Advanced Calculus Sequence Convergence

[MathSquareRoo]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

 

[jbunniii]

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?

 

[jbunniii]

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.

 

[jbunniii]

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.

 

[HallsofIvy]

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 $a_n- A$!

Remember that definitions in mathematics are "working definitions"- you use the precise words of definitions in problems and proofs.