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Real Analysis Convergence Question

  1. Oct 22, 2013 #1
    1) Use mathematical induction to prove that for any k ∈ N, lim (1+k/n)^n = e^k.

    I already used monotone Convergence Thm to prove k=1 case. Do I just need to go through the same process to show k? If not, could you please help?


    2) Suppose that ( x_n ) is a sequence of real numbers, ( y_n ) is a bounded sequence of non-zero real numbers, and that lim x_n/ y_n = 1. Prove that lim x_n - y_n = 0.

    Since y_n is bounded, there exist M such that |y_n| <= M for all n in N. Then what should I do?

    Thanks
     
  2. jcsd
  3. Oct 22, 2013 #2

    UltrafastPED

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    1) Next you show that if it is true for some k, then it is true for k+1. You are anchored on k=1, so the glide
    from k -> k+1 takes care of the rest.
     
  4. Oct 22, 2013 #3

    BTP

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    2) To solve the problem you have to show for every ε>0 there exists an N .st. if n>N then |x_n - y_n|<ε. Now since x_n/y_n-->1 , given an ε>0 there exist N .st. if n>N then 1-ε< x_n/y_n <1 +ε. And now given that y_n>0 makes the next step easier. (When proving limits always go back to the basic definition to see where you need to go). Anyway now you are in business...
     
    Last edited: Oct 22, 2013
  5. Oct 22, 2013 #4

    Dick

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    BTP deleted the response to post 1) but I'll echo it. It's insane to do this by induction if you are anchored on k=1. I don't even see how you would do it. It's just a change of variables. 1+k/n=1+1/(n/k). Change the limiting variable to n'=n/k.
     
  6. Oct 23, 2013 #5

    BTP

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    I wasn't sure of etiquette so I pulled my insane comment. But now I know.
     
  7. Oct 23, 2013 #6

    Dick

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    Calling a person insane is one thing. Calling a question strategy insane is another.
     
  8. Oct 23, 2013 #7

    BTP

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    Ha, I got the not calling a person insane part. I wasn't sure about calling a problem insane. Cheers!
     
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