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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.

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    @ Hagen von Eitzen Thanks! You are right, my mistake, but why the choice of sequences $a_n=\frac{1}{n}$ and $b_n=(-1)^{n}$ is wrong?2012-11-14

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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.)

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    @Tina: No I can't, for it isn't wrong. (But mine is even simpler.)2012-11-14
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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.

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    This is answer to the question about $a_n\to L$ and $b_n\to\infty$. Then th OP [edited their question](http://math.stackexchange.com/posts/237033/revisions) to $a_n$ convergent and $b_n$ not convergent. I thought it was worth pointing out, just in case that somebody will read this and will wonder why there is a difference between the current version of the question and your answer.2016-01-30