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Showing a sequence is monotonically increasing

  1. Oct 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that if ##0<c<1## then llim##c^{\frac{1}{n}}=1## using the monotone convergence theorem.

    2. Relevant equations



    3. The attempt at a solution
    I let ##c_n=c^{\frac{1}{n}}## and it follows since ##0<c<1 \implies 0<c^{\frac{1}{n}}<1## Thus ##c_n## is bounded above by 1. Now I want to show that ##c_n## is monotonically increasing by induction but im not sure how to do it. So for my base case I know I need to show ##c_1<c_2## And for my inductive case I suppose that ##c_k<c_{k+1}## and show ##c_{k+1}<c_{k+2}## which is what im stuck on.
     
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  3. Oct 30, 2013 #2

    Dick

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    I don't think you need induction at all. Just look at ##\frac{c_{k+1}}{c_k}##. Is it greater than 1 or less than 1?
     
  4. Oct 30, 2013 #3
    Well ##\frac{c_{k+1}}{c_{k}}≥ 1##. But I'm not sure how to show that. I know that ##c_k=c^\frac{1}{k}\implies \frac{1}{c_k}=\frac{1}{c^\frac{1}{n}}≥1##. Then im not sure from here.
     
    Last edited: Oct 30, 2013
  5. Oct 30, 2013 #4

    Dick

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    It's ##\frac{c^{1/(k+1)}}{c^{1/k}}##. Use the laws of exponents and combine them.
     
  6. Oct 30, 2013 #5
    I see ##\frac{c_{k+1}}{c_{k}}=\frac{c^\frac{1}{k+1}}{c^\frac{1}{k}}=c^{\frac{1}{k+1}-\frac{1}{k}}=c^\frac{-1}{k(k+1)}## But since ##c<1## it follows ##\frac{1}{c^\frac{1}{k(k+1)}}>1##. Thanks I got it.
     
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