Limit of $\sqrt[3]{(n+1)(n+2)(n+3)}-n$

Question

I came up with a Solution for $\displaystyle\lim_{n\to\infty}\sqrt[3]{(n+1)(n+2)(n+3)}-n$

That's my Solution: $\sqrt[3]{(n+1)(n+2)(n+3)}-n$
\begin{aligned} &=\frac{\left(\sqrt[3]{n^3+6 n^2+11 n+6}-n\right)\left(n^2+n \sqrt[3]{n^3+6 n^2+11 n+6}+\left(n^3+6 n^2+11 n+6\right)^{2 / 3}\right)}{\left(n^2+n \sqrt[3]{n^3+6 n^2+11 n+6}+\left(n^3+6 n^2+11 n+6\right)^{2 / 3}\right)}\\ &=\frac{6 n^2+11 n+6}{n^2+n \sqrt[3]{n^3+6 n^2+11 n+6}+\left(n^3+6 n^2+11 n+6\right)^{2 / 3}}\\ &\lim_{n\to \infty}\frac{n^2(6+\frac{11}{n}+\frac{6}{n^2})}{n^2\left(1+\frac{n \sqrt[3]{n^3+6 n^2+11 n+6}+\left(n^3+6 n^2+11 n+6\right)^{2 / 3}}{n^2}\right)}\\ &=\lim_{n\to \infty}\frac{n^2(6+\frac{11}{n}+\frac{6}{n^2})}{n^2}\lim_{n\to \infty}\frac{1}{1+\frac{n \sqrt[3]{n^3+6 n^2+11 n+6}+\left(n^3+6 n^2+11 n+6\right)^{2 / 3}}{n^2}}\\ &=6\frac{1}{\displaystyle{\lim_{n\to \infty}}1+\frac{n \sqrt[3]{n^3+6 n^2+11 n+6}+\left(n^3+6 n^2+11 n+6\right)^{2 / 3}}{n^2}}\\ &=6\frac{1}{1+\displaystyle{\lim_{n\to \infty}}\frac{n \sqrt[3]{n^3+6 n^2+11 n+6}+\left(n^3+6 n^2+11 n+6\right)^{2 / 3}}{n^2}}\\ &=\frac{6}{1+\displaystyle{\lim_{n\to \infty}}\frac{\sqrt[3]{n^3+6 n^2+11 n+6}}{n}+\frac{\left(n^3+6 n^2+11 n+6\right)^{2 / 3}}{n^2}}\\ &=\frac{6}{1+\left(\displaystyle{\lim_{n\to \infty}}\frac{\sqrt[3]{n^3+6 n^2+11 n+6}}{n}\right)+\left(\displaystyle{\lim_{n\to \infty}}\frac{\left(n^3+6 n^2+11 n+6\right)^{2 / 3}}{n^2}\right)}\\ &=\frac{6}{1+\left(\displaystyle{\lim_{n\to \infty}}\sqrt[3]{\frac{n^3+6 n^2+11 n+6}{n^3}}\right)+\left(\displaystyle{\lim_{n\to \infty}}\frac{\left(n^3+6 n^2+11 n+6\right)^{2 / 3}}{n^2}\right)}\\ &=\frac{6}{1+\left(\displaystyle{\lim_{n\to \infty}}\sqrt[3]{\frac{1+\frac{6}{n}+\frac{11}{n^2}+\frac{6}{n^3}}{1}}\right)+\left(\displaystyle{\lim_{n\to \infty}}\frac{\left(n^3+6 n^2+11 n+6\right)^{2 / 3}}{n^2}\right)}\\ &=\frac{6}{1+1+\left(\displaystyle{\lim_{n\to \infty}}\frac{\left(n^3+6 n^2+11 n+6\right)^{2 / 3}}{n^2}\right)}\\ &=\frac{6}{1+1+\left(\displaystyle{\lim_{n\to \infty}}\left(\frac{n^3+6 n^2+11 n+6}{n^3}\right)^{2 / 3}\right)}\\ &=\frac{6}{1+1+\left(\displaystyle{\lim_{n\to \infty}}\frac{1+\frac{6}{n}+\frac{11}{n^2}+\frac{6}{n^3}}{1}\right)^{2 / 3}}\\ &=\frac{6}{1+1+1}=2\\ \end{aligned}

The question I wanna ask is there a more elegant way to solve this problem?

Answer 1

It is best to avoid writing complicated expressions. Let us write $$A_n=\sqrt[3]{(n+1)(n+2)(n+3)}$$ so that $A_n/n\to 1$ and $$A_n^3=n^3+6n^2+11n+6$$ Next we have $$A_n-n=n((A_n/n) - 1)=n\cdot\frac{(A_n/n)-1}{(A_n/n)^3-1}\cdot\frac{A_n^3-n^3}{n^3}$$ The middle fraction tends to $1/3$ (by virtue of standard limit formula $$\lim_{x\to a} \frac{x^m-a^m} {x-a} =ma^{m-1}$$ with $a=1$ and $A_n/n$ playing the role of $x$ and $m=3$) and hence desired limit is equal to the limit of $$\frac{A_n^3-n^3}{3n^2}$$ which is $2$.

Answer 2

You can use Taylor expansion to solve this in a faster way; as $x \to 0$ you have
$(1+x)^{\alpha} \sim 1+\alpha x$ $\quad(*)$
Now: $$(n+1)(n+2)(n+3)=n^3+6n^2+11n+6=n^3\left(1+\dfrac{6}{n}+\dfrac{11}{n^2}+\dfrac{6}{n^3}\right)$$ Moreover: $$\sqrt[3]{(n+1)(n+2)(n+3)}=n\left(1+\dfrac{6}{n}+\dfrac{11}{n^2}+\dfrac{6}{n^3}\right)^{1/3}$$ Applying $(*)$ to $\left(1+\dfrac{6}{n}+\dfrac{11}{n^2}+\dfrac{6}{n^3}\right)^{1/3}$ we have: $$\left(1+\dfrac{6}{n}+\dfrac{11}{n^2}+\dfrac{6}{n^3}\right)^{1/3}\sim 1+\dfrac{1}{3}\left(\dfrac{6}{n}+\dfrac{11}{n^2}+\dfrac{6}{n^3} \right) $$ Finally: $$ \sqrt[3]{(n+1)(n+2)(n+3)}-n = n\left(1+\dfrac{6}{n}+\dfrac{11}{n^2}+\dfrac{6}{n^3}\right)^{1/3}-n \sim 2+\dfrac{11}{3n}+\dfrac{2}{n^2} \to 2$$

Answer 3

I would start the same as you did and obtain that we want to find the limit of $$\frac{6 n^2+11 n+6}{n^2+n \sqrt[3]{n^3+6 n^2+11 n+6}+\left(n^3+6 n^2+11 n+6\right)^{2 / 3}}$$ as $n$ goes to infinity. But then I would notice that this expression is $\frac{6n^2 + o(n^2)}{3n^2 + o(n^2)}$ using the little-o notation. Now, since $\lim_{n \rightarrow \infty} o(n^2)/n^2 = 0$, we get that $$\lim_{n \rightarrow \infty} \frac{6n^2 + o(n^2)}{3n^2 + o(n^2)} = 2.$$

Answer 4

Let's substitute $n = t^{3}$ for simplification . The limit is then :

$\displaystyle\lim_{t\to0}t^3 [[{(1 + t^{-3})(1 + 2t^{-3})(1 + 3t^{-3})]^{1/3}}-1]$

Multiply numerator and denominator by $[{(1 + t^{-3})(1 + 2t^{-3})(1 + 3t^{-3})]^{1/3}}+1]$

Substitute $u = t^{-3}$ and limit becomes :

$(1/2)$$\displaystyle\lim_{u\to0}\frac{(6u^3 + 11u^2 + 6u + 1)^{2/3}-1}{u}$

The limit at right is equal to $4$ and is easy to calculate.(apply L hospital rule once!)

Hence, $\displaystyle\lim_{n\to\infty}\sqrt[3]{(n+1)(n+2)(n+3)}-n = 2$

Answer 5

$$\lim_{n\rightarrow \infty} \sqrt[3]{(n+1)(n+2)(n+3)}-n$$ Factor $n$ $$\lim_{n\rightarrow \infty} n\left(\sqrt[3]{\frac{(n+1)(n+2)(n+3)}{n^3}}-1\right)$$

$$\lim_{n\rightarrow \infty} n\left(\sqrt[3]{\left(1+\frac1n\right)\left(1+\frac2n\right)\left(1+\frac3n\right)}-1\right)$$

Set $x = \frac{1}{n}$

$$\lim_{x\rightarrow 0} \frac{\sqrt[3]{(1+x)(1+2x)(1+3x)}-1}{x}=g'(0)$$ Where $g(x) = \sqrt[3]{(1+x)(1+2x)(1+3x)}$

Answer 6

To simplify the algebra a little, let $m = n + 2.$ Then we want to find $$ \lim_{m\to\infty} \left[\sqrt[3]{m(m - 1)(m + 1)} - (m - 2)\right] = 2 - \lim_{m\to\infty} \left[m - \sqrt[3]{m^3 - m}\right], $$ if the latter limit exists.

To see that the limit does exist, and that it is $0,$ let $f(m) = \sqrt[3]{m^3 - m}.$ Then $$ m^2(m - f(m)) < (m - f(m))(m^2 +mf(m) + f(m)^2) = m^3 - f(m)^3 = m, $$ therefore $0 < m - f(m) < 1/m,$ therefore $$ 2 - \lim_{m\to\infty} \left[m - \sqrt[3]{m^3 - m}\right] = 2 - \lim_{m\to\infty} (m - f(m)) = 2 - 0 = 2. $$

Answer 7

Admittedly, I'm late to the party, but my prezzie is conceptionally different.

Let's apply the Harmonic-Geometric-Arithmetic mean inequality to the given expression to obtain $$\sqrt[3]{(n+1)(n+2)(n+3)}-n \:<\: \frac{(n+1)+(n+2)+(n+3)}3-n\:=\: 2\tag{1}$$ and $$2\cdot\frac{3n^2+11n+9}{3n^2+12n+11} \:=\:\frac3{\frac1{n+1}+\frac1{n+2}+\frac1{n+3}}-n \:<\: \sqrt[3]{(n+1)(n+2)(n+3)}-n\,.$$

As the variables $\,n+1, n+2,$ and $\,n+3\,$ are not equal, the stronger "$<$" holds instead of "$\leqslant$"$\,$.
And as an extra, the inequality chain shows that the limit which is $2$ is approached from below.

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Note: Inequality $(1)$ also follows, if not even quicker, from $\,(n+1)(n+2)(n+3)<(n+2)^3.$