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Math Help - Find the sum...?

  1. #1
    Super Member fardeen_gen's Avatar
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    Find the sum...?

    Find the sum of arcsin 1/√2 + arcsin (√2 - 1)/√6 + ... + arcsin {√n - √(n - 1)}/√{n(n + 1)}, for n approaching infinity?
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  2. #2
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    I suggest first checking that the series converges.

    Calcuating a_n for a few large n suggests that the ratio test would be inconclusive (ratio = 1).

    Using Raabe's test for few large n suggests that the series may be divergent. However, while Raabe's test can be used to show that a series is convergent I am not sure that the converse is true.

    I suspect that one of the convergence tests in your toolbox may show that this series diverges.
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  3. #3
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    Quote Originally Posted by fardeen_gen View Post
    Find the sum of arcsin 1/√2 + arcsin (√2 - 1)/√6 + ... + arcsin {√n - √(n - 1)}/√{n(n + 1)}, for n approaching infinity?
    let a_n=\arcsin \left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}}\right). then: \sin a_n=\frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}} and hence: \cos a_n=\frac{1+\sqrt{n(n-1)}}{\sqrt{n(n+1)}}. therefore: \tan a_n=\frac{\sqrt{n}-\sqrt{n-1}}{1+\sqrt{n(n-1)}}. thus:

    a_n=\arctan \left(\frac{\sqrt{n} - \sqrt{n-1}}{1+\sqrt{n(n-1)}}\right)=\arctan \sqrt{n} - \arctan \sqrt{n-1}. hence: \sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}(\arctan \sqrt{n} - \arctan \sqrt{n-1})=\lim_{n\to\infty} \arctan \sqrt{n}=\frac{\pi}{2}. \ \ \ \ \ \square
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