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Thread: find the sum of a series q

  1. #1
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    find the sum of a series q

    find the sum of sigma from n =1 to infinity of [arctan(n+1)- arctann]
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  2. #2
    MHF Contributor chisigma's Avatar
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    $\displaystyle \sum_{n=1}^{\infty} \{\tan^{-1} (n+1) - \tan^{-1} (n) \} = \sum_{n=1}^ {\infty} \int_{n}^{n+1} \frac{dx}{1+x^{2}} = \int_{1}^{\infty} \frac{dx}{1+x^{2}} = \frac{\pi}{4}$

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
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    how did u get 1/(1-x)^2?
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    MHF Contributor chisigma's Avatar
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    Remember that...

    $\displaystyle \int \frac{dx}{1+x^{2}} = \tan^{-1} x + c $

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  5. #5
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    Hello, twilightstr!

    Find the sum of: .$\displaystyle S \;=\;\sum^{\infty}_{n=1}\bigg[\arctan(n+1)- \arctan(n)\bigg]$
    We have:

    . . $\displaystyle S \;=\;\bigg[{\color{red}\rlap{////////}}\arctan(2) - \arctan(1)\bigg] +$ $\displaystyle \bigg[{\color{green}\rlap{////////}}\arctan(3) - {\color{red}\rlap{////////}}\arctan(2)\bigg] +$ $\displaystyle \bigg[{\color{blue}\rlap{////////}}\arctan(4) - {\color{green}\rlap{////////}}\arctan(3)\bigg] + \hdots $

    . . $\displaystyle S \;=\;-\arctan(1)$

    . . $\displaystyle S\;=\;-\frac{\pi}{4}$

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  6. #6
    Senior Member Spec's Avatar
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    Aren't you forgetting the term $\displaystyle \arctan \infty$

    $\displaystyle S=-\frac{\pi}{4}+\lim_{n \to \infty}\arctan (n+1)=-\frac{\pi}{4}+\frac{\pi}{2}=\frac{\pi}{4}$
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  7. #7
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    Hello, Spec!

    Aren't you forgetting the term $\displaystyle \arctan \infty$ ?

    $\displaystyle S\:=\:-\frac{\pi}{4}+\lim_{n \to \infty}\arctan (n+1)=-\frac{\pi}{4}+\frac{\pi}{2}=\frac{\pi}{4}$

    Yes, I did!

    I forgot to look at the other end of the series . . . *blush*

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