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Math Help - PLEASE help with 2 tricky series problems!

  1. #1
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    PLEASE help with 2 tricky series problems!

    1) Verify that the series:

    1-(1/2)+(1/2)-(1/3)+(1/2)-(1/3)-(1/4)+(1/3)-(1/4)+(1/3)-(1/4)+...

    diverges and explain how this does not violate the basic theorem on alternating series.

    [To my knowledge, the basic theorem they are referring to is that the summation from 0 to infinity of (-1)^k ak converges IFF, if and only if, ak-->0.]



    2) Let a0, a1, a2,... be a nonincreasing sequence of positive numbers that converges to 0. Does the alternating series SUMMATION [(-1)^k]ak necessarily converge?

    Thanks in advance for the help!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by seth55 View Post
    1) Verify that the series:

    1-(1/2)+(1/2)-(1/3)+(1/2)-(1/3)-(1/4)+(1/3)-(1/4)+(1/3)-(1/4)+...

    diverges and explain how this does not violate the basic theorem on alternating series.

    [To my knowledge, the basic theorem they are referring to is that the summation from 0 to infinity of (-1)^k ak converges IFF, if and only if, ak-->0.]

    No, if I recall corectly it requires that eventualy the series of absolute values be decreasing as well as the limit of the absolute value of the terms be 0(and this series fails that requirement).

    RonL
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  3. #3
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    Yes you're right. I just checked with my textbook:
    Let a0,a1,a2 be a decreasing sequence of positive numbers...
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by seth55 View Post
    1) Verify that the series:

    1-(1/2)+(1/2)-(1/3)+(1/2)-(1/3)-(1/4)+(1/3)-(1/4)+(1/3)-(1/4)+...

    diverges and explain how this does not violate the basic theorem on alternating series.
    If I have interpreted your intention correctly:

    After the third term the terms may be grouped into threes so:

     <br />
S=1+\sum_{n=3}^{\infty} -\frac{2}{n}+\frac{1}{n+1}=1+\sum_{n=3}^{\infty} -\frac{1}{n}+\frac{1}{n(n+1)}<br />

    But

     <br />
\frac{1}{n(n+1)}<\frac{1}{4n}<br />

    So:

     <br />
\frac{1}{n}-\frac{1}{n(n+1)}>\frac{3}{4n}<br />

    So the series must diverge as the partial sums are greater than a constant plus the partial sums of a multiple of the harmonic series which diverges.

    RonL
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