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:
$\displaystyle |a_n| = |a_n - L + L| \leq |a_n - L| + |L|$

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

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

Is this enough to show that it is bounded?

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

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

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

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

And if $\displaystyle n>N$ then:

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

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


This is more elegant proof.

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

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

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

Quote:

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Originally Posted by chiph588@

I may be getting into semantics here, but wouldn't this sequence be a counterexample?
$\displaystyle a_n = \frac1{(n-1)^2} $ since $\displaystyle a_1 $ isn't defined?

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Most competent mathematicians would not be that picky.
Now granted that is outside the standard class in real analysis, or some such course.