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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 $n>n_0 \, \, |a_n|\lt M$.

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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.

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  • $\begingroup$ 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 $a_n\, \, \forall n\gt n_0$? $\endgroup$ – user Jun 12 '19 at 17:24
  • $\begingroup$ That's the reason that we use the definition of convergence, if that's what you mean. Using the definition, we can bound all but finitely many terms. There are only finitely many terms left, so we can just add them to the max (if we want to take a max of a bunch of quantities, we want to have a finite number of them). $\endgroup$ – cmk Jun 12 '19 at 17:28
  • $\begingroup$ 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? $\endgroup$ – user Aug 13 '19 at 19:05
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    $\begingroup$ @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 $\leq.$ $\endgroup$ – cmk Aug 13 '19 at 19:52

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