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Math Help - Problem finding the limit of the remainder of a series

  1. #1
    Senior Member x3bnm's Avatar
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    Problem finding the limit of the remainder of a series

    Suppose there is a series

    \frac{1}{1+t^2} = 1 -t^2 + t^4 - t^6 + ... + (-1)^n t^{2n} + \frac{(-1)^{(n+1)} t^{(2n+2)}}{1+t^2}

    in which the last term comes from adding the remaining terms as a
    geometric series with the first term a = (-1)^{(n+1)} t^{(2n+2)}
    and ratio r= -t^2 Integrating both sides of equation from
    t = 0 to t = x gives

    tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ... + (-1)^n \frac{x^{(2n+1)}}{2n+1} + R_n(x)<br />

    where

    <br />
R_n(x) = \int_0^x \frac{(-1)^{(n+1)} t^{(2n+2)}}{1+t^2}dt<br />

    The denominator of integrand is greater than or equal to 1;hence

    <br />
\left|R_n(x)\right| \le \int_0^{\left|x\right|} t^{(2n+2)} dt = \frac{\left|x\right|^{(2n+3)}}{2n+3}<br />

    if \left|x\right| \le 1 , the right side of this inequality approaches zero as n \to \infty
    Therefore

    <br />
\lim _{n \to \infty} R_n(x) = 0\,\, if \left|x\right| \le 1<br />


    Now my question is how the book found the limit of R_n(x) as 0?
    Book showed and I understand that the limit of \frac{\left|x\right|^{(2n+3)}}{2n+3} is 0 when n
    approaches \infty. But that's only half of it? The author did not calculate
    the limit of integral R_n(x) as a whole. I don't understand how he
    reached the conclusion? Obviously I'm missing something simple. Is
    there any theorem that i'm unaware of? Anyone knows how
    he reached the conclusion?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by x3bnm View Post
    Suppose there is a series

    \frac{1}{1+t^2} = 1 -t^2 + t^4 - t^6 + ... + (-1)^n t^{2n} + \frac{(-1)^{(n+1)} t^{(2n+2)}}{1+t^2}

    in which the last term comes from adding the remaining terms as a
    geometric series with the first term a = (-1)^{(n+1)} t^{(2n+2)}
    and ratio r= -t^2 Integrating both sides of equation from
    t = 0 to t = x gives

    tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + ... + (-1)^n \frac{x^{(2n+1)}}{2n+1} + R_n(x)<br />

    where

    <br />
R_n(x) = \int_0^x \frac{(-1)^{(n+1)} t^{(2n+2)}}{1+t^2}dt<br />

    The denominator of integrand is greater than or equal to 1;hence

    <br />
\left|R_n(x)\right| \le \int_0^{\left|x\right|} t^{(2n+2)} dt = \frac{\left|x\right|^{(2n+3)}}{2n+3}<br />

    if \left|x\right| \le 1 , the right side of this inequality approaches zero as n \to \infty
    Therefore

    <br />
\lim _{n \to \infty} R_n(x) = 0\,\, if \left|x\right| \le 1<br />


    Now my question is how the book found the limit of R_n(x) as 0?
    Book showed and I understand that the limit of \frac{\left|x\right|^{(2n+3)}}{2n+3} is 0 when n
    approaches \infty. But that's only half of it? The author did not calculate
    the limit of integral R_n(x) as a whole. I don't understand how he
    reached the conclusion? Obviously I'm missing something simple. Is
    there any theorem that i'm unaware of? Anyone knows how
    he reached the conclusion?
    I don't understand your question.

    The author showed that the remainder in the series for arctan goes to zero as n becomes arbitrarily large for all x such that |x|\le 1, what more do you need?

    CB
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