fibonacci and the golden ratio proof

**gpenguin** · February 8th 2011, 11:55 AM

any help on getting started would be great. i really don't know where to go with this as i'm useless with limits.

alpha is the golden ration ((1+sqrt5)/2)

f2k etc are part of the fibonacci sequence

prove lim f2k/f2k+1 = 1/alpha
k->infinity

**Ackbeet** · February 8th 2011, 12:13 PM

The Fibonacci sequence can be written as a recurrence relation. Solve that relation for the nth term, and then form the desired ratio and take the limit. Does that give you some ideas?

**pickslides** · February 8th 2011, 12:19 PM

Here's a hint: The golden ratio is the solution to the equation $x^2-x-1=0$

Now look at the recurrence relationship of Fibonacci numbers, can you find this equation?

**chisigma** · February 9th 2011, 12:44 PM

Originally Posted by gpenguin

any help on getting started would be great. i really don't know where to go with this as i'm useless with limits.

alpha is the golden ration ((1+sqrt5)/2)

f2k etc are part of the fibonacci sequence

prove lim f2k/f2k+1 = 1/alpha
k->infinity

The Fibonacci's sequence is the solution of the 'recursive relation'...

$\displaystyle y_{n} = y_{n-1} + y_{n-2}\\, y_{0}=0\\, \\y_{1}=1$ (1)

The (1) is linear and homogenous and its general solution is...

$\displaystyle y_{n} = c_{1}\ \varphi^{n} + c_{2}\ (-\varphi)^{-n}$ (2)

... where $\varphi = \frac{1+\sqrt{5}}{2}$ is the 'golden ratio'. Now in (2) is...

$\displaystyle \lim_{n \rightarrow \infty} (-\varphi)^{-n}=0$ (3)

... so that is...

$\displaystyle \lim_{n \rightarrow \infty} \frac{y_{n}}{y_{n-1}} = \varphi$ (4)

Kind regards

$\chi$ $\sigma$

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