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Math Help - Showing the following sum is finite

  1. #1
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    Showing the following sum is finite

    Hi guys,

    I'm having real trouble proving the following

    -\sum^{\infty}_{n=1}\log \left(1 - \frac{x^2}{n^2}\right) < \infty

    Every avenue I take reaches a dead end!

    Thank you all very much in advance,

    HTale
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  2. #2
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    Quote Originally Posted by HTale View Post
    Hi guys,

    I'm having real trouble proving the following

    -\sum^{\infty}_{n=1}\log \left(1 - \frac{x^2}{n^2}\right) < \infty

    Every avenue I take reaches a dead end!

    Thank you all very much in advance,

    HTale
    If you write out the terms

     - \ln \left( 1 - \frac{x^2}{1^2}\right) - \ln \left( 1 - \frac{x^2}{2^2}\right) - \ln \left( 1 - \frac{x^2}{3^2}\right) - \ln \left( 1 - \frac{x^2}{4^2}\right) \cdots

    Then we can combine the log's giving

    - \ln \prod_{i = 1}^\infty \left( 1 - \frac{x^2}{n^2}\right)

    The product has a nice closed form

    \sin \pi x = \pi x \prod_{i = 1}^\infty \left( 1 - \frac{x^2}{n^2}\right)

    so what you have is

    - \ln \frac{\sin \pi x}{\pi x} .

    It might be easier to deal with now.
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  3. #3
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    I think you should assume that |x| < 1 (otherwise the first term of the series will not be defined, being the log of a negative number).

    Now use the power series expansion for –log(1–t) (valid when 0<t<1):

    \begin{aligned}-\log(1-t) &= t + \frac{t^2}2 + \frac{t^3}3 + \frac{t^4}4 + \ldots \\<br />
&\leqslant t + t^2 + t^3 + t^4 + \ldots = \frac t{1-t}.\end{aligned}

    Substitute  t = x^2/n^2 to see that -\log\Bigl(1-\frac{x^2}{n^2}\Bigr) \leqslant \frac{x^2}{n^2-x^2} and use the comparison test with \sum\frac1{n^2}.
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  4. #4
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    Quote Originally Posted by Opalg View Post
    I think you should assume that |x| < 1 (otherwise the first term of the series will not be defined, being the log of a negative number).

    Now use the power series expansion for –log(1–t) (valid when 0<t<1):

    \begin{aligned}-\log(1-t) &= t + \frac{t^2}2 + \frac{t^3}3 + \frac{t^4}4 + \ldots \\<br />
&\leqslant t + t^2 + t^3 + t^4 + \ldots = \frac t{1-t}.\end{aligned}

    Substitute  t = x^2/n^2 to see that -\log\Bigl(1-\frac{x^2}{n^2}\Bigr) \leqslant \frac{x^2}{n^2-x^2} and use the comparison test with \sum\frac1{n^2}.
    Thanks for that Opalg. This was incidently one of the routes I took. But, I convinced myself otherwise as my problem here is that the comparision test is set for positive series. So, we have to set x < n (which is what we already have). However, for some fixed x, it's not always the case that

    \frac{x^2}{n^2-x^2} \leq \frac{1}{n^2}

    what do I do/say in my argument to fix this?

    Thanks in advance,

    HTale
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  5. #5
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    Quote Originally Posted by HTale View Post
    Thanks for that Opalg. This was incidentally one of the routes I took. But, I convinced myself otherwise as my problem here is that the comparison test is set for positive series. So, we have to set x < n (which is what we already have). However, for some fixed x, it's not always the case that

    \frac{x^2}{n^2-x^2} \leq \frac{1}{n^2}

    what do I do/say in my argument to fix this?
    Convergence (or otherwise) of a series is unaffected by what happens to finitely many terms at the start of the series. So to apply the comparison test, for some fixed value of x, there is no loss in assuming that n>2x. Then \frac{x^2}{n^2-x^2} will be less than some multiple of \frac{1}{n^2}.
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