Prove that if {$a_n$ } is a sequence that converges to A and $a_n$ ≥ 0 for all n then A ≥ 0.

Question

Prove that if {$a_n$} is a sequence that converges to A and $a_n$ $\geq$ 0 for all n, then A $\geq$ 0.

I have assumed on that contrary that A < 0. Pick $\epsilon$ = |A| > 0. Now, |$a_n$ - A| = $a_n$ - A $\geq$ -A = |A|. This is a contradiction because by definition of convergence, we have to have for all $\epsilon$ > 0, there exists a natural number N such that for all n $\geq$ N, |$a_n$ - A| < $\epsilon$=|A|. Therefore, A $\geq$ 0.

Is this correct?

Answer 1

I believe that your proof is valid, but I also think that there are a few things you could have left out..

So your conclusion is that $A<0$ implies the contradiction $a_n+|A|<|A|$ and $a_n+|A|>|A|$. This means that the assumption that $A<0$ is wrong. To get these to inequality we use the property of a the converging sequence. Here is what i mean:

First the easy inequality:

$a_n+|A|>|A|$

Where we used the $a_n\ge0$ for all $n$.

And the converging sequence property. Choose $\epsilon=|A|$

$|a_n-A|<|A|$

But since $-A$ is always negative we have the equality:$a_n+|A|=|a_n-A|$ and therefore:

$a_n+|A|=|a_n-A|<|A|$

And here we have the contradiction. (This is of course just for $n>N$ for some $N$, but i hope this is implied)