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We were given this exercise in class with the solution but some things about it are unclear for me.

Assume that the sequence $(a_n)$ converges. Show that the set $\{a_n | > n\in \mathbb{N}\}$ is bounded.

The solution goes as follows.

Since $(a_n)$ converges, there is a number $L$ such that $\lim_{n\rightarrow\infty}a_n=L$. By definition, for $\epsilon=1$ we can find $n_{\epsilon}\in\mathbb{N}$ such that

$$n\gt n_{\epsilon} \Rightarrow |a_n-L|\lt\epsilon \Rightarrow > |a_n|-|L|\lt\epsilon \Rightarrow |a_n|\lt 1+|L|$$

and by the definition of absolute value

$$-1-|L|\lt a_n\lt 1+|L|.$$

We have found an index of the sequence from which onward its members are bounded from below by $-1-|L|$ and above by $1+|L|$.

Now we can choose $M=max\{|a_0|,|a_1|,|a_2|,...,|a_{n_{\epsilon}}|, > 1+|L|\}$ to bound all members of the sequence. Now $|a_n|\lt M$ for all $n\in\mathbb{N}$ hence the set $\{a_n | n\in \mathbb{N}\}$ is bounded.

The solution makes sense to me until the $M=max\{|a_0|,|a_1|,|a_2|,...,|a_{n_{\epsilon}}|, > 1+|L|\}$ part. Didn't we already prove the set was bounded by $-1-|L|$ and $1+|L|$ so why was this necessary? I also don't understand the phrase "we have found an index of the sequence from which onward its members are bounded -". Didn't we prove that the whole sequence was bounded?

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  • $\begingroup$ You may found the proof in any text of sequences. $\endgroup$ – MANI Oct 24 '19 at 11:40
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No, you have that $|a_n|\leq 1+L$ only if $n>n_\varepsilon $. So, if you set $M=\max\{|a_0|,...,|a_{n_\varepsilon }|, 1+L\}$, then obviously $|a_n|\leq M$ for all $n$.

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