Convergence of a seq

Question

I would like to know if I can say (and how to justify it) that :

Let $(u_n)_{n\in\mathbb{N}}$, and let $(v_n)_{n\in\mathbb{N}}$ be a sequence defined by $v_n = | u_n - k|$ with $k$ a real. If $v_n$ is a strictly decreasing and convergent sequence, then $u_n$ also converges.

I know that this statement is false if $v_n$ is not strictly decreasing (example of $(-1)^n$ it seems), but since I can't find any counterexamples to my statement, I was wondering if it is right.

If it is, could going back to the definition of convergence be a good idea?

Answer 1

The sequence $$u_n=(-1)^n$$ is not convergent, but if $k=0$ then $$v_n=1$$ is convergent. You already found this sequence. It is not decreasing, as you also stated in your post, but it can be easily modified that is is a decreasing sequence. Take $$u_n=(-1)^n\left(1+\frac 1 n\right)$$ Then this is still not converging but $$v_n=1+\frac 1 n$$ is converging and strictly decreasing.

Answer 2

The statement doesn't need to be true. Take u as a sequence, that oscillates, such that the absolute value of $u_k$ gets closer to 1, but u_k is exchangingly positive and negative, but with absolutely decreasing absolute value. Then for $k=0$ $v_k$ as defined converges and decreases strictly, but $u_k$ does not by construction.