discrete-mathematics-prove-that-n-1-1-F-2n-1-1-5-5-2-F-n-fibonacci-sequence-

Question Number 142723 by mnjuly1970 last updated on 04/Jun/21

\dots \dots . d i s c r e t e \dots . . m a t h e m a t i c s \dots \dots .

p r o v e t h a t :

\underset{n = 1}{\sum^{\infty}} \frac{1}{F_{2 n + 1} - 1} \overset{?}{=} \frac{5 - \sqrt{5}}{2}

F_{n} :: f i b o n a c c i s e q u e n c e \dots

Answered by Olaf_Thorendsen last updated on 05/Jun/21

S = \underset{n = 1}{\sum^{\infty}} \frac{1}{F_{2 n + 1} - 1}

S = \underset{n = 1}{\sum^{\infty}} \frac{1}{\frac{1}{\sqrt{5}} (φ^{2 n + 1} - φ^{' 2 n + 1}) - 1}

φ = \frac{1 + \sqrt{5}}{2} and φ^{'} = \frac{1 - \sqrt{5}}{2} = - \frac{1}{φ}

S = \underset{n = 1}{\sum^{\infty}} \frac{1}{\frac{1}{\sqrt{5}} (φ^{2 n + 1} - {(- \frac{1}{φ})}^{2 n + 1}) - 1}

S = \underset{n = 1}{\sum^{\infty}} \frac{1}{\frac{1}{\sqrt{5}} (φ^{2 n + 1} + \frac{1}{φ^{2 n + 1}}) - 1}

S = \sqrt{5} \underset{n = 1}{\sum^{\infty}} \frac{φ^{2 n + 1}}{φ^{4 n + 2} - \sqrt{5} φ^{2 n + 1} + 1}

S = \sqrt{5} \underset{n = 1}{\sum^{\infty}} \frac{φ^{2 n + 1}}{(φ^{2 n + 1} - \frac{\sqrt{5} + 1}{2}) (φ^{2 n + 1} - \frac{\sqrt{5} - 1}{2})}

S = \sqrt{5} \underset{n = 1}{\sum^{\infty}} \frac{φ^{2 n + 1}}{(φ^{2 n + 1} - φ) (φ^{2 n + 1} - \frac{1}{φ})}

S = \sqrt{5} \underset{n = 1}{\sum^{\infty}} \frac{φ^{2 n + 1}}{(φ^{2 n} - 1) (φ^{2 n + 2} - 1)}

S = \sqrt{5} \underset{n = 1}{\sum^{\infty}} φ^{2 n + 1} \frac{1}{φ - \frac{1}{φ}} [\frac{\frac{1}{φ}}{φ^{2 n} - 1} - \frac{φ}{φ^{2 n + 2} - 1}]

φ - \frac{1}{φ} = \frac{1 + \sqrt{5}}{2} + \frac{1 - \sqrt{5}}{2} = 1

S = \sqrt{5} \underset{n = 1}{\sum^{\infty}} [\frac{φ^{2 n}}{φ^{2 n} - 1} - \frac{φ^{2 n + 2}}{φ^{2 n + 2} - 1}]

S = \sqrt{5} \underset{n = 1}{\sum^{\infty}} [(1 + \frac{1}{φ^{2 n} - 1}) - (1 + \frac{1}{φ^{2 n + 2} - 1})]

S = \sqrt{5} \underset{n = 1}{\sum^{\infty}} [\frac{1}{φ^{2 n} - 1} - \frac{1}{φ^{2 n + 2} - 1}]

S = \sqrt{5} (\frac{1}{φ^{2} - 1})

S = \sqrt{5} (\frac{1}{{(\frac{1 + \sqrt{5}}{2})}^{2} - 1})

S = \sqrt{5} (\frac{1}{\frac{3 + \sqrt{5}}{2} - 1})

S = \sqrt{5} (\frac{1}{\frac{1 + \sqrt{5}}{2}}) = \sqrt{5} (\frac{\sqrt{5} - 1}{2}) = \frac{5 - \sqrt{5}}{2}

Commented by mnjuly1970 last updated on 05/Jun/21

v e r y n i c e m r o l a f \dots

Commented by Dwaipayan Shikari last updated on 05/Jun/21

N i c e s i r!