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Use recurrence relations to show that the convergents for $(\sqrt{5}-1)/2$ are ratios of successive Fibonacci numbers.

Here the recurrence relation is: $f_n=0$ if $n=0$, $f_n=1$ if $n=1$, and $f_n=f_{n-1}+f_{n-2}$ if $n\geq2$.

Here the convergents are $c_0=0$, $c_1=1$, $c_2=1/2$, $c_3=2/3$, etc. This pattern is found by looking at the infinite continued fraction of $(\sqrt{5}-1)/2$, which can easily be derived from the infinite continued fraction of the golden ratio.

So far I have tried induction, but got stuck in the middle of my inductive step. I'm not sure if it is a simple algebraic trick I am missing, or if I should approach the proofe from a different angle.

TIA

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    The numerators and denominators of continued fractions satisfy $r_k=a_kr_{k-1}+r_{k-2}$, where the $a_k$ are the partial quotients. For the golden ratio, $a_k=1$ for all $k$, so there's your Fibonacci recurrence. – 2017-02-14
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    is it really that simple? One does not need to do proof my induction to show for all k+1? – 2017-02-14
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    It's really that simple. – 2017-02-14

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