1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Three problems

  1. Mar 2, 2004 #1
    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
  2. jcsd
  3. Mar 3, 2004 #2


    User Avatar
    Science Advisor

    "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.

    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?

    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook