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Math Help - Prove that the series 1/n^2 converges

  1. #1
    EMR
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    Prove that the series 1/n^2 converges

    Prove that the series whose terms are 1/n^2 converges by showing that the partial sums form a Cauchy sequence.

    I've tried to start this as follows: Assuming that m>n, we have

    |a_n-a_m|=1/m^2+1/(m+1)^2+...+1/(n+1)^2 <= (m-n)/(n+1)^2.

    So to show it's Cauchy, I need to find N such that m,n>N implies |a_n-a_m|<epsilon for any epsilon. But how can I find this? Don't I have to somehow eliminate m from the expression? I don't know how to do this. Help!
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  2. #2
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    Series not sequence

    Quote Originally Posted by EMR View Post
    Prove that the series whose terms are 1/n^2 converges by showing that the partial sums form a Cauchy sequence.

    I've tried to start this as follows: Assuming that m>n, we have

    |a_n-a_m|=1/m^2+1/(m+1)^2+...+1/(n+1)^2 <= (m-n)/(n+1)^2.

    So to show it's Cauchy, I need to find N such that m,n>N implies |a_n-a_m|<epsilon for any epsilon. But how can I find this? Don't I have to somehow eliminate m from the expression? I don't know how to do this. Help!


    I think what you need to show is this is less than epsilon


    S_n=\sum_{i=1}^{n}\frac{1}{i^2} and S_m=\sum_{i=1}^{m}\frac{1}{i^2}

    then
    |S_m-S_n|=|\sum_{i=1}^{m}\frac{1}{i^2}-\sum_{i=1}^{n}\frac{1}{i^2}|=|\sum_{i=n}^{m}\frac{  1}{i^2}|

    See what you can do from here.
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  3. #3
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    Quote Originally Posted by EMR View Post
    Prove that the series whose terms are 1/n^2 converges by showing that the partial sums form a Cauchy sequence.

    I've tried to start this as follows: Assuming that m>n, we have

    |a_n-a_m|=1/m^2+1/(m+1)^2+...+1/(n+1)^2 <= (m-n)/(n+1)^2.

    So to show it's Cauchy, I need to find N such that m,n>N implies |a_n-a_m|<epsilon for any epsilon. But how can I find this? Don't I have to somehow eliminate m from the expression? I don't know how to do this. Help!
    You might be interested to know that the series converges to the value \frac{\pi^2}{6}. A (non-rigorous) proof was first given by Euler. It can be proved rigorously and easily using Fourier analysis.
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  4. #4
    EMR
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    Quote Originally Posted by TheEmptySet View Post
    I think what you need to show is this is less than epsilon


    S_n=\sum_{i=1}^{n}\frac{1}{i^2} and S_m=\sum_{i=1}^{m}\frac{1}{i^2}

    then
    |S_m-S_n|=|\sum_{i=1}^{m}\frac{1}{i^2}-\sum_{i=1}^{n}\frac{1}{i^2}|=|\sum_{i=n}^{m}\frac{  1}{i^2}|

    See what you can do from here.
    Yes, that's what I meant...couldn't figure out how to do the Latex code here.

    Okay, so I need to show that
    |S_m-S_n||\sum_{i=1}^{m}\frac{1}{i^2}-\sum_{i=1}^{n}\frac{1}{i^2}|=|\sum_{i=n+1}^{m}\fra  c{1}{i^2}|\le \frac{m-n}{(n+1)^2}<\frac{m}{n^2}<\epsilon.

    But again I am faced with finding an N such that m,n>N implies that this is less than any given epsilon. But to do that, I should somehow eliminate m, yes?

    Thanks again!
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  5. #5
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    You have to show that \sum_{k=m+1}^n \frac{1}{k^2}  < \epsilon for n>m. Note, \frac{1}{k^2} \leq \frac{1}{(m+1)^2} for m+1\leq k \leq n. Thus, \sum_{k=m+1}^n \frac{1}{k^2} \leq \sum_{k=m+1}^n \frac{1}{(m+1)^2} = \frac{n-m}{(m+1)^2} \leq \frac{n}{m^2}. The problem is this is a bad estimate, you cannot make this quanity sufficiently small.
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  6. #6
    EMR
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    So is it even possible to directly show that the sequence of partial sums form a Cauchy sequence directly? Clearly they do, because the series itself converges, but I can't show it directly from the definition.
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  7. #7
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    Quote Originally Posted by EMR View Post
    So is it even possible to directly show that the sequence of partial sums form a Cauchy sequence directly? Clearly they do, because the series itself converges, but I can't show it directly from the definition.
    It is possible to prove this using an integral estimate.

    Consider the function, f(x) = \frac{1}{x^2} defined on the positive reals. We will place a good bound on the quantity \sum_{k=m+1}^n \frac{1}{k^2} where n>m\geq 1. Consider the function f(x) on the interval [m,n]. The area is equal to \int_m^n \frac{dx}{x^2} = \frac{1}{m} - \frac{1}{n}. Now consider the right-endpoint rectangular approximation of the area equally spaced at integer points between m and n (each sub-rectangle has width 1) this approximation is clearly smaller then the area, furthermore its quantity is \sum_{k=m+1}^n \frac{1}{k^2}.

    Thus, \sum_{k=m+1}^n \frac{1}{k^2} \leq \frac{1}{m} - \frac{1}{n} \leq \frac{1}{m}. Note, 1/m can be made arbitrary small if we pick N so large that n>m>N.
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    \sum_{k=n}^m \frac{1}{k^2} < \sum_{k=n}^m \frac{1}{k(k-1)} = \sum_{k=n}^m \left( \frac{1}{k-1} - \frac{1}{k} \right) = \frac{1}{n-1} - \frac{1}{m} < \frac{1}{n-1}
    Last edited by awkward; March 27th 2008 at 07:46 PM. Reason: changed i to k just for the heck of it
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