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Advanced Calculus Sequence Convergence

  • #1

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!
 

Answers and Replies

  • #2
jbunniii
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Try writing each statement in epsilon-delta form, and compare.
 
  • #3
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?
 
  • #4
jbunniii
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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?
 
  • #5
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.
 
  • #6
jbunniii
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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:

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

versus what you wrote earlier:

([itex]\lim_{n \rightarrow \infty} a_n = A)[/itex]: "Given epsilon>0, there exists N>0 s.t. lan-Al<epsilon for all n>N."
 
  • #7
I'm not sure. I don't know if I have it written correctly. I feel like I'm working in circles.
 
  • #8
HallsofIvy
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Look at the definitions! "[itex]a_n[/itex] converges to A" means "Given [itex]\epsilon> 0[/itex], there exist an integer N such that if n> N, [itex]|a_n- A|< \epsilon[/itex]".

Applying exactly the same definition, "[itex](a_n- A)[/itex] converges to 0" means "Given [itex]\epsilon> 0[/itex], there exist N such that if n> N, [itex]|(a_n-A)- 0|< \epsilon[/itex]".

But [itex](a_n- A)- 0[/itex] is just [itex]a_n- A[/itex]!

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

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