Prove: An, En, On --> L iff An --> L

[stunner5000pt]

1. From a sequence An, collect even numbered terms En = A2n, and Odd terms O = A2n-1. SHow that An --> L iff En --> L and On --> L
Im not sure if this is true while thinking about the proof but a sequence can behave in any way and therefore the even and odd terms may not necessairly come one after another so we can jus say for some n large enough En will approach the same thing as On. I'm clueless, please give me some idea

2. well if that wasn't enough here's another one( |x| means absolute value)
If An --> L then |An| --> L, is the converse true? That is If |An| -- > |L| then An --> L. Prove or give a counterexample. First of all is this even true? If it isn't then maybe a sequence like (-1/n)-1, maybe? Or does counterexample mean something else?

3. If An --> L and Bn --> L then show that
a1,b1,a2,b2,a3,b3,... converges to L. SO somehow we have to make sure that every term of B is greater than A somehow so obviously Bn> An but An+1 > Bn. I can think of a function that does this but how would you prove it?

Any sort of guidance on ANY of these questions would be greatly appreciated

 

[HallsofIvy]

Originally posted by stunner5000pt
1. From a sequence An, collect even numbered terms En = A2n, and Odd terms O = A2n-1. SHow that An --> L iff En --> L and On --> L
Im not sure if this is true while thinking about the proof but a sequence can behave in any way and therefore the even and odd terms may not necessairly come one after another so we can jus say for some n large enough En will approach the same thing as On. I'm clueless, please give me some idea

"iff" works two ways. To show IF An--> L then En-->L and On-->L, you need to use the definition of limit: since An-->L, given &epsilon;>0, there exist N such that if n> N then |An-L|< &epsilon;
Okay, now look at En and On. In order to get |En-L|< &epsilon; how large does n have to be? Same thing for On.

To show IF En-->L and On-->L then An-->L, use the fact that "En-->L" means: for &epsilon;> 0, there exist N1 such that ...
"On-->L" means: for &epsilon;> 0, there exist N2 such that... Take N=larger of N1, N2 and use the fact that every member of An is in either En or On.

2. well if that wasn't enough here's another one( |x| means absolute value)
If An --> L then |An| --> L, is the converse true? That is If |An| -- > |L| then An --> L. Prove or give a counterexample. First of all is this even true? If it isn't then maybe a sequence like (-1/n)-1, maybe? Or does counterexample mean something else?

What if L> 0? What if L< 0?! The "theorem" stated is clearly incorrect. It should be "If An-->L then |An|--> |L|". Did you copy it correctly? What would the converse be?

3. If An --> L and Bn --> L then show that
a1,b1,a2,b2,a3,b3,... converges to L. SO somehow we have to make sure that every term of B is greater than A somehow so obviously Bn> An but An+1 > Bn. I can think of a function that does this but how would you prove it?

No, you DON'T need to do that. Nothing was said about this being an increasing sequence! Once again: An-->L means: given &epsilon;> 0, there exist N1 such that... Bn-->L means: given &epsilon;> 0, there exist N2 such that...
When you "fold" An and Bn together, you will need to look at N= larger of N1, N2.

Do you see the similarity between this and #1?

 

[M98Ranger]

. Thank you!

1. To prove the statement, we can use the definition of convergence. Let An --> L, which means that for any ε > 0, there exists an N such that for all n > N, |An - L| < ε.

Now, let's consider the even numbered terms En = A2n. Since n > N, 2n > N as well. This means that for all n > N, |A2n - L| < ε. But this is the same as |En - L| < ε, which means that En --> L.

Similarly, for the odd terms On = A2n-1, we have 2n-1 > N for all n > N. Therefore, for all n > N, |A2n-1 - L| < ε, which is the same as |On - L| < ε. This shows that On --> L.

Conversely, if En --> L and On --> L, then for any ε > 0, there exists an N such that for all n > N, |En - L| < ε and |On - L| < ε. But since En = A2n and On = A2n-1, this means that for all n > N, |A2n - L| < ε and |A2n-1 - L| < ε. This is exactly the definition of An --> L. Therefore, An --> L iff En --> L and On --> L.

2. The statement is not true. A counterexample would be the sequence An = (-1)^n. This sequence does not converge, but |An| = 1 for all n. So |An| --> 1, but An does not converge to 1.

3. To prove that a1,b1,a2,b2,a3,b3,... converges to L, we can use the definition of convergence again. Let ε > 0 be given. Since An --> L and Bn --> L, there exists an N1 such that for all n > N1, |An - L| < ε/2. Similarly, there exists an N2 such that for all n > N2, |Bn - L| < ε/2.

Now, let N = max(N1, N2). For all n > N, we have n > N1 and n > N2