MathSquareRoo

## Homework Statement

Prove that the sequence {a_n} converges to A if and only if lim n--->∞ (a_n-A)=0.

## The Attempt at a Solution

It's an if and only if proof, but I'm not sure how to prove it. Please help!

Homework Helper
Gold Member
Try writing each statement in epsilon-delta form, and compare.

MathSquareRoo
I'm not good at writing proofs. So far I have:
Let {an} converge to A. Given epsilon>0, there exists N>0 s.t. lan-Al<epsilon for all n>N.
So l((a_n)-A)l<epsilon for all n>n.
Thus, we can write lim n--->infinity (a_n-A)=0.

Then, I'm not sure how to prove the statement's converse. Can someone help?

Homework Helper
Gold Member
I'm not good at writing proofs. So far I have:
Let {an} converge to A. Given epsilon>0, there exists N>0 s.t. lan-Al<epsilon for all n>N.
So l((a_n)-A)l<epsilon for all n>n.
Thus, we can write lim n--->infinity (a_n-A)=0.

Then, I'm not sure how to prove the statement's converse. Can someone help?

Well, how did you prove it in the forward direction? Can you simply reverse the reasoning?

MathSquareRoo
I'm not sure how to write the reverse. Would I start with: lim n--->infinity(a_n -A)=0. So given epsilon>0, there exists N>0 s.t. la_n-Al<epsilon for all n>N.

I don't know if this is correct, and I don't know where to go after that.

Homework Helper
Gold Member
I'm not sure how to write the reverse. Would I start with: lim n--->infinity(a_n -A)=0. So given epsilon>0, there exists N>0 s.t. la_n-Al<epsilon for all n>N.

I don't know if this is correct, and I don't know where to go after that.

Yes, that's correct. But what's the difference between this:

($\lim_{n \rightarrow \infty} (a_n - A) = 0)$: "So given epsilon>0, there exists N>0 s.t. la_n-Al<epsilon for all n>N."

versus what you wrote earlier:

($\lim_{n \rightarrow \infty} a_n = A)$: "Given epsilon>0, there exists N>0 s.t. lan-Al<epsilon for all n>N."

MathSquareRoo
I'm not sure. I don't know if I have it written correctly. I feel like I'm working in circles.

Look at the definitions! "$a_n$ converges to A" means "Given $\epsilon> 0$, there exist an integer N such that if n> N, $|a_n- A|< \epsilon$".
Applying exactly the same definition, "$(a_n- A)$ converges to 0" means "Given $\epsilon> 0$, there exist N such that if n> N, $|(a_n-A)- 0|< \epsilon$".
But $(a_n- A)- 0$ is just $a_n- A$!