$ \lim_{n \to \infty} \cos^2(\pi \sqrt[\leftroot{-2} \uproot{1} 3]{n^3+n^2+2n}), \space n \in N$

Question

$$\lim_{n \to \infty} \cos^2(\pi \sqrt[\leftroot{-2} \uproot{1} 3]{n^3+n^2+2n}), \space n \in N$$

I think that the expression under the root will resolve into a $(n+a)^3$ term, which means that the limit will eventually be of the form $\cos^2(n\pi +a\pi)$, which is solvable. However, I am unable to follow through with this strategy. Any hints/helps are appreciated!

Answer 1

First we have

$\forall n>0, \pi\sqrt[3]{n^3+n^2+2n}=\pi n\sqrt[3]{1+\frac{1}{n}+\frac{2}{n^2}}$

This shows that: $\pi\sqrt[3]{n^3+n^2+2n} \sim \pi n$ ($n\rightarrow +\infty$)

Hence the idea of adding and subtracting $\pi n$ in the Cosine:

$\cos^2\left(\pi n\sqrt[3]{1+\frac{1}{n}+\frac{2}{n^2}}\right)=\cos^2\left(\left(\pi n\sqrt[3]{1+\frac{1}{n}+\frac{2}{n^2}}-\pi n\right)+\pi n\right)$

$=\cos^2\left(\pi n \left(\sqrt[3]{1+\frac{1}{n}+\frac{2}{n^2}}-1\right)+\pi n\right)$

$=\cos^2\left(\pi n \left(\sqrt[3]{1+\frac{1}{n}+\frac{2}{n^2}}-1\right)\right)$

($(-1)^n$ appears but disappears thanks to the square)

However

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

The limit is equal to $\cos^2\left(\frac{\pi}{3}\right)=\frac14$

Note: To conclude we can use the limit

$\lim_{h\rightarrow 0} \frac{\sqrt[3]{1+h}-1}{h}=\frac{1}{3}$

(derivative of $x\mapsto \sqrt[3]{1+x}$ at 0)

Then

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

Answer 2

$$L=\lim_{n \rightarrow \infty} \cos^2(\pi \sqrt[3]{n^3+n^2+2n} = \cos^2 (n\pi [1+1/n+2/n^2]^{1/3}- 1])$$ $$\implies L= \lim_{n \rightarrow \infty}\cos^2 n\pi[1/(3n)+2/(3n^2)+...]=\lim_{n \rightarrow \infty} \cos^2 (\pi/3+O(1/n))]=\frac{1}{4}.$$ In above we have used $\cos^2 (x-n\pi)=\cos^2x$ and $(1+h)^{\nu}=1+\nu h+O(h^2)$ if $|h|<<1.$

Answer 3

Rewrite the cubic root using the binomial expansion at order $1$: $$n\biggl(1+\frac 1n+\frac 2{n^2}\biggr)^{\!\rlap{1/3}}= n\biggl(1+\frac1{3n} +o\Bigl(\frac1n\Bigr)\biggr)=n+\frac13+o(1)$$ and do some trigonometry: \begin{align} \cos^2(\pi \sqrt[\leftroot{2} \uproot{1} 3]{n^3+n^2+2n})&=\cos^2\Bigl(n\pi+\frac\pi3+o(1)\Bigr)=\frac{1+\cos\bigl(2\pi+\frac{2\pi}3+o(1)\bigr)}{2}=\frac{1+\cos\bigl(\frac{2\pi}3+o(1)\bigr)}{2}\\ &\to\frac{1+\cos\bigl(\frac{2\pi}3\bigr)}2=\frac14. \end{align}

Answer 4

It could be interesting to make the problem more general considering $$\cos^2(\pi \sqrt[3]{n^3+an^2+bn})=\frac 12\left(1+\cos(2\pi \sqrt[3]{n^3+an^2+bn})\right)$$

Now, using the binomial expansion or Taylor expansion $$\sqrt[3]{n^3+an^2+bn}=n+\frac{a}{3}-\frac{a^2-3 b}{9 n}+O\left(\frac{1}{n^2}\right)$$ making $$\cos\left(2\pi \sqrt[3]{n^3+an^2+bn}\right)=\cos\left(2n\pi+\frac{2\pi a}{3}-\frac{2\pi(a^2-3 b)}{9 n}+O\left(\frac{1}{n^2}\right)\right)$$ that is to say $$\cos\left(\frac{2\pi a}{3}-\frac{2\pi(a^2-3 b)}{9 n}+O\left(\frac{1}{n^2}\right)\right)=\cos \left(\frac{2 \pi a}{3}\right)+\frac{2 \pi \left(a^2-3 b\right) }{9 n}\sin \left(\frac{2 \pi a}{3}\right)+O\left(\frac{1}{n^2}\right)$$ So, for the whole expression $$\frac 12\left(1+\cos(2\pi \sqrt[3]{n^3+an^2+bn})\right)=\frac 12\left(1+\cos \left(\frac{2 \pi a}{3}\right)+\frac{2 \pi \left(a^2-3 b\right) }{9 n}\sin \left(\frac{2 \pi a}{3}\right)\right)+O\left(\frac{1}{n^2}\right)$$ which shows the limit and also how it is approached.

Edit

The above shows that the case where $b=\frac {a^2} 3$ is particular. If wa want to see how is the limit is approached, we need one ore term in the expansion to get $$\frac{1}{2} \left(1+\cos \left(\frac{2 \pi a}{3}\right)\right)+\frac{\pi a^3 }{81 n^2}\sin \left(\frac{2 \pi a}{3}\right)+O\left(\frac{1}{n^3}\right)$$