Evened Fibonacci sum

Calculus Level 4

Evaluate n = 0 ( 1 ) n F 2 n 9 n \sum_{n=0}^{\infty}(-1)^n\frac{F_{2n}}{9^n}

Notation: F n F_n denotes the n th n^\text{th} Fibonacci number .


Source of the problem : Prof. Dan .


The answer is -0.0825688.

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2 solutions

Naren Bhandari
Oct 24, 2020

Since the generating function of Fibonacci series is F n ( x ) = n 0 F n x n = x 1 x x 2 , x < ϕ 1 \mathcal{F}_n(x)= \sum_{n\geq 0} F_n x^n=\frac{x}{1-x-x^2}, \;\;x<\phi^{-1} where ϕ \phi is golden ratio and since it is easy to derive by it recursive relation so it is easy to note that F 2 n ( x ) \displaystyle \mathcal{F}_{2n}(x) n 0 F 2 n x 2 n = 1 2 ( F n ( x ) + F n ( x ) ) \sum_{n\geq 0} F_{2n} x^{2n}=\frac{1}{2}\left(\mathcal{F}_n(x)+\mathcal{F}_n(-x)\right) giving us the last expression for F 2 n ( x ) = 1 2 ( x 1 x x 2 x 1 + x x 2 ) = x 2 x 4 3 x 2 + 1 \mathcal{F}_{2n}(x)=\frac{1}{2}\left(\frac{x}{1-x-x^2}-\frac{x}{1+x-x^2}\right)=\frac{x^2}{x^4-3x^2+1} now replacing x x by x \sqrt{x} and them x = 1 9 x=-\frac{1}{9} and simplification gives us n 0 ( 1 ) n F 2 n 9 n = x x 2 3 x + 1 x = 1 9 = 9 109 \sum_{n\geq 0} (-1)^n\frac{F_{2n}}{9^n}=\frac{x}{x^2-3x+1}\Bigg|_{x=-\frac{1}{9}}=-\frac{9}{109}


We can note that, for all x < ϕ 1 x<\phi^{-1} n 0 F 2 n x n = x x 2 3 x + 1 \sum_{n\geq 0} F_{2n} x^n=\frac{x}{x^2-3x+1} and n 0 F 2 n + 1 x n = 1 x x 2 3 x + 1 \sum_{n\geq 0} F_{2n+1} x^n=\frac{1-x}{x^2-3x+1}

Another great question. +1+1+1

I recommend that you change your question to...

... This series is equal to a b -\frac ab , where a a and b b are coprime positive integers. What is a + b a+b ?

With an answer of 9 + 109 = 118 9 + 109 = \boxed{118} .


Also, is this FB group any good? I might need to get an FB account for this.

Pi Han Goh - 7 months, 2 weeks ago

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Thank you sir.

Firstly I thought like of your recommendation however, I thought to share the same problem as it is the linked one. So I didn't make any changes.

Yes, it is a facebook group of Romanian mathematical Magazine which share problem on facebook.

Naren Bhandari - 7 months, 2 weeks ago

I entered in the right answer but for some reason Brilliant didn't accept the negative sign. I figured I had done something incorrectly but I guess not...

Alexander McDowell - 7 months, 2 weeks ago

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Since I can't give a solution because of an error with Brilliant (even though my answer was correct), here is my solution:

The fibonacci numbers can be generated using the function F(n) = (phi1^n - phi2^n)/sqrt(5) where phi1 = (1 + sqrt(5)) / 2 and phi2 = (1 - sqrt(5)) / 2. Thus, the given summation is equivalent to (sum((-phi1^2 / 9)^n) - sum((-phi2^2 / 9)^n))/sqrt(5). Each of the sums is an infinite geometric series with r < 1 so we can use the formula 1/1-r to calculate the infinite sum. The sum therefore equals (1/(1 + phi1^2/9) + 1/(1 + phi2^2/9))/sqrt(5). I would show all the steps I took to get the final answer but it would make the comment look very messy. Through a lot of manipulation, you should get -9/109 as the final answer.

Alexander McDowell - 7 months, 2 weeks ago

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Sorry, I cannot be helpful in case of technical issues. I hope @Brilliant Mathematics can help here.

Naren Bhandari - 7 months, 2 weeks ago

What is ϕ 1 \phi^{-1} ?

Chew-Seong Cheong - 7 months, 2 weeks ago

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reciprocal of the golden ratio. Yeah, the solution should have mentioned it.

Pi Han Goh - 7 months, 2 weeks ago

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Thanks. Yes, the solution should have mentioned it.

Chew-Seong Cheong - 7 months, 2 weeks ago

I have mentioned about it. Please check it.

Naren Bhandari - 7 months, 2 weeks ago

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Sorry, I didn't check

Chew-Seong Cheong - 7 months, 2 weeks ago
Arindam Ghosh
Oct 24, 2020

I know this is the cheating way but ...

I can see the sum is converging to a value -0.0825688

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def fibo(n):
    if n < 1:
        return 0
    elif n == 1:
        return 1
    else:
        return fibo(n-1) + fibo(n-2)

sum = 0
for i in range(1,100):
    sum += ((-1/9)**i)*fibo(2*i)

    print(sum)

You could greatly reduce the runtime if you add the following two lines at the top:

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from functools import lru_cache
@lru_cache(maxsize = 100)
def fibo(n):
    if n < 1:
        return 0
    elif n == 1:
        return 1
    else:
        return fibo(n-1) + fibo(n-2)

sum = 0
for i in range(1,100):
    sum += ((-1/9)**i)*fibo(2*i)

    print(sum)

For more information, check out this video .

Pi Han Goh - 7 months, 2 weeks ago

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