# Sum of series- Fibonacci numerator, geometric denominator

1. Jul 6, 2012

### Apollonian

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. Jul 6, 2012

### tiny-tim

Welcome to PF!

Hi Apollonian! Welcome to PF!

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

3. Jul 6, 2012

### AlephZero

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 $4a_{n+1} = 2a_n + a_{n-1}$.

4. Jul 10, 2012

### phillip1882

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?

5. Jul 10, 2012

### Mentallic

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

6. Jul 10, 2012

### coolul007

replace the numerator with the Binet formula for the nth term of a Fibonacci series gives each term in terms of n.
$\frac{(1+\sqrt{5})^{n}-(1-\sqrt{5})^{n}}{2^{n} \sqrt{5}}$
Multiplying that by $\frac{1}{2^{n}}$ yields each term as:
$\frac{(1+\sqrt{5})^{n}-(1-\sqrt{5})^{n}}{2^{2n} \sqrt{5}}$

7. Jul 16, 2012

### SumThePrimes

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? :)

8. Jul 17, 2012

### Millennial

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.

9. Jul 19, 2012

### Apollonian

I have never used binets formula before so how would I use it to prove the convergence?

10. Jul 19, 2012

### Millennial

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.