TITLE: If $a_n$ and $b_n$ are equivalent sequences and $a_n$ is bounded then so is $b_n$.
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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 $\max \{|b_1|,\dots,|b_{N-1}|, M+1\}$