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Let $a_n$ and $b_n$ two sequences such as: $\lim_{n\rightarrow \infty }\left | a_n - b_n\right |=1 $.

Is the following always true?

1) if $a_n$ converges to the real limit, then the $b_n$ converges to the real limit too.

2) if $a_n$ is bounded then the $b_n$ bounded also.

I think, that it isn't true, but i can't to find two sequences $a_n$ and $b_n$ which contradict to these sentences Could you help me please?

Thanks!

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  • $\begingroup$ Let $a_{n}\equiv 0$ and $b_{n} \equiv i$. $\endgroup$
    – JessicaK
    Commented Nov 14, 2014 at 9:54
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    $\begingroup$ Second proposition is true. $\endgroup$
    – mvggz
    Commented Nov 14, 2014 at 9:56
  • $\begingroup$ For the first proposition let $a_n=0$ and $b_n=(-1)^n$. $\endgroup$
    – bof
    Commented Nov 14, 2014 at 9:59

3 Answers 3

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For the first (which is not true), take $$ a_n = 0, \quad b_n = (-1)^n $$ For the second (which is true), note that $$ |b_n| \leq |a_n| + |b_n - a_n| $$

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An example for 1)

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

As for 2), suppose $a_n$ is bounded. Then $|a_n| \leq L$ for some constant $L$.

Eventually you have $||a_n - b_n| -1 | \leq 1$. So, eventually you have $$|b_n| = |-a_n + b_n -1 + a_n +1| \leq ||a_n - b_n| -1 | + L +1 \leq L+2$$

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For (1), consider $a_n = 0$ and $b_n = (-1)^n$ to see it is false.

For (2), suppose $|a_n| \leq M$.

Pick $N \in \mathbb{N}$ s.t. for $n \geq N$, $|a_n - b_n| \leq 2$.

Then for $n \leq N$, $|b_n| \leq |a_n| + 2 \leq M + 2$.

Hence $|b_n| \leq max\{|b_1|, |b_2|, \ldots |b_{N-1}|, M + 2\}$

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