Whether the sequence $\{x_n-y_n\}$ converge or not.

Solution 1:

Suppose that $\lim_n(x_n-y_n)\ne 0$; then there are subsequences $\langle x_{n_k}:k\in\Bbb N\rangle$ and $\langle y_{n_k}:k\in\Bbb N\rangle$ and an $\epsilon>0$ such that $|x_{n_k}-y_{n_k}|\ge\epsilon$ for all $k\in\Bbb N$. Infinitely many of the differences $x_{n_k}-y_{n_k}$ must have the same sign, so without loss of generality we may assume that $x_{n_k}-y_{n_k}\ge\epsilon$ for all $k\in\Bbb N$. Then $x_{n_k}\ge y_{n_k}+\epsilon$ for all $k\in\Bbb N$, and hence

$$\begin{align*} x_{n_k}^3-y_{n_k}^3&\ge(y_{n_k}+\epsilon)^3-y_{n_k}^3\\ &=\epsilon\left(3y_{n_k}^2+3\epsilon y_{n_k}+\epsilon^2\right)\\ &=\epsilon\left(3\left(y_{n_k}+\frac{\epsilon}2\right)^2+\frac{\epsilon^2}4\right)\\ &\ge\frac{\epsilon^3}4 \end{align*}$$

for all $k\in\Bbb N$, contradicting the hypothesis that $\lim_n(x_n^3-y_n^3)=0$.

Solution 2:

Yes, $x_n^3 -y_n^3 \to 0 $ implies $ x_n - y_n \to 0$ for real-valued sequences. This follows from the following estimate:

For all $x, y \in \Bbb R$ is $|x-y| \le \sqrt[3]{4|x^3 - y^3|}$.

(In other words: the function $x \mapsto x^{1/3}$ is “Hölder continuous” with exponent $1/3$.)

Due to the symmetry, it suffices to prove the inequality for the case $x \ge y$: $$ \begin{align} 4(x^3-y^3) &= 4(x-y)(x^2+xy+y^2) \\ &= (x-y)\left( (x-y)^2 + 3(x+y)^2\right) \\ &\ge (x-y)^3 \, . \end{align} $$

An alternative proof (which generalizes to other odd exponents) would be to compute the maximum of the function $$ f(u) = \frac{(u-1)^3}{u^3-1} = \frac{(u-1)^2}{u^2+u+1} $$ which turns out to be $f(-1) = 4$.