4
$\begingroup$

This is what i know;

If $(a_n)$ is an infinite sequence of which is bounded then we can say;

$|a_i| < M $ for all $i \geq 0.$

since $a_n$ and $b_n$ are equivalent sequences, we can say that for every rational $\epsilon >0$, there exists $ N \geq 0$ such that

$|a_i-b_i| < \epsilon $ for all $i \geq N$.

I would like to show that $b_n$ is bounded.

now using the triangle inequality we obtain,

$|b_i| = | b_i - a_i + a_i| \leq |b_i - a_i| + |a_i| \leq \epsilon + M$

and hence the sequence is bounded above by $M' = max\{\epsilon, M\}$.

I feel like there is something missing at the end, could someone lend a helping hand please.

$\endgroup$
  • 5
    $\begingroup$ You're definitely on the right path. However, what you've proven is that from some point on, the sequence $b_n$ is bounded by $M+\epsilon$. There might be billions of terms before you get there. You have to point out that all the $b_n$ before that point has a bound as well. $\endgroup$ – Arthur Jan 24 '15 at 18:35
  • 1
    $\begingroup$ If we take $M' = max\{|b_0|,|b_1|,....,|b_{i-1}|, \epsilon + M\}$, then $|b_i| < M'$ for all $i>o$...Is that what you mean? or am i off :|. $\endgroup$ – user197848 Jan 24 '15 at 18:41
  • $\begingroup$ Yes, that's exactly what I'm talking about. $\endgroup$ – Arthur Jan 24 '15 at 20:04
1
$\begingroup$

Since $a_n$ and $b_n$ are equivalent sequences, we can say that for every rational $1 \geq \epsilon >0$, there exists $ N \geq 0$ such that

$|a_i-b_i| < \epsilon $ for all $i \geq N$.

...Then $b_n$ for $n \geq N$ will be bounded by $M' := M+1$.

So $b_n$ is bounded by $\max \{|b_1|,\dots,|b_{N-1}|, M+1\}$

$\endgroup$
  • $\begingroup$ I like, but whats with the $ 1 \geq \epsilon > 0$ inequality? i understand that if they are equivilent then this will eventually be true, is there any harm in saying $\epsilon > 0$?. $\endgroup$ – user197848 Jan 24 '15 at 18:47
  • $\begingroup$ As you know the sequences are equivalent you are able choose epsilon to make things easier $\endgroup$ – Permian Jan 24 '15 at 18:48
  • 1
    $\begingroup$ I see, thanks very much indeed. $\endgroup$ – user197848 Jan 24 '15 at 18:50

Your Answer

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