Limit is less than the supremum

Question

Suppose I have a convergent sequence $\{a_n\}$ with the property that $$ \sup_{n \in \mathbb{N}} |a_n| < P $$ for some $P > 0$. My question is: does $$ \lim_{n \to \infty} |a_n| \leq \sup_{n \in \mathbb{N}} |a_n| $$ hold and, if yes, can I immediately write $$ \lim_{n \to \infty} |a_n| < P $$ without proving anything? My answers would be 'yes' and 'yes', but maybe the inequality needs some justification.

(Note: Almost all 'similar questions' are about the limes superior. I'm not about asking anything about $\limsup$!)

Answer 1

Since the sequence $\{a_n\}$ is assumed to be convergent, also $\{|a_n|\}$ is convergent, so we can assume $a_n\ge0$.

Let's show that, if $P$ is an upper bound for $\{a_n\}$, then $l=\lim_{n\to\infty}a_n\le P$.

If $l>P$, then there exists $n$ with $|a_n-l|<l-P$, which is the same as $$ P-l<a_n-l<l-P $$ and this implies $a_n>P$, a contradiction.

Now the supremum is by definition an upper bound for the sequence, so $$ l\le\sup_{n\in\mathbb{N}}a_n<P $$

Answer 2

If $(a_n)$ is convergent then so is $(|a_n|)$. Since $$ |a_k| \leq \sup_n |a_n|< P$$ holds for every $k$ we have $$ \lim_k |a_k| \leq \sup_n |a_n| < P$$ Depending on the context in which you are supposed to present this you may or may not pass directly to the conclusion.

Answer 3

Your first statement is false, if $(|a_n|)$ is convergent then $(a_n)$ is absolutely convergent, not the other way around.