# Thread: unsure of a proof

1. ## unsure of a proof

Ok so the question is:
Prove carefully that a convergent sequence is bounded.
Here is my proof thus far, but i'm not really sure if it is correct\finished:
$|a_n| = |a_n - L + L| \leq |a_n - L| + |L|$

$a_n$ is convergent, say to a limit L so:
Let $\epsilon > 0$
$\exists N \in \mathbb{N}$ such that $\forall n \geq N |a_n - L| < \epsilon$.

$|a_n| \leq |a_n - L| + |L| < \epsilon + |L|$ $\forall n \geq N$.

Is this enough to show that it is bounded?

2. Yes, but I will wrote a better proof...

$\{a_n\}^{\infty}_{n=1}$ convergent say to $L$ hence by definition:

Exist $N\in\mathbb{N}$ so for all $n>N$: $|a_n-L|<1$.

Let us mark $M=max\{|a_1|,|a_2|,...,|a_N|,|L|+1\}$ and then if $n\leq N$ then we have: $|a_n|\lwq M$

And if $n>N$ then:

$|a_n|\leq |L|+|a_n-L|<|L|+1\leq M$

Hence $M$ is upper bound to sequence $\{|a_n|\}^{\infty}_{n=1}$, the conclusion is that $\{a_n\}^{\infty}_{n=1}$ bounded.

This is more elegant proof.

3. Here is the standard proof.
If for every $\epsilon>0$ it is true that $|a|<|b|+\epsilon$ then $|a|\le|b|.$

4. I may be getting into semantics here, but wouldn't this sequence be a counterexample?

$a_n = \frac1{(n-1)^2}$ since $a_1$ isn't defined?

5. Originally Posted by chiph588@
I may be getting into semantics here, but wouldn't this sequence be a counterexample?
$a_n = \frac1{(n-1)^2}$ since $a_1$ isn't defined?
Most competent mathematicians would not be that picky.
Now granted that is outside the standard class in real analysis, or some such course.