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Interval of Convergence for acrtan(x) series Watch

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    So I derived the general formula for each term in the power series of arctan(x) :



    \sum_{n=0}^{n=\infty}\frac{(-1)^n x^{2n+1}}{2n+1}



    I then used the ratio test and simplified to get :

    \lim_{n \to\infty } \left | \frac{2n+1}{2n+3} \right |\left | x^2 \right |

    Giving just lx^2l which means the series only converges for :

    -1<x^2 <1

    At this point I'm stuck on what to do. How do I proceed ? I cant simply take a square root of each side....
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    (Original post by Ari Ben Canaan)
    So I derived the general formula for each term in the power series of arctan(x) :



    \sum_{n=0}^{n=\infty}\frac{(-1)^n x^{2n+1}}{2n+1}



    I then used the ratio test and simplified to get :

    \lim_{n \to\infty } \left | \frac{2n+1}{2n+3} \right |\left | x^2 \right |

    Giving just lx^2l which means the series only converges for :

    -1<x^2 <1

    At this point I'm stuck on what to do. How do I proceed ? I cant simply take a square root of each side....
    There is not x^2 at the limit when you calculate the radius.
    So your radius is 1.
    The series is given around 0
    the range of convergence |x-0|<1
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    (Original post by Ari Ben Canaan)
    Giving just lx^2l which means the series only converges for :

    -1<x^2 <1

    At this point I'm stuck on what to do. How do I proceed ? I cant simply take a square root of each side....
    Why not?

    Spoiler:
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    You might also want to observe that it is always true that 0<= x^2.
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    (Original post by DFranklin)
    Why not?

    Spoiler:
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    You might also want to observe that it is always true that 0<= x^2.
    Sorry, I meant to say taking the square root would mean getting i on the LHS.

    Hmmm, so essentially the interval of convergence is just 0<x<1 ?
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    (Original post by Ari Ben Canaan)
    Hmmm, so essentially the interval of convergence is just 0<x<1 ?
    You're forgetting the "other half"
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    (Original post by Lord of the Flies)
    You're forgetting the "other half"
    I don't quite follow...
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    (Original post by ztibor)
    There is not x^2 at the limit when you calculate the radius.
    So your radius is 1.
    The series is given around 0
    the range of convergence |x-0|<1
    What do you mean there is no x^2 ? Of course there is....
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    (Original post by Ari Ben Canaan)
    I don't quite follow...
    I was hinting at negative values of x (also, don't forget to check endpoints!).
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    (Original post by Lord of the Flies)
    I was hinting at negative values of x (also, don't forget to check endpoints!).
    -1<x<1 ?

    Mu question as to how to deal with the x^2 still hasn't been answered or at least I haven't understood it yet. Could you please just spell out for me ?
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    (Original post by Ari Ben Canaan)
    -1<x<1 ?

    Mu question as to how to deal with the x^2 still hasn't been answered or at least I haven't understood it yet. Could you please just spell out for me ?
    There's nothing to deal with really: 0\leq x^2&lt;1\Rightarrow -1&lt;x&lt;1

    Since the ratio test is inconclusive when x=\pm 1 you need to check if the series converges at these values (it does).
 
 
 
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