and converge. Prove converges.
Let and converge to , respectively,
Then there is an and ,
and
Therefore, converges.
Correct?
It looks good except for this part:
You need to say something like "Let be given."Then there is an
Otherwise you are saying that there exists an epsilon and N for which the following is true, which is not what you want to prove. By saying that epsilon is given and arbitrary, you essentially show that it is true for any epsilon (greater than zero, of course).
As Joanna said, it needs to be a "for all epsilon", not a "there exists an epsilon". You need to work with
See where that leads. Your proof is incorrect, because you're inserting items in the inequality chain: that's not valid.
What is "not valid"?
We know converges. So if we let it converge to , we have
And by the triangle inequality,
Since we also know converges, I can let it converge to some A.
Because it converges
Since , we can substitute for and make the inequality strictly less than.
That yields:
But we also know that .
So
I don't see what is wrong in the inequality structure.
I don't agree with you two and here is why:
So if we let . Then
Now, let
Since those two sequences converge,
And
By the triangle inequality,
Now, we have
Let's subtract epsilon over 2.
I started with something giving to be true {a_n+b_n} converges, used another true piece {a_n} converges, used the triangle inequality, and made substitution that is true.
If , it is which is by definition less than epsilon.
Here's a specific example where your argument fails.
As Plato pointed out, you need an N for each sequence; let be such that implies and let be such that implies . Then we can obtain both inequalities with .
Then n>N implies and . This is a particular N.
Suppose that we already know that converges to . Also suppose that and (which is certainly less than ).
Then we have
but
.
The point is that there exists an such that and . You would need to find a larger N to get .