# Range of a convergent sequence is bounded.

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?

• You may found the proof in any text of sequences. – MANI Oct 24 '19 at 11:40

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 $$n$$.