Consider the sequence ${x_n},x_n = n^{1/3}$
a) Show that for any $\varepsilon \gt 0$ there is $n_0$such that for all $n\ge n_0 , |x_{n+1}-x_n|\lt\varepsilon $?
Sol: $|(n+1)^{1/3}-n^{1/3}|$=$\frac {n+1-n}{(n+1)^{2/3} +((n+1)(n))^{1/3}+n^{2/3}}$$\lt \frac{1}{n^{2/3}}\lt \varepsilon$
$n\gt \varepsilon^{3/2}$ Choose $n_0\ge \varepsilon^{3/2}$
I use this formal$ (a^3-b^3)=(a-b)(a^2+ab+b^2)$
Can proof this Cauchy seq. by this way
Thanks
Your proof for
"for any $ \varepsilon >0$ there is $n_0$ such that $|x_{n+1}-x_n|\lt\varepsilon$ for $n \ge n_0$"
is correct.
But $(x_n)$ is not Cauchy, since $(x_n)$ is unbounded.