Thread: Analysis: Please check my proof on the arithmetic of CONVERGENT SEQUENCES

I am new to analysis and am having a lot of trouble with this proof. I am not sure if what I have written is valid. Please let me know. Thank you.

Theorem: If {an} converges to A and {an + bn} converges to A + B, then {bn} converges.


Proof:

Since {an + bn} converges to A + B, for every ε> 0 there exists an N such that for
n > N, |(an + bn) - (A + B)| < ε

=>

|(an - A) + (bn - B)| < ε

It is true that

|(an - A) + (bn - B)| ≤ |an - A| + |bn - B|

It is also true that

|(an - A) + (bn - B)| - |an - A| < ε for n > N

Thus it is true that

|(an - A) + (bn - B)| - |an - A| ≤ |an - A| + |bn - B| - |an - A| < ε
for n > N

Therefore

|an - A| + |bn - B| - |an - A| < ε for n > N

so

|bn - B| < ε for n > N

Since for ε there exists an N such that n > N => |bn - B| < ε, then {bn} converges.■





I used the same N for {bn} that I used for {an +bn} because if it works for
{an + bn} and {bn} is inside {an + bn}, then I reason it will work for {bn} too.

First, one can never assume that same N works in all cases.

Now you know that .
Let .
You can controle both making each less than .

Do you already know the theorems that if and converges, then converges for any constant , and converges?

In that case, you can take a short-cut. Since converges, you know that converges. Then, since converges, you know that converges.

Originally Posted by HappyJoe

Do you already know the theorems that if and converges, then converges for any constant , and converges?

In that case, you can take a short-cut. Since converges, you know that converges. Then, since converges, you know that converges.

Is that essentially letting = , and = in the sequence , which is known by theorem to converge?

Originally Posted by Plato

First, one can never assume that same N works in all cases.

Now you know that .
Let .
You can controle both making each less than .

My question about this is that I already know that in the proof where you're given

converges to A and converges to B, prove that converges to A + B

you use the exact same thing:

< ε/2

and

< ε/2


So if in this proof < ε/2, how can also be less than ε/2 for the proof I'm trying to understand in the original post? Because isn't that essentially saying that all three sequences are less than ε/2?

Well, what I mean is, I see how all three could, since epsilon is arbitrary, but I don't see how it helps me. (I do realize this is because of me, not you, so don't get the wrong idea. I'm just a little lost here.)





Thanks for both of your respective input!

If then we have.
and

Now let if then so both of the above conditions are true.

Can you finish?

Originally Posted by Plato

If then we have.
and

Now let if then so both of the above conditions are true.

Can you finish?

I don't know. I'll try. If you don't mind, criticism would be nice. I know that I suck, so hopefully I won't infuriate you with my denseness.


Since , then

+

So then

+ ≤ + +

So

+ ≤ +

This should be satisfied when



and




but since we already know


then



for




therefore

converges.

Originally Posted by SunRiseAir

Is that essentially letting = , and = in the sequence , which is known by theorem to converge?

Yeah, that's exactly it.