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Math Help - Partial sums of alternating Harmonic series satisfy Cauchy Criterion?

  1. #1
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    Partial sums of alternating Harmonic series satisfy Cauchy Criterion?

    Hello

    I am a little stuck in seeing how the partial sums of an alternating Harmonic series satisfy the cauchy condition.

    i.e. The partial sums are defined as s_n=\underset{i=0}{\overset{n}{\sum }}\frac{(-1)^i}{i}

    This yields for n<m

    \left|s_m-s_n\right|=\left|\frac{1}{n+1}-\frac{1}{n+2}+\text{...} \pm \frac{1}{m}\right|<\frac{1}{n}

    Its this last inequality I am having trouble verifying. Any ideas?
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  2. #2
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    Quote Originally Posted by aukie View Post
    Hello

    I am a little stuck in seeing how the partial sums of an alternating Harmonic series satisfy the cauchy condition.

    i.e. The partial sums are defined as s_n=\underset{i=0}{\overset{n}{\sum }}\frac{(-1)^i}{i}

    This yields for n<m

    \left|s_m-s_n\right|=\left|\frac{1}{n+1}-\frac{1}{n+2}+\text{...} \pm \frac{1}{m}\right|<\frac{1}{n}

    Its this last inequality I am having trouble verifying. Any ideas?

    You can try induction on m\,,\,m>n:

    First, it is not hard to see that \left|\frac{1}{n+1}-\frac{1}{n+2}+\text{...} \pm \frac{1}{m}\right|=\frac{1}{n+1}-\frac{1}{n+2}+\text{...} \pm \frac{1}{m} .

    Now we have two cases: either m = n+2k\,,\,k\in \mathbb{N} , or m=n+2(k+1)\,,\,k\in\mathbb{N} .
    In the first case it is trivial, whereas in the second case you can use inductive hypothesis not with m-1 but with m-2, since

    if \frac{1}{n+1}-\frac{1}{n+2}+\text{...} \pm \frac{1}{m-1}<\frac{1}{n} , then \frac{1}{n+1}-\frac{1}{n+2}+\text{...} \pm\frac{1}{m-1}\mp \frac{1}{m}\pm\frac{1}{m+1}<\frac{1}{n}\mp\frac{1}  {m}\pm\frac{1}{m+1}

    Tonio
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  3. #3
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    Hello Tonio

    In working out why the absolute value can be ommitted and what you meant by trivial. I seen an argument that doesn't require induction. If you're interested: Assuming m = n + k,

    \frac{1}{n+1} - \frac{1}{n+2} + \text{...} \pm \frac{1}{n+k}

    Group the terms as follows (for even k)

     \frac{1}{n+1} - (\frac{1}{n+2} - \frac{1}{n+3}) - \text{...} - (\frac{1}{n+(k-1)} - \frac{1}{n+k})

    (and for odd k)

     \frac{1}{n+1} - (\frac{1}{n+2} - \frac{1}{n+3}) - \text{...} - ( \frac{1}{n+k})

    Then we have a series of positive terms subtracted from  \frac{1}{n+1} so

    \frac{1}{n+1} - \frac{1}{n+2} + \text{...} \pm \frac{1}{n+k} < \frac{1}{n+1} < \frac{1}{n}

    Thanks for your input
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  4. #4
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    Quote Originally Posted by aukie View Post
    Hello Tonio

    In working out why the absolute value can be ommitted and what you meant by trivial. I seen an argument that doesn't require induction. If you're interested: Assuming m = n + k,

    \frac{1}{n+1} - \frac{1}{n+2} + \text{...} \pm \frac{1}{n+k}

    Group the terms as follows (for even k)

     \frac{1}{n+1} - (\frac{1}{n+2} - \frac{1}{n+3}) - \text{...} - (\frac{1}{n+(k-1)} - \frac{1}{n+k})

    (and for odd k)

     \frac{1}{n+1} - (\frac{1}{n+2} - \frac{1}{n+3}) - \text{...} - ( \frac{1}{n+k})

    Then we have a series of positive terms subtracted from  \frac{1}{n+1} so

    \frac{1}{n+1} - \frac{1}{n+2} + \text{...} \pm \frac{1}{n+k} < \frac{1}{n+1} < \frac{1}{n}

    Thanks for your input


    I think it works just fine! Yet even in this case we have what's sometimes called "concealed or hidden" induction: it is a weak form and it just could be stressed that what you did actually works no matter what n in \frac{1}{n+k} you used...

    Tonio
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