0
$\begingroup$

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!

$\endgroup$
3
  • $\begingroup$ Let $a_{n}\equiv 0$ and $b_{n} \equiv i$. $\endgroup$ – JessicaK Nov 14 '14 at 9:54
  • 1
    $\begingroup$ Second proposition is true. $\endgroup$ – mvggz Nov 14 '14 at 9:56
  • $\begingroup$ For the first proposition let $a_n=0$ and $b_n=(-1)^n$. $\endgroup$ – bof Nov 14 '14 at 9:59
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $$|b_n| = |-a_n + b_n -1 + a_n +1| \leq ||a_n - b_n| -1 | + L +1 \leq L+2$$

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.