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Math Help - Infinite Series Convergence or Divergence

  1. #1
    Junior Member EliteAndoy's Avatar
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    Infinite Series Convergence or Divergence

    Hi everyone. Today we are asked to do determine if this infinite series, \sum_{n=1}^{\infty}\left [ \arctan (x+2)-\arctan(x) \right ], converges or diverges and if it converges, to find it's value. Divergence test does not help in determining the series' behavior, so what I did is I did the integral test since the sequence can be represented as a continuous, positive, and decreasing function at the interval [1,\infty). Turns out this series is convergent and I have to evaluate this ugly series. By the looks of it, it looks like a telescopic series so by making a diagram of the partial sums up to n=n, everything cancels out except \left [ -\arctan(1)-\arctan(2) \right ]+\left [ \arctan(n+1)+\arctan(n+2) \right ]. So I've figured that taking the \lim_{n\to{\infty}}[\left [ -\arctan(1)-\arctan(2) \right ]+\left [ \arctan(n+1)+\arctan(n+2) \right ]] would give me the value for the series. So I ended up with \sum_{n=1}^{\infty}\left [ \arctan (x+2)-\arctan(x) \right ]=\frac{3\pi}{4}-\arctan(2). Is this solution correct? What do you guys think? Thanks everyone for your help!
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    Re: Infinite Series Convergence or Divergence

    I agree with your solution. Well done
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    Junior Member EliteAndoy's Avatar
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    Re: Infinite Series Convergence or Divergence

    Thanks man. I was really pretty confident with my solution, but the thing that made me a bit skeptical is, if you plot the sequence of partial sums as an element of real numbers that is not exclusive to natural numbers, it would look like it is converging at 1.2 or something close to that. Hmm, got any idea as to why?
    Last edited by EliteAndoy; March 26th 2013 at 05:30 PM.
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    Re: Infinite Series Convergence or Divergence

    Hello, EliteAndoy!

    S \;=\;\sum_{n=1}^{\infty}\lbig[ \arctan (x+2)-\arctan(x) \big ]

    We have:

    S \;=\;({\color{red}\rlap{///////}}\arctan3-\arctan1) + ({\color{green}\rlap{///////}}\arctan4 - \arctan2) + ({\color{blue}\rlap{///////}}\arctan5 - {\color{red}\rlap{///////}}\arctan3)

    . . . . .  + ({\color{red}\rlap{///////}}\arctan6 - {\color{green}\rlap{///////}}\arctan4) + (\arctan7 - {\color{blue}\rlap{///////}}\arctan 5) + (\arctan 8 - {\color{red}\rlap{///////}}\arctan 6) + \hdots


    Therefore: . S \;=\;-\arctan1 - \arctan2


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    If you want to impress/surprise/terrify your teacher,
    . . you can demonstrate that: . S\,=\,\arctan3
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    Junior Member EliteAndoy's Avatar
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    Re: Infinite Series Convergence or Divergence

    But isn't there supposed to be a nth term evaluated in the series? Like if you think about it, as n approaches n in the sequence of partial sums, what is left is [-arctan(1)-arctan(2)]+[arctan(n+1)+arctan(n+2)]. Right? Correct me if I'm wrong but isn't there supposed to be that (arctan(n+1)+arctan(n+2) at the end because those wouldn't cancel themselves out? Thanks by the way man.
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  6. #6
    Junior Member EliteAndoy's Avatar
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    Re: Infinite Series Convergence or Divergence

    Quote Originally Posted by EliteAndoy View Post
    Thanks man. I was really pretty confident with my solution, but the thing that made me a bit skeptical is, if you plot the sequence of partial sums as an element of real numbers that is not exclusive to natural numbers, it would look like it is converging at 1.2 or something close to that. Hmm, got any idea as to why?
    Oops on my last formal calculation I accidentally entered arctan(2) as tan(2), that's why my graph seem to converge somewhere else lol. It's all resolved now. Thank you guys!
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