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

$$.......discrete.....mathematics.......$$$$provethat:$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{F_{2n+1}-1}\stackrel{?}{=}\frac{5-\sqrt{5}}{2}$$$$F_n::fibonaccisequence...$$

Answered by Olaf_Thorendsen last updated on 05/Jun/21

$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{F_{2n+1}-1}$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\frac{1}{\sqrt{5}}({\phi }^{2n+1}-{\phi }^{\prime 2n+1})-1}$$$$\phi =\frac{1+\sqrt{5}}{2}\mathrm{and}{\phi }^{\prime }=\frac{1-\sqrt{5}}{2}=-\frac{1}{\phi }$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\frac{1}{\sqrt{5}}({\phi }^{2n+1}-{(-\frac{1}{\phi })}^{2n+1})-1}$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\frac{1}{\sqrt{5}}({\phi }^{2n+1}+\frac{1}{{\phi }^{2n+1}})-1}$$$$\mathrm{S}=\sqrt{5}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\phi }^{2n+1}}{{\phi }^{4n+2}-\sqrt{5}{\phi }^{2n+1}+1}$$$$\mathrm{S}=\sqrt{5}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\phi }^{2n+1}}{({\phi }^{2n+1}-\frac{\sqrt{5}+1}{2})({\phi }^{2n+1}-\frac{\sqrt{5}-1}{2})}$$$$\mathrm{S}=\sqrt{5}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\phi }^{2n+1}}{({\phi }^{2n+1}-\phi )({\phi }^{2n+1}-\frac{1}{\phi })}$$$$\mathrm{S}=\sqrt{5}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\phi }^{2n+1}}{({\phi }^{2n}-1)({\phi }^{2n+2}-1)}$$$$\mathrm{S}=\sqrt{5}\underset{n=1}{\sum ^{\mathrm{\infty }}}{\phi }^{2n+1}\frac{1}{\phi -\frac{1}{\phi }}[\frac{\frac{1}{\phi }}{{\phi }^{2n}-1}-\frac{\phi }{{\phi }^{2n+2}-1}]$$$$\phi -\frac{1}{\phi }=\frac{1+\sqrt{5}}{2}+\frac{1-\sqrt{5}}{2}=1$$$$\mathrm{S}=\sqrt{5}\underset{n=1}{\sum ^{\mathrm{\infty }}}[\frac{{\phi }^{2n}}{{\phi }^{2n}-1}-\frac{{\phi }^{2n+2}}{{\phi }^{2n+2}-1}]$$$$\mathrm{S}=\sqrt{5}\underset{n=1}{\sum ^{\mathrm{\infty }}}[(1+\frac{1}{{\phi }^{2n}-1})-(1+\frac{1}{{\phi }^{2n+2}-1})]$$$$\mathrm{S}=\sqrt{5}\underset{n=1}{\sum ^{\mathrm{\infty }}}[\frac{1}{{\phi }^{2n}-1}-\frac{1}{{\phi }^{2n+2}-1}]$$$$\mathrm{S}=\sqrt{5}\left(\frac{1}{{\phi }^2-1}\right)$$$$\mathrm{S}=\sqrt{5}\left(\frac{1}{{\left(\frac{1+\sqrt{5}}{2}\right)}^2-1}\right)$$$$\mathrm{S}=\sqrt{5}\left(\frac{1}{\frac{3+\sqrt{5}}{2}-1}\right)$$$$\mathrm{S}=\sqrt{5}\left(\frac{1}{\frac{1+\sqrt{5}}{2}}\right)=\sqrt{5}\left(\frac{\sqrt{5}-1}{2}\right)=\frac{5-\sqrt{5}}{2}$$

Commented by mnjuly1970 last updated on 05/Jun/21

$$verynicemrolaf...$$

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

$$Nicesir!$$

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