So, there is a theorem that states that if sequence is convergent, then it's bounded. That's because:
According to the definition of the limit of real sequence,
$$\forall \epsilon \gt 0,\, e.g. \, \epsilon=1 \, \exists n_0 : \forall n>n_0 \, |a_n-a| \lt 1 \, e.g. \, a-1\lt|a_n|\lt a+1 $$
Sequence is bounded, because $\forall n\gt n_0 \, \, |a_n|\lt M$,
where $M = \max\{|a_1|, ..., |a_{n_0}|, |a-1|, |a+1|\}$.
What I don't understand is that it's already stated that $a-1\lt|a_n|\lt a+1, $ why M should be chosen between all the sequence members. It is said, that sequence is bounded if for only $n>n_0 \, \, |a_n|\lt M$.
A sequence is bounded if all of its terms are bounded by one $M$. If we know that $|a_n-a|<1$ for all $n>n_0$, then the reverse triangle inequalty says that $|a_n|<1+|a|.$ This tells us how to bound all terms $a_n$ for $n>n_0$. This doesn't bound the whole sequence; we're still missing bounds on the terms $a_n$ for $n\leq n_0.$ Since there are finitely many of these, we just append their magnitude to the maximum (this is a sequence of real numbers, so their magnitude is finite), and this gives us a uniform bound on all of the terms.