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Math Help - [SOLVED] Checking my result (infinite series)

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Checking my result (infinite series)

    Just want to be sure about my results.
    Say whether it converges or not :
    1) 1-\frac{1}{2}+\frac{2}{3}-\frac{1}{3}+\frac{2}{4}-\frac{1}{4}+\frac{2}{5}-\frac{1}{5}+.... I wrote it as \sum_{n=2}^{+\infty} \frac{2}{n}+\sum_{n=2}^{+\infty} -\frac{1}{n}= The harmonic series, so divergent.
    2) 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+... I wrote it as \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}{2n-1}. Taking the limit when n tends to +\infty of \frac{1}{2n+1} and seeing it's equal to 0, the series converges.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by arbolis View Post
    Just want to be sure about my results.
    Say whether it converges or not :
    1) 1-\frac{1}{2}+\frac{2}{3}-\frac{1}{3}+\frac{2}{4}-\frac{1}{4}+\frac{2}{5}-\frac{1}{5}+.... I wrote it as \sum_{n=2}^{+\infty} \frac{2}{n}+\sum_{n=2}^{+\infty} -\frac{1}{n}= The harmonic series, so divergent.
    If this series is convergent it is conditionaly convergent. But a conditionaly convergent series can be rearranged to sum to anything, so you cannot rearrange the terms.

    Now the sum of pairs of terms gives the harmonic series, so the partial sum of 2n terms is equal to the partial sum of n terms of the harmonic series, so the series cannot converge (since if it did every subsequence of the sequence of partial sums would have to have the same finite limit).

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by arbolis View Post
    2) 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+... I wrote it as \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}{2n-1}. Taking the limit when n tends to +\infty of \frac{1}{2n+1} and seeing it's equal to 0, the series converges.
    Your argument is either invalid, or you have missed out some of the explanation that would make it valid.

    This series satisfies the conditions of the altenating series test, the limit of the terms is zero, and the absolute values form a decreasing sequence. Hence it converges.

    RonL
    Last edited by CaptainBlack; July 12th 2008 at 10:07 AM.
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  4. #4
    Super Member PaulRS's Avatar
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    Quote Originally Posted by arbolis View Post
    Just want to be sure about my results.
    Say whether it converges or not :
    1) 1-\frac{1}{2}+\frac{2}{3}-\frac{1}{3}+\frac{2}{4}-\frac{1}{4}+\frac{2}{5}-\frac{1}{5}+.... I wrote it as \sum_{n=2}^{+\infty} \frac{2}{n}+\sum_{n=2}^{+\infty} -\frac{1}{n}= The harmonic series, so divergent.
    2) 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+... I wrote it as \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}{2n-1}. Taking the limit when n tends to +\infty of \frac{1}{2n+1} and seeing it's equal to 0, the series converges.
    1) We may also write that series as: <br />
\sum\limits_{k = 0}^\infty  {a_k } <br />
where <br />
a_n  = \tfrac{{3 + \left( { - 1} \right)^n }}<br />
{{2 \cdot \left( {\left\lfloor {\tfrac{n}<br />
{2}} \right\rfloor  + 2} \right)}} \cdot \left( { - 1} \right)^n <br />
and <br />
\left\lfloor x \right\rfloor <br />
is the floor function

    So: <br />
\sum\limits_{k = 0}^m {a_k }  = \tfrac{3}<br />
{2} \cdot \sum\limits_{n = 0}^m {\tfrac{{\left( { - 1} \right)^n }}<br />
{{\left\lfloor {\tfrac{n}<br />
{2}} \right\rfloor  + 2}}}  + \tfrac{1}<br />
{2} \cdot \sum\limits_{n = 0}^m {\tfrac{1}<br />
{{\left\lfloor {\tfrac{n}<br />
{2}} \right\rfloor  + 2}}} <br />

    <br />
\sum\limits_{n = 0}^\infty  {\tfrac{{\left( { - 1} \right)^n }}<br />
{{\left\lfloor {\tfrac{n}<br />
{2}} \right\rfloor  + 2}}} <br />
converges by Leibniz Test, whereas <br />
\sum\limits_{n = 0}^\infty  {\tfrac{1}<br />
{{\left\lfloor {\tfrac{n}<br />
{2}} \right\rfloor  + 2}}} <br />
diverges by comparison with the Harmonic Series

    2) Converges by Leibniz test, the sequence <br />
\tfrac{1}<br />
{{2n + 1}}<br />
is decreasing and converges to 0, so the alternating series converges
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  5. #5
    MHF Contributor arbolis's Avatar
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    If this series is convergent it is conditionaly convergent. But a conditionaly convergent series can be rearranged to sum to anything, so you cannot rearrange the terms.
    Thanks to precise it. I was not sure I could do this. That's mainly why I wanted a check for my answer.
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