Results 1 to 5 of 5

Math Help - sum of a series

  1. #1
    aston
    Guest

    sum of a series

    hi, i have a series 1-1/3+1/5-1/7+..., which i think is also called the Leibniz formula, i know that to find the nth term you can can use (-1)^n/(2n+1), but i need help trying to find a formula or a way to find the sum of the series for a certain number of terms i.e. the sum of the first 10 terms, can anyone help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Your serie is equal to \frac\pi4.

    Consider \arctan x = \int_0^x {\frac{1}<br />
{{1 + u^2 }}\,du} . So,

    \arctan x = \int_0^x {\frac{1}<br />
{{1 + u^2 }}\,du}  = \int_0^x {\left\{ {\sum\limits_{k = 0}^\infty  {( - 1)^k u^{2k} } } \right\}\,du} .

    Finally

    \arctan x = \sum\limits_{k = 0}^\infty  {( - 1)^k \left\{ {\int_0^x {u^{2k} \,du} } \right\}}  = \sum\limits_{k = 0}^\infty  {\frac{{( - 1)^k x^{2k + 1} }}<br />
{{2k + 1}}} ,\,\left| x \right| \le 1.

    Now set x=1 and we're done.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Krizalid View Post
    Your serie is equal to \frac\pi4.

    Consider \arctan x = \int_0^x {\frac{1}<br />
{{1 + u^2 }}\,du} . So,

    \arctan x = \int_0^x {\frac{1}<br />
{{1 + u^2 }}\,du}  = \int_0^x {\left\{ {\sum\limits_{k = 0}^\infty  {( - 1)^k u^{2k} } } \right\}\,du} .

    Finally

    \arctan x = \sum\limits_{k = 0}^\infty  {( - 1)^k \left\{ {\int_0^x {u^{2k} \,du} } \right\}}  = \sum\limits_{k = 0}^\infty  {\frac{{( - 1)^k x^{2k + 1} }}<br />
{{2k + 1}}} ,\,\left| x \right| \le 1.

    Now set x=1 and we're done.
    You cannot do that. Because \frac{1}{1+u^2} = \sum_{k\geq 0} (-1)^k u^{2k} only when |u|<1. What you are doing is letting u=1 and it does not converge there.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by aston View Post
    hi, i have a series 1-1/3+1/5-1/7+..., which i think is also called the Leibniz formula, i know that to find the nth term you can can use (-1)^n/(2n+1), but i need help trying to find a formula or a way to find the sum of the series for a certain number of terms i.e. the sum of the first 10 terms, can anyone help?
    Proof 1: First note that \frac1{1+x^2} = 1 - x^2 + x^4 -... for |x|<1. This tells us that (integrate both sides) \tan^{-1} x = x - \frac{x^3}{3}+\frac{x^5}{5}-... for |x|<1. Now to justify what Krizalid did is to use Abel's Theorem which tells us that the series f(x) is continous at x=1 because it converges there (by alternating series test). Thus, if x_k\to 1 is a sequence in (-1,1) then \lim ~ f(x_k) = f(\lim x_k) thus f(1) = \lim ~ f(x_k) = \lim ~ \tan^{-1} (x_k) = \frac{\pi}{4}.

    Abel's Theorem: If f(x) = \sum_{n=0}^{\infty}a_nx^n is a power series of radius of convergence of radius 0<R<\infty such that the series converges at x=R then f is continous at x=R.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Here is a more elementary proof.

    Proof 2: For any x we have \sum_{k=0}^{n}(-1)^k x^{2k} = \frac{1 +(-1)^{n+1} x^{2n+1}}{1+x^2}. Thus, \int_0^1 \sum_{k=0}^n (-1)^k x^{2k} dx = \int_0^1 \frac{dx}{x^2+1} \pm \int_0^1 \frac{x^{2n+1}}{x^2+1} dx where \pm depends on the parity of n. This tells us that \sum_{k=0}^n \frac{(-1)^k x^{2k+1}}{2k+1} = \tan^{-1} 1 \pm \int_0^1 \frac{x^{2n+1}}{x^2+1} dx. Thus, \left| \sum_{k=0}^n \frac{(-1)^k x^{2k+1}}{2k+1}  - \frac{\pi}{4}\right| = \int_0^1 \frac{x^{2n+1}}{x^2+1} dx. Now \frac{x^{2n+1}}{x^2+1} \leq x^{2n+1}. This means, \left| \sum_{k=0}^n \frac{(-1)^k x^{2k+1}}{2k+1}  - \frac{\pi}{4}\right| \leq \int_0^1 x^{2n+1} dx = \frac{1}{2n+2} \leq \frac{1}{n}. This means for any \epsilon > 0 if we chose N> \frac{1}{\epsilon} it will mean that \sum_{k=0}^N \frac{(-1)^kx^{2k+1}}{2k+1} is within \epsilon of \frac{\pi}{4}. And so, by definition, this series converges to \frac{\pi}{4}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: October 3rd 2011, 02:12 AM
  2. Replies: 3
    Last Post: September 29th 2010, 07:11 AM
  3. Replies: 0
    Last Post: January 26th 2010, 09:06 AM
  4. Replies: 2
    Last Post: December 1st 2009, 01:45 PM
  5. Replies: 1
    Last Post: May 5th 2008, 10:44 PM

Search Tags


/mathhelpforum @mathhelpforum