Question

Show that it is possible to have a sequence (x_sub(n)) diverge while the sequence (|x_sub(n)|) converges

Answer 1

Question 1. Show that it is possible to have a sequence $(x_n)$ diverge while the sequence $(|x_n|)$ converges.

Solution. Consider $x_n = (-1)^n$. Then $(x_n)$ diverges, because there are two subsequences $x_{2n} = 1$ and $x_{2n-1} = -1$, which have different limits (1 and -1, respectively). But $|x_n| = 1$ for all $n$, so $(|x_n|)$ obviously converges to 1.