and converge. Prove converges.

Let and converge to , respectively,

Then there is an and ,

and

Therefore, converges.

Correct?

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- Jun 24th 2011, 12:55 PMdwsmith{a_n} and {a_n+b_n} converge. Prove {b_n} converges.
and converge. Prove converges.

Let and converge to , respectively,

Then there is an and ,

and

Therefore, converges.

Correct? - Jun 24th 2011, 01:06 PMJoannaRe: {a_n} and {a_n+b_n} converge. Prove {b_n} converges.
It looks good except for this part:

Quote:

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). - Jun 24th 2011, 01:09 PMAckbeetRe: {a_n} and {a_n+b_n} converge. Prove {b_n} converges.
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. - Jun 24th 2011, 01:20 PMdwsmithRe: {a_n} and {a_n+b_n} converge. Prove {b_n} converges.
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. - Jun 24th 2011, 01:23 PMJoannaRe: {a_n} and {a_n+b_n} converge. Prove {b_n} converges.
- Jun 24th 2011, 01:41 PMdwsmithRe: {a_n} and {a_n+b_n} converge. Prove {b_n} converges.
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. - Jun 24th 2011, 01:56 PMAckbeetRe: {a_n} and {a_n+b_n} converge. Prove {b_n} converges.

You're essentially saying that because A < B and A < C, that therefore A < B < C. But 4 < 5, and 4 < 6; that does not imply 4 < 6 < 5. - Jun 24th 2011, 02:07 PMPlatoRe: {a_n} and {a_n+b_n} converge. Prove {b_n} converges.
- Jun 24th 2011, 03:02 PMJoannaRe: {a_n} and {a_n+b_n} converge. Prove {b_n} converges.
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 .