# Any convergent sequence is bounded. Don’t we need to use the absolute value in this proof?

We have the following elementary result on real sequences.

Any convergent sequence is bounded.

This is basically the proof given in my notes:

Suppose that $$a_n \to a \in \mathbb{R}$$. Now choose $$\epsilon = 1$$. From the $$\epsilon-N$$ definition of convergence, $$\exists N \in \mathbb{N}: n > N \implies |a_n - a| < 1$$. The triangle inequality gives us $$|a_n| \leq |a| + 1$$ for $$n > N$$. Therefore, choosing $$C = \sup \{a_n | n \leq N\} \cup \{|a| + 1\}$$ we obtain $$|a_n| < C$$, so $$a_n$$ is bounded. $$\Box$$

It seems to me that it is possible that, for $$n < N$$, $$\sup \{a_n\} < -a_n$$ and therefore$$\sup\{a_n\} < |a_n|$$. This will be the case if, for $$n < N$$, $$a_n$$ is very negative but bounded above by a relatively small positive quantity. Therefore, we actually need $$C = \sup \{|a_n| | n \leq N\} \cup \{|a| + 1\}$$ for this proof to work.

Am I correct?

• Yes, you are right. – Mark Apr 17 at 21:32

Yes, you are correct. Everything in their proof is correct until they choose $$C$$.

$$|a_n|<|a|+1$$ for all $$n>N$$ and $$|a_n|\leq\sup\{|a_n|:n\leq N\}$$, so $$|a_n|\leq \sup(\{|a_n|:n\leq N\}\cup\{|a|+1\})$$ for all $$n$$.

Take any sequence $$(a_n)$$ such that $$a_n<0$$ and $$|a_n|>|a|+1$$ for all $$n\leq N$$, and it will not be bounded in the way that their proof asserts.