Question Number 142723 by mnjuly1970 last updated on 04/Jun/21
$$\:\:\:\:\:\:\:\:…….\:{discrete}\:\:…..\:\:{mathematics}……. \\ $$$$\:\:\:\:{prove}\:{that}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{F}_{\mathrm{2}{n}+\mathrm{1}} −\mathrm{1}}\overset{?} {=}\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:{F}_{{n}} \:::\:{fibonacci}\:\:{sequence}… \\ $$
Answered by Olaf_Thorendsen last updated on 05/Jun/21
![S = Σ_(n=1) ^∞ (1/(F_(2n+1) −1)) S = Σ_(n=1) ^∞ (1/((1/( (√5)))(ϕ^(2n+1) −ϕ′^(2n+1) )−1)) ϕ = ((1+(√5))/2) and ϕ′ = ((1−(√5))/2) = −(1/ϕ) S = Σ_(n=1) ^∞ (1/((1/( (√5)))(ϕ^(2n+1) −(−(1/ϕ))^(2n+1) )−1)) S = Σ_(n=1) ^∞ (1/((1/( (√5)))(ϕ^(2n+1) +(1/ϕ^(2n+1) ))−1)) S = (√5)Σ_(n=1) ^∞ (ϕ^(2n+1) /(ϕ^(4n+2) −(√5)ϕ^(2n+1) +1)) S = (√5)Σ_(n=1) ^∞ (ϕ^(2n+1) /((ϕ^(2n+1) −(((√5)+1)/2))(ϕ^(2n+1) −(((√5)−1)/2)))) S = (√5)Σ_(n=1) ^∞ (ϕ^(2n+1) /((ϕ^(2n+1) −ϕ)(ϕ^(2n+1) −(1/ϕ)))) S = (√5)Σ_(n=1) ^∞ (ϕ^(2n+1) /((ϕ^(2n) −1)(ϕ^(2n+2) −1))) S = (√5)Σ_(n=1) ^∞ ϕ^(2n+1) (1/(ϕ−(1/ϕ)))[((1/ϕ)/(ϕ^(2n) −1))−(ϕ/(ϕ^(2n+2) −1))] ϕ−(1/ϕ) = ((1+(√5))/2)+((1−(√5))/2) = 1 S = (√5)Σ_(n=1) ^∞ [(ϕ^(2n) /(ϕ^(2n) −1))−(ϕ^(2n+2) /(ϕ^(2n+2) −1))] S = (√5)Σ_(n=1) ^∞ [(1+(1/(ϕ^(2n) −1)))−(1+(1/(ϕ^(2n+2) −1)))] S = (√5)Σ_(n=1) ^∞ [(1/(ϕ^(2n) −1))−(1/(ϕ^(2n+2) −1))] S = (√5)((1/(ϕ^2 −1))) S = (√5)((1/((((1+(√5))/2))^2 −1))) S = (√5)((1/(((3+(√5))/2)−1))) S = (√5)((1/((1+(√5))/2))) = (√5)((((√5)−1)/2)) = ((5−(√5))/2)](https://www.tinkutara.com/question/Q142766.png)
$$\mathrm{S}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{F}_{\mathrm{2}{n}+\mathrm{1}} −\mathrm{1}} \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\varphi^{\mathrm{2}{n}+\mathrm{1}} −\varphi'^{\mathrm{2}{n}+\mathrm{1}} \right)−\mathrm{1}} \\ $$$$\varphi\:=\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\mathrm{and}\:\varphi'\:=\:\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:−\frac{\mathrm{1}}{\varphi} \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\varphi^{\mathrm{2}{n}+\mathrm{1}} −\left(−\frac{\mathrm{1}}{\varphi}\right)^{\mathrm{2}{n}+\mathrm{1}} \right)−\mathrm{1}} \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\varphi^{\mathrm{2}{n}+\mathrm{1}} +\frac{\mathrm{1}}{\varphi^{\mathrm{2}{n}+\mathrm{1}} }\right)−\mathrm{1}} \\ $$$$\mathrm{S}\:=\:\sqrt{\mathrm{5}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\varphi^{\mathrm{2}{n}+\mathrm{1}} }{\varphi^{\mathrm{4}{n}+\mathrm{2}} −\sqrt{\mathrm{5}}\varphi^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}} \\ $$$$\mathrm{S}\:=\:\sqrt{\mathrm{5}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\varphi^{\mathrm{2}{n}+\mathrm{1}} }{\left(\varphi^{\mathrm{2}{n}+\mathrm{1}} −\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right)\left(\varphi^{\mathrm{2}{n}+\mathrm{1}} −\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\mathrm{S}\:=\:\sqrt{\mathrm{5}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\varphi^{\mathrm{2}{n}+\mathrm{1}} }{\left(\varphi^{\mathrm{2}{n}+\mathrm{1}} −\varphi\right)\left(\varphi^{\mathrm{2}{n}+\mathrm{1}} −\frac{\mathrm{1}}{\varphi}\right)} \\ $$$$\mathrm{S}\:=\:\sqrt{\mathrm{5}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\varphi^{\mathrm{2}{n}+\mathrm{1}} }{\left(\varphi^{\mathrm{2}{n}} −\mathrm{1}\right)\left(\varphi^{\mathrm{2}{n}+\mathrm{2}} −\mathrm{1}\right)} \\ $$$$\mathrm{S}\:=\:\sqrt{\mathrm{5}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\varphi^{\mathrm{2}{n}+\mathrm{1}} \frac{\mathrm{1}}{\varphi−\frac{\mathrm{1}}{\varphi}}\left[\frac{\frac{\mathrm{1}}{\varphi}}{\varphi^{\mathrm{2}{n}} −\mathrm{1}}−\frac{\varphi}{\varphi^{\mathrm{2}{n}+\mathrm{2}} −\mathrm{1}}\right] \\ $$$$\varphi−\frac{\mathrm{1}}{\varphi}\:=\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}+\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:\mathrm{1} \\ $$$$\mathrm{S}\:=\:\sqrt{\mathrm{5}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\varphi^{\mathrm{2}{n}} }{\varphi^{\mathrm{2}{n}} −\mathrm{1}}−\frac{\varphi^{\mathrm{2}{n}+\mathrm{2}} }{\varphi^{\mathrm{2}{n}+\mathrm{2}} −\mathrm{1}}\right] \\ $$$$\mathrm{S}\:=\:\sqrt{\mathrm{5}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\left(\mathrm{1}+\frac{\mathrm{1}}{\varphi^{\mathrm{2}{n}} −\mathrm{1}}\right)−\left(\mathrm{1}+\frac{\mathrm{1}}{\varphi^{\mathrm{2}{n}+\mathrm{2}} −\mathrm{1}}\right)\right] \\ $$$$\mathrm{S}\:=\:\sqrt{\mathrm{5}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{\varphi^{\mathrm{2}{n}} −\mathrm{1}}−\frac{\mathrm{1}}{\varphi^{\mathrm{2}{n}+\mathrm{2}} −\mathrm{1}}\right] \\ $$$$\mathrm{S}\:=\:\sqrt{\mathrm{5}}\left(\frac{\mathrm{1}}{\varphi^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$$\mathrm{S}\:=\:\sqrt{\mathrm{5}}\left(\frac{\mathrm{1}}{\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$$\mathrm{S}\:=\:\sqrt{\mathrm{5}}\left(\frac{\mathrm{1}}{\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}−\mathrm{1}}\right) \\ $$$$\mathrm{S}\:=\:\sqrt{\mathrm{5}}\left(\frac{\mathrm{1}}{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\right)\:=\:\sqrt{\mathrm{5}}\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 05/Jun/21
$${very}\:{nice}\:{mr}\:{olaf}… \\ $$
Commented by Dwaipayan Shikari last updated on 05/Jun/21
$${Nice}\:{sir}! \\ $$