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Math Help - Convergence of sin(1/n) - sin(1/(n+1)) and arctg(n)

  1. #1
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    Convergence of sin(1/n) - sin(1/(n+1)) and arctg(n)

    Hi,

    Still me with some limit problems...

    Now I just don't any clues about what to do with trigonometric functions...

    I have this :

    \sum_{n=1}^{\infty}sin(\frac{1}{n}) - sin(\frac{1}{n+1})

    I know it becomes this :

    \sum_{n=1}^{\infty}sin(\frac{1}{n}) - \sum_{n=1}^{\infty}sin(\frac{1}{n+1})

    I know that the first one will tend towards 0 since sin(1/infinity) = sin(0) = 0

    The second one, believe it or not, I'm not so sure about how to find its limit.

    But I know that the limits of those two blocks isn't the answer of the question, that I have to look at their "sn"'s, but how would I be able to estimate sin(1) + sin(1/2) + sin(1/2) + ... and the other one that I won't write here ?

    Same thing for the following, I'm clueless...

    \sum_{n=1}^{\infty}arctg(n)

    Please help, I'm kind of lost right now.
    Last edited by NZAU1984; October 12th 2012 at 02:11 PM.
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  2. #2
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    Re: Convergence of sin(1/n) - sin(1/(n+1)) and arctg(n)

    Quote Originally Posted by NZAU1984 View Post
    I have this :
    \sum_{n=1}^{\infty}sin(\frac{1}{n}) - sin(\frac{1}{n+1})

    \sum_{n=1}^{\infty}arctg(n)
    Like the other you posted this is a collapsing sum.
    If S_N=\sum_{n=1}^{N}sin(\frac{1}{n}) - sin(\frac{1}{n+1}) then S_N  = \left[ {\sin (1) - \sin \left( {\frac{1}{{n + 1}}} \right)} \right].

    So \left( {S_N } \right) \to ?


    Because (\arctan(n))\to\frac{\pi}{2} can \sum_{n=1}^{\infty}\arctan(n) converge?
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