Two sequences such that $\lim_{n \to \infty }| a_n-b_n |=1$ and only one of them is convergent

Question

I should to find $a_n$ which converges to L (real number) and any sequence $b_n$ (which not converges to real number) so that they holds:

$\lim_{n \to \infty }\left | a_n-b_n \right |=1$

Thanks for your help!

Updated.

Answer 1

Put $a_n=0$, $b_n=(-1)^n$.

(I noticed after posting this that something like this is alreay in the comments, but the question needs an answer, so I let it stand.)

Answer 2

It's not possible to do this. Proof: Suppose that $\lim_{n\rightarrow\infty}\vert a_n-b_n\vert=1$. Then consider $\vert b_n-L\vert$: adding and subtracting $a_n$, we can use the triangle inequality to see that:

$$\vert b_n-L\vert=\vert b_n-a_n+a_n-L\vert\leq\vert b_n-a_n\vert+\vert a_n-L\vert$$

as $n\rightarrow\infty$, the latter would converge to $0$ while the former would (by assumption) converge to 1. Thus $\vert b_n-L\vert\rightarrow C<1$ as $n\rightarrow \infty$, and so $b_n$ cannot diverge to infinity.