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Math Help - real analysis

  1. #1
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    real analysis

     {  For  n \in N \ , let x_n = 1+ \frac{1}{2^3} +\frac{1}{3^3} + ...+\frac{1}{n^3} , and \ let \ n_k = 2^k - 1 , k \in N }

    a- prove  \ that  \ x_{n_k} < \ 1 + (\frac {1}{4})+(\frac {1}{4})^2+...+(\frac {1}{4})^{k-1} , k=2,3,4,...

    b- use \ (a )  \ to \ prove  \ that \ x_n \ is  \ convergent
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  2. #2
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    Hi,
    we can prove a) by induction:
    base case: k=2 .... n_2=3 .... x_{n_2}=1+\frac{1}{2^3}+\frac{1}{3^3}<1+\frac{1}{8  }+\frac{1}{8} = 1+\frac{1}{4}

    inductive step: suppose x_{n_k} = 1+ \frac{1}{2^3} +\frac{1}{3^3} + ...+\frac{1}{(2^k-1)^3}< 1 + \frac {1}{4}+\frac {1}{4^2}+...+\frac {1}{4^{k-1}}=y_k and let's prove the statement holds for k+1:

    x_{n_{k+1}}= 1+ \frac{1}{2^3} +\frac{1}{3^3} + ...+\frac{1}{(2^{k+1}-1)^3} =
    =\underbrace{1+ \frac{1}{2^3} +\frac{1}{3^3} + ...+\frac{1}{(2^k-1)^3}}_{<y_k}+\frac{1}{(2^k)^3}+\frac{1}{(2^k+1)^3  }+\frac{1}{(2^k+2)^3}+...+\frac{1}{(2^{k+1}-1)^3}<

    <y_k +\underbrace{\frac{1}{(2^k)^3}}_{=\frac{1}{2^{3k}}  }+\underbrace{\frac{1}{(2^k+1)^3}}_{<\frac{1}{2^{3  k}}}+\underbrace{\frac{1}{(2^k+2)^3}}_{<\frac{1}{2  ^{3k}}}+...+\underbrace{\frac{1}{(2^{k+1}-1)^3}}_{<\frac{1}{2^{3k}}}<

    <y_k+ \left( (2^{k+1}-1)-2^k+1 \right)\frac{1}{2^{3k}} = y_k+(2\cdot 2^k-2^k)\frac{1}{2^{3k}} = y_k+\frac{2^k}{2^{3k}}=

    = y_k+\frac{1}{2^{2k}}= y_k+\frac{1}{4^k}= 1 + \frac {1}{4}+\frac {1}{4^2}+...+\frac {1}{4^{k-1}}+\frac{1}{4^k} = y_{k+1}

    so the inductive step is completed and a) holds.

    As for b), from a) we see that that the sequence x_{n_k} is bounded from above by y_k.
    y_k is the sum of first k terms of a geometric progression, y_k converges to 4/3.
    So since x_n is increasing, it is convergent.
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  3. #3
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    real analysis

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