Fibonacci is Golden

Algebra Level 4

$\large \lim_{n\to\infty}\sum_{k=1}^{2n} \left [ (-1)^k\frac{F_k}{\phi^k} \right ]$ Find the limit above, where $F_k$ is the $k$ -th Fibonacci number, $\phi$ is the Golden Ratio, and $n$ is a positive integer.

Answers in decimal form are accepted. If you find that the limit fails to exist, enter 0.666 as your answer.

Inspiration .

The answer is -0.276.

1 solution

Abhishek Sinha
May 29, 2015

We have $F_k=\frac{1}{\sqrt{5}} \big(\phi^k-\frac{(-1)^k}{\phi^k}\big)$ Hence the $k$ th term in the given series simplifies to $\frac{1}{\sqrt{5}} \big((-1)^k-\frac{1}{\phi^{2k}}\big)$ In the first $2n$ partial sum, the sum of $(-1)^k$ vanishes. Hence the result follows by the infinite sum of the latter geometric term.

Yes exactly, very nice!

We have the geometric series $\sum_{k=1}^{\infty}\frac{1}{\phi^{2k}}=\frac{1}{\phi}$ so the limit is $-\frac{1}{\sqrt{5}\phi}\approx\boxed{-0.276}$ It's always good to explore what happens at the radius of convergence.

Greetings from Harvard Square!

Otto Bretscher - 6 years ago

Here $\infty = \sup \mathbb{Z}$ , yes? I think that should probably be specified.

Jake Lai - 6 years ago

Yes, that's a good point, Jake. When dealing with sequences and series, the meaning of $2n$ , $2n+1$ etc. is usually implied (just look at a table of Taylor polynomials!), but one can never be too explicit when posting a problem on Brilliant ;)

Otto Bretscher - 6 years ago

Yes, the geometric series converges as $\phi^2>1$ . Thanks from Kendall Square :)

Abhishek Sinha - 6 years ago

Fibonacci is Golden

The answer is -0.276.

1 solution

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