Vrftsjtryk

Define the following sequence:
$$x_n = sqrt[3]{n+1}cos{sqrt{n+1}} - sqrt[3]{n}cos{sqrt{n}},forall nin mathbb{N} $$

Does the limit
$$lim_{n to infty} x_n$$
exist?

I believe that it does not, mainly because (i think that) $cossqrt{n+1}$ and $cos{sqrt{n}}$ have opposite signs an infinite number of times.

More exactly, I tried showing that there exists an infinite sequence $a_n$ of natural numbers such that
$$ sqrt{a_n} text{ mod } 2pi in left(frac{3pi}{2} + frac{pi}{4} ,2piright) text{ and } sqrt{a_n + 1} text{ mod } 2pi in left( 0, frac{pi}{4} right)$$

and a sequence $b_n$ such that
$$sqrt{b_n} text{ mod } 2pi in left(frac{pi}{2} + frac{pi}{4}, pi right) text{ and } sqrt{b_n + 1} text{ mod }in left(pi, pi + frac{pi}{4} right),$$

because then the subsequences $x_{a_n}$ and $x_{b_n}$ would both be bounded and of different signs, showing that $x_n$ is divergent.

Would this approach work? Is the sequence actually somehow convergent?

By binomial series

$$sqrt[3]{n+1}=sqrt[3]{n}left(1+frac1nright)^frac13=sqrt[3]{n}+Oleft(frac1{n^frac23}right)$$

then

$$sqrt[3]{n+1}cos{sqrt{n+1}} - sqrt[3]{n}cos{sqrt{n}}=$$$$=sqrt[3]{n}(cos{sqrt{n+1}}-cos{sqrt{n}})+Oleft(frac{cos{sqrt{n+1}}}{n^frac23}right)$$

and by sum to product formula

$$cos{sqrt{n+1}}-cos{sqrt{n}}=-2sinleft(frac{sqrt{n+1}+sqrt{n}}{2}right)sinleft(frac{sqrt{n+1}-sqrt{n}}{2}right)=$$

$$=-2sinleft(frac{sqrt{n+1}+sqrt{n}}{2}right)sinleft(frac{1}{2(sqrt{n+1}+sqrt{n})}right)sim -frac{sin(sqrt n)}{sqrt n}$$

therefore the given sequence converges to zero.