Thread: How do I prove a theorem about the limit of the absolute value of a sequence?

I'm suppose to use the definition of a limit of a sequence to prove the theorem that states:If limit(n approaches Infinity) of |A_n| = 0 then limit(n approaches Infinity) of A_n = 0.

If I showed that the "then" statement follows from the "if" statement by manipulating the "if" statement to arrive at the "then" statement, will I be proving the theorem? How would this prove the theorem to be true or am I not trying to prove the theorem to be true or false but rather that it is valid? (I'm sorry if I confused you with my questions ... I'm pretty sure I'm having trouble with the logic behind proofs and understanding what they're trying to ask me to do.)

Also, please give me a hint as to how to go about proving the theorem. Here is what I've done so far (and my accompanying notes for each step):

1. If lim|A_n| = 0 then lim A_n = 0 ; Theorem to be "proven".
2. lim|A_n| = 0 lim -A_n = 0 or lim A_n = 0 ; This is where I stopped since I don't know how to prove that the equivalent 2 limits equal zero and i'm not sure how proving them to be zero would prove that lim A_n = 0;

Originally Posted by Elusive1324

I'm suppose to use the definition of a limit of a sequence to prove the theorem that states:If limit(n approaches Infinity) of |A_n| = 0 then limit(n approaches Infinity) of A_n = 0.

If then .

We know that
Squeeze away.

Originally Posted by Elusive1324

I'm suppose to use the definition of a limit of a sequence to prove

Limit of a sequence - Wikipedia, the free encyclopedia

Notice that the definition uses an absolute value anyway: it says that x is the limit if the terms of the sequence eventually settle and stay inside of any specified window of deviation around x, however small. It assumes the window is always symmetrical about x, so we'll only ever specify it by a number that allows the same deviation above x as below. Hence the modulus brackets in the condition: . Which is equivalent to .

So if all terms A_n (past some value N of n) are within the window (of however small a radius epsilon) when they've all been turned positive, they are just as well going to be within the window (but maybe use both halves of it) when left with their original signs.

If I showed that the "then" statement follows from the "if" statement by manipulating the "if" statement to arrive at the "then" statement, will I be proving the theorem?

Yes. The "if" statement implies and which together imply

How would this prove the theorem to be true or am I not trying to prove the theorem to be true or false but rather that it is valid?

Yes you're right that we are proving the theorem to be valid rather than true, because the theorem doesn't commit to the truth of the 'if' statement.

Theorem - Wikipedia, the free encyclopedia

Thank you. I really appreciate that you considered the questions I had regarding the logic of proving theorems.