^{1/3}+(\infty+0)^{2/3}} = 0. \end{align}

Wolfram confirms.

Now that I think about it, it would have probably been much simpler to use asymptotics.

Answer 2

As an alternative by binomial series we have

• $(n^2 + 1)^{1/3}=n^{2/3}\left(1+\frac1{n^2}\right)^{1/3}=n^{2/3}\left(1+\frac1{3n^2}+o\left(\frac1{n^2}\right)\right)=n^{2/3}+o\left(\frac1/{n^{2/3}}\right)$

• $(n^2 +n)^{1/3}=n^{2/3}\left(1+\frac1n\right)^{1/3}=n^{2/3}\left(1+\frac1{3n}+o\left(\frac1n\right)\right)=n^{2/3}+\frac1{3n^{1/3}}+o\left(\frac1{n^{1/3}}\right)$

therefore

$$(n^2 + 1)^{1/3} - (n^2 + n)^{1/3}=\frac{-1}{3n^{1/3}}+o\left(\frac{1}{n^{1/3}}\right)\to 0$$

Answer 3

Factor out $n^{2/3}$ and do a Taylor expansion:

$$n^{2/3}\left(\left(1+\frac 1{n^2}\right)^{1/3} - \left(1+\frac 1{n}\right)^{1/3} \right) = n^{2/3}\left(1+\frac 1{3n^2} - 1 - \frac 1{3n} + o(\frac 1n)\right) \sim -\frac1{3n^{1/3}} \to 0 $$

Answer 4

Formally substitute $n=1/x^{3/2}$ to get $$ f(x)=\sqrt[3]{\frac{1}{x^3}+1}-\sqrt[3]{\frac{1}{x^3}+\frac{1}{x^{3/2}}}= \frac{\sqrt[3]{1+x^3}-\sqrt[3]{1+x^{3/2}}}{x}= \frac{1-1-x^{3/2}/3+o(x^{3/2})}{x}=-\frac{x^{1/2}}{3}+o(x^{1/2}) $$ Then $\lim_{x\to0^+}f(x)=0$. Since $$ \sqrt[3]{n^2+1}-\sqrt[3]{n^2+n}=f\left(\frac{1}{n^{2/3}}\right) $$ you have $$ \lim_{n\to\infty}f\left(\frac{1}{n^{2/3}}\right)=0 $$