# Thread: xn + ym is a Cauchy sequence

1. ## xn + ym is a Cauchy sequence

Let xn and yn be Cauchy sequences.
Give a direct arguement that xn+yn is a Cauchy sequence that does not use the Cauchy Criterion or the Algebraic Limit Theorem.

given epsilon>0 there exists an N in the natural numbers such that whenever m,n>N, it follows that:
abs value(xm-xn),epsilon and abs value(ym-yn) <epsilon
I'm not sure where to go next.

2. I think you're going to need to use some sort of $\displaystyle \epsilon/2$ argument here. You have to control the size of

$\displaystyle |(x_{n}+y_{n})-(x_{m}+y_{m})|=|(x_{n}-x_{m})+(y_{n}-y_{m})|.$

Any ideas?

3. but doesn't epsilon/2 imply part of the Cauchy criterion?

4. Thats say xn t yn is a Cauchy sequence by the triangle inequality

5. No, the Cauchy criterion is something you apply to see if a particular series is convergent. But in this problem, all you're worried about is showing that the Cauchy property survives term-by-term sequence addition. Convergence doesn't have to enter into it anywhere. Epsilons are used in the definition of a sequence being Cauchy, so I think they'll end up in your final proof.

The triangle inequality is certainly useful. Now you can control the quantity

$\displaystyle |(x_{n}-x_{m})+(y_{n}-y_{m})|\le |x_{n}-x_{m}|+|y_{n}-y_{m}|.$

Does this suggest anything to you?

Incidentally, you may need to use separate sequence index variables for the $\displaystyle x_{n}$ versus the $\displaystyle y_{n}$ sequence, and then maybe take the maximum or something like that.

How would you start writing your proof?

6. It's less than Epsilon. Thus xn+yn is Cauchy!

7. Well, that's the goal. I'm not sure we've quite gotten there yet. For one thing, according to the way you wrote out the Cauchy property for each sequence, we can only get that

$\displaystyle |x_{n}-x_{m}|+|y_{n}-y_{m}|\le 2\epsilon.$

How can you get rid of that pesky 2?

For another, I'm not sure it's allowed to use the same m and n in both statements of the Cauchy property (that is, one for x and one for y).

How can you fix these two problems?

8. epsilon/2?