Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sum of series- Fibonacci numerator, geometric denominator

  1. Jul 6, 2012 #1
    I have a series presented to me that goes something like this- 1/1+1/2+2/4+3/8+5/16+8/32+... I am aware that it is a sum to inifinity problem and the common ratio of the bottom is 2 but I don't know what to do with the numerator.
     
  2. jcsd
  3. Jul 6, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi Apollonian! Welcome to PF! :smile:

    Since the numerator is the fibonacci sequence, it obeys the recurrence relation an+2 - an+1 -an = 0,

    so solve that to get an = ABn + CDn :wink:
     
  4. Jul 6, 2012 #3

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    If the terms were 1, 1, 2, 3, 5, 8 (a Fibonacci sequence) it satisfies ##a_{n+1} = a_n + a_{n-1}##.

    Comparing this with your sequence, you need to compensate for the powers of 2, so ##4a_{n+1} = 2a_n + a_{n-1}##.
     
  5. Jul 10, 2012 #4
    the numerator increases at the rate of roughly 1.618, or more precicely, (1+sqrt(5))/2.
    the general formula for finding the sum of such a continued fraction is
    1/(1 -r)
    so since r is (1+sqrt(5))/4....
    can you solve from there?
     
  6. Jul 10, 2012 #5

    Mentallic

    User Avatar
    Homework Helper

    There are many things wrong with this.

    Firstly, the rate is "roughly" r, not exactly r.
    Secondly, that formula only works for infinite sums when |r|<1
     
  7. Jul 10, 2012 #6
    replace the numerator with the Binet formula for the nth term of a Fibonacci series gives each term in terms of n.
    [itex]\frac{(1+\sqrt{5})^{n}-(1-\sqrt{5})^{n}}{2^{n} \sqrt{5}}[/itex]
    Multiplying that by [itex]\frac{1}{2^{n}}[/itex] yields each term as:
    [itex]\frac{(1+\sqrt{5})^{n}-(1-\sqrt{5})^{n}}{2^{2n} \sqrt{5}}[/itex]
     
  8. Jul 16, 2012 #7
    If f(x)=1+1x+2x^2+3x^3+5x^4+8x^5......
    f(x)*x=1x+1x^2+2x^3+3x^4+5x^5......
    f(x)*x^2 =1x^2+1x^3+2x^4+3x^5.......
    Then f(x)-x*f(x)-x^2*f(x)=1 , so f(x)(1-x-x^2)=1 , so f(x)=1/(1-x-x^2)
    Your series is this function evaluated at x=1/2 , which has a value of 4.
    Isn't that something? :)
     
  9. Jul 17, 2012 #8
    The series has a value of 4 if it converges. To establish the convergence, I suggest using the integral test along with Binet's formula.
     
  10. Jul 19, 2012 #9
    I have never used binets formula before so how would I use it to prove the convergence?
     
  11. Jul 19, 2012 #10
    Binet's formula is basically an expression with an unknown variable n, which, when you plug in a value for n, gives the nth Fibonacci number. Replacing the Fibonacci number in your series with Binet's formula puts the series in the condition of a perfect integral test.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Sum of series- Fibonacci numerator, geometric denominator
  1. Fibonacci sums (Replies: 5)

  2. Fibonacci series (Replies: 10)

Loading...