Results 1 to 2 of 2

Math Help - Prove the summation

  1. #1
    Junior Member
    Joined
    Jun 2008
    Posts
    54

    Prove the summation

    \sum_{n=1} ^\infty \; \frac{\theta}{2^n}\;\tan\; \frac{\theta}{2^n} \;=\; \boxed{1 - \theta \;\cot\;\theta}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by mathwizard View Post
    \sum_{n=1} ^\infty \; \frac{\theta}{2^n}\;\tan\; \frac{\theta}{2^n} \;=\; \boxed{1 - \theta \;\cot\;\theta}
    start with this identity: \tan \alpha = \cot \alpha - 2 \cot(2\alpha). so: \frac{\theta}{2^j} \tan \left(\frac{\theta}{2^j} \right)=\frac{\theta}{2^j} \cot \left(\frac{\theta}{2^j} \right) - \frac{\theta}{2^{j-1}} \cot \left(\frac{\theta}{2^{j-1}} \right). thus:

    I_n=\sum_{j=1}^n \frac{\theta}{2^j} \tan \left(\frac{\theta}{2^j} \right)=\sum_{j=1}^n \left[\frac{\theta}{2^j} \cot \left(\frac{\theta}{2^j} \right)-\frac{\theta}{2^{j-1}} \cot \left(\frac{\theta}{2^{j-1}} \right) \right] \ \ \ \ \ \ \text{(telescoping sum)}

    =\frac{\theta}{2^n} \cot \left(\frac{\theta}{2^n} \right) - \theta \cot \theta. therefore: \lim_{n\to\infty} I_n= \lim_{n\to\infty} \frac{\theta}{2^n} \cot \left(\frac{\theta}{2^n} \right)- \theta \cot \theta=1 - \theta \cot \theta. \ \ \ \square
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Summation
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 20th 2011, 11:55 PM
  2. how do i prove this summation?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 15th 2011, 03:28 PM
  3. Summation of summation?
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 3rd 2010, 06:42 PM
  4. Induction to prove bounding summation
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 30th 2010, 03:42 AM
  5. Summation question - Prove
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 5th 2009, 12:42 PM

Search Tags


/mathhelpforum @mathhelpforum