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Math Help - Help with 2 sequences

  1. #1
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    Help with 2 sequences

    Show a_n = 1 + 1/3 + 1/5 + ... +1/(2n-1) diverges

    and

    show b_n = 1 + 2x + 3x^2 + ... + nx^(n-1) converges to (1-x)^-2 for an x with absolute value less than 1.

    Thanks!
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  2. #2
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    I was able to prove part A, any help with part B please?
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  3. #3
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    Sorry, should have posted my answer to the first one. Still need help on second one.


    a2 = 1 + 1/3
    a5 = 1 + 1/3 + 1/5 + 1/7 + 1/9 > 1 + 1/3 + 1/9 + 1/9 + 1/9 = 1 + 1/3 + 1/3
    a14 = 1 + 1/3 + (1/5 + 1/7 + 1/9) + (1/11 + 1/13 + 1/15 + 1/17 + 1/19 + 1/21 + 1/23 + 1/25 + 1/27) > 1 + 1/3 + 1/3 + (1/27 + 1/27 + 1/27 + 1/27 + 1/27 + 1/27 + 1/27 + 1/27 + 1/27) = 1 + 1/3 + 1/3 + 1/3

    and you can continue to find a series of elements that are greater than 1/3. Since 1 + 1/3 + 1/3 + 1/3 + ... diverges, and a_n is greater than that sequence by the above, then a_n diverges.
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Jazz10 View Post
    Sorry, should have posted my answer to the first one. Still need help on second one.


    a2 = 1 + 1/3
    a5 = 1 + 1/3 + 1/5 + 1/7 + 1/9 > 1 + 1/3 + 1/9 + 1/9 + 1/9 = 1 + 1/3 + 1/3
    a14 = 1 + 1/3 + (1/5 + 1/7 + 1/9) + (1/11 + 1/13 + 1/15 + 1/17 + 1/19 + 1/21 + 1/23 + 1/25 + 1/27) > 1 + 1/3 + 1/3 + (1/27 + 1/27 + 1/27 + 1/27 + 1/27 + 1/27 + 1/27 + 1/27 + 1/27) = 1 + 1/3 + 1/3 + 1/3

    and you can continue to find a series of elements that are greater than 1/3. Since 1 + 1/3 + 1/3 + 1/3 + ... diverges, and a_n is greater than that sequence by the above, then a_n diverges.
    Easier way to prove divergence of the first sum: Observe that

    \frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k}<\sum_{k=  1}^{\infty}\frac{1}{2k-1}

    so the comparison test says that the sum diverges.

    For the second one, we know that the geometric series converges:

    1+x+x^2+...+x^n+...=\frac{1}{1-x} for |x|<1

    Differentiate both sides with respect to x to achieve the desired result.
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