Results 1 to 5 of 5

Thread: Sum of the series

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    8

    Sum of the series

    How do I find the sum of the series: (sum. n=1, infinity) [2^(n+1)]/pi^n
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2009
    Posts
    186
    Quote Originally Posted by schnek View Post
    How do I find the sum of the series: (sum. n=1, infinity) [2^(n+1)]/pi^n

    This looks to be a geometric series, and you are trying to find the sum as it goes to infinite. So lets break it down

    $\displaystyle \sum_{n=1}^{\infty}\frac{2^{n+1}}{\pi^n}$

    Now If you can remember properties of exponents we can transform the numerator to help us

    $\displaystyle \sum_{n=1}^{\infty}\frac{2^{n}\cdot2^1}{\pi^n}$

    We can pull that out as a constant multiplicative of this geometric series

    $\displaystyle \sum_{n=1}^{\infty}2\cdot\frac{2^{n}}{\pi^n}$


    $\displaystyle \sum_{n=1}^{\infty}2\cdot\frac{2^{n}}{\pi^n}$

    We can use another property of exponents here


    $\displaystyle \sum_{n=1}^{\infty}2\cdot(\frac{2}{\pi})^n$

    Now the rest is easy, Our constant multiple lets call it a which is the first term in the series, is $\displaystyle a = 2\cdot\frac{2}{\pi}$
    and your common ration r is $\displaystyle r=\frac{2}{\pi}$, now if we recall the formula for a sum of the geometric series as it goes to infinity

    $\displaystyle \frac{a}{1-r}$

    You should be able to get the rest
    Last edited by RockHard; Nov 23rd 2009 at 06:58 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    One small caveat...the sum starts at 1, so $\displaystyle a$ should be $\displaystyle 2\cdot\frac{2}{\pi}$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    Hello, schnek!

    Find the sum of the series: .$\displaystyle S \;=\;\sum^{\infty}_{n=1} \frac{2^{n+1}}{\pi^n} $
    We have: .$\displaystyle S \;=\;\frac{2^2}{\pi} + \frac{2^3}{\pi^2} + \frac{2^4}{\pi^3} + \hdots$

    This is a geometric series with first term $\displaystyle a = \frac{4}{\pi}$ and common ratio $\displaystyle r = \frac{2}{\pi}$

    The sum is: .$\displaystyle S \;=\;\frac{a}{1-r} \;=\;\frac{\frac{4}{\pi}}{1 - \frac{2}{\pi}} \;=\;\frac{4}{\pi-2}$

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2009
    Posts
    186
    Aw, I knew I was gonna make a careless mistake, that the numerator is the first term in the series. , Thanks Sox, learned from this mistake
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Oct 3rd 2011, 01:12 AM
  2. Replies: 3
    Last Post: Sep 29th 2010, 06:11 AM
  3. Replies: 0
    Last Post: Jan 26th 2010, 08:06 AM
  4. Replies: 1
    Last Post: May 5th 2008, 09:44 PM
  5. Replies: 11
    Last Post: Apr 1st 2008, 12:06 PM

Search Tags


/mathhelpforum @mathhelpforum