# Question about convergent sequences' boundedness

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.
• Ok, then the only reason why we can't find the $M$ for all sequence using maximum is that there are infinitely many members of $a_n\, \, \forall n\gt n_0$? – user Jun 12 '19 at 17:24
• But on the proof it is written that $\forall n > n_0 // |a_n|< M$ not for all $n$. The sequence is not bounded for all n in the proof, is it? Also, if it were said that $\forall n |a_n|< M$ then in case the max would be one of the members of the sequence that one member wouldn't be strictly smaller than itself, should we use smaller or equal to $M$ sign or am I missing something? – user Aug 13 '19 at 19:05
• @levabrakmane what it should say is that, for all $n$, $|a_n|\leq M.$ Indeed, if $n\leq n_0$, then it is bounded by $\max\{a_1,a_2,\cdots, a_{n_0}\},$ and if $n>n_0,$ then $|a_n|<|a|+1.$ So, the whole sequence is bounded by the max of these quantities. And, to be safe, we should have $\leq.$ – cmk Aug 13 '19 at 19:52