# Existence of limit of two sequences

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:

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

Updated.

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This should look quite implausible. After a while, $a_n$ is very close to $L$. After a while, $b_n$ is really big. Surely $|a_n-b_n|$ cannot be close to $1$. You might want to check for a typo. –  André Nicolas Nov 14 '12 at 7:17
@André Nicolas Thanks! Can I say that if $a_n$ converges to L, $b_n$ also must to converge to M and M-L=1 ? –  Tina Nov 14 '12 at 7:31
Are you sure that $b_n$ should diverge to infinity and not simply diverge? For example $a_n=L$, $b_n=(-1)^n+L$ has $a_n\to L$, $|a_n-b_n|\to 1$ and $b_n$ divergent. –  Hagen von Eitzen Nov 14 '12 at 7:40
If $a_n$ and $b_n$ each converge to a number, say $L$ and $M$, then you can say that $|a_n-b_n|$ converges to $|L-M|$. Just knowing that $a_n$ converges can tell you nothing about $b_n$. The question you were asked must be different from what is written above. Maybe they want you to say that the claimed situation is impossible, and to explain why. If so, the answer that was given by isurasy1 can be your guide. Added: The suggestion of Hagen von Eitzen looks quite plausible. –  André Nicolas Nov 14 '12 at 7:43
@ 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? –  Tina Nov 14 '12 at 8:07

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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Thank you! But can you please to explain why my choice of sequences $a_n=\frac{1}{n}$ $b_n=(-1)^{n}$ is wrong? Thanks! –  Tina Nov 14 '12 at 8:39
@Tina: No I can't, for it isn't wrong. (But mine is even simpler.) –  Harald Hanche-Olsen Nov 14 '12 at 8:40

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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