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Math Help - Help getting started in these particular series questions

  1. #1
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    Help getting started in these particular series questions



    I'm supposed to determine whether the above series are convergent or divergent by using the telescoping sum. If it's convergent, I'm supposed to calculate the sum. Using my calculator I know the first one is diverges and the second one converges.

    I've expanded the fraction in the first series and showed how when added, the telescoping sums do not cancel out. Is this adequate to show that the series is divergent?

    The second series, however, I have no clue how to go about. I can't seem to find any way to expand the series into canceling telescoping sums.
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  2. #2
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    Hello,
    Quote Originally Posted by diablo2121 View Post


    I'm supposed to determine whether the above series are convergent or divergent by using the telescoping sum. If it's convergent, I'm supposed to calculate the sum. Using my calculator I know the first one is diverges and the second one converges.

    I've expanded the fraction in the first series and showed how when added, the telescoping sums do not cancel out. Is this adequate to show that the series is divergent?

    The second series, however, I have no clue how to go about. I can't seem to find any way to expand the series into canceling telescoping sums.
    For the second one...

    \sum_{n=1}^\infty \left(\cos \tfrac{1}{n^2}-\cos \tfrac{1}{(n+1)^2}\right)=\lim_{k \to \infty} \sum_{n=1}^k \left(\cos \tfrac{1}{n^2}-\cos \tfrac{1}{(n+1)^2}\right)

    But this is a telescoping series

    \sum_{n=1}^k \left(\cos \tfrac{1}{n^2}-\cos \tfrac{1}{(n+1)^2}\right)=\cos \tfrac{1}{1^2}-\cos \tfrac{1}{2^2}+\cos \tfrac{1}{2^2}-\cos \tfrac{1}{3^2}+ \dots +\cos \tfrac{1}{(k-1)^2}-\cos \tfrac{1}{k^2}+\cos \tfrac{1}{k^2}-\cos \tfrac{1}{(k+1)^2}

    _______________________________________________
    For the first one :

    \sum_{n=1}^\infty=\lim_{k \to \infty} \sum_{n=1}^k \ln \left(\frac{n}{n+1}\right)

    Now, remember that : from \ln(a)+\ln(b)=\ln(ab), we can generalize it :

    \sum_{p=1}^q \ln(p)=\ln \left(\prod_{p=1}^q p\right)

    this will simplify much your series
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  3. #3
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    Quote Originally Posted by Moo View Post

    But this is a telescoping series

    \sum_{n=1}^k \left(\cos \tfrac{1}{n^2}-\cos \tfrac{1}{(n+1)^2}\right)=\cos \tfrac{1}{1^2}-\cos \tfrac{1}{2^2}+\cos \tfrac{1}{2^2}-\cos \tfrac{1}{3^2}+ \dots +\cos \tfrac{1}{(k-1)^2}-\cos \tfrac{1}{k^2}+\cos \tfrac{1}{k^2}-\cos \tfrac{1}{(k+1)^2}
    I can't believe I missed that!

    Thanks for the clarification.
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