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Question Number 135757 by Dwaipayan Shikari last updated on 15/Mar/21

((1.1)/2)+(((1+(1/2))1)/2^2 )+(((1+(1/2)+(1/3))2)/2^3 )+(((1+(1/2)+(1/3)+(1/4))3)/2^4 )+(((1+(1/2)+(1/3)+(1/4)+(1/5))5)/2^5 )+...

$$\frac{\mathrm{1}.\mathrm{1}}{\mathrm{2}}+\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right)\mathrm{2}}{\mathrm{2}^{\mathrm{3}} }+\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right)\mathrm{3}}{\mathrm{2}^{\mathrm{4}} }+\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}\right)\mathrm{5}}{\mathrm{2}^{\mathrm{5}} }+... \\ $$

Commented by Dwaipayan Shikari last updated on 15/Mar/21

My way was Σ_(n=1) ^∞ H_n x^n =S S=H_1 x+H_2 x^2 +H_3 x^3 +... (H_n −H_(n−1) =(1/n)) S(1−x)=H_1 x+(H_2 −H_1 )x^2 +(H_3 −H_2 )x^3 +... S(1−x)=x+(x^2 /2)+(x^3 /3)+...⇒S=((−log(1−x))/(1−x)) So Σ_(n=1) ^∞ H_n F_n x^n =(1/( (√5)))Σ_(n=1) ^∞ H_n (((1+(√5))/2)x)^n −H_n (((1−(√5))/2)x)^n =−(1/( (√5))).((log(1−((1+(√5))/2)x))/(1−((1+(√5))/2)x))+(1/( (√5))).((log(1−((1−(√5))/2)x))/(1−((1−(√5))/2)x)) x=(1/2)

$${My}\:{way}\:{was} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{{n}} {x}^{{n}} ={S} \\ $$$${S}={H}_{\mathrm{1}} {x}+{H}_{\mathrm{2}} {x}^{\mathrm{2}} +{H}_{\mathrm{3}} {x}^{\mathrm{3}} +...\:\:\:\:\:\:\left({H}_{{n}} −{H}_{{n}−\mathrm{1}} =\frac{\mathrm{1}}{{n}}\right) \\ $$$${S}\left(\mathrm{1}−{x}\right)={H}_{\mathrm{1}} {x}+\left({H}_{\mathrm{2}} −{H}_{\mathrm{1}} \right){x}^{\mathrm{2}} +\left({H}_{\mathrm{3}} −{H}_{\mathrm{2}} \right){x}^{\mathrm{3}} +... \\ $$$${S}\left(\mathrm{1}−{x}\right)={x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+...\Rightarrow{S}=\frac{−{log}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}} \\ $$$${So} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{{n}} {F}_{{n}} {x}^{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{{n}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{x}\right)^{{n}} −{H}_{{n}} \left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}{x}\right)^{{n}} \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}.\frac{{log}\left(\mathrm{1}−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{x}\right)}{\mathrm{1}−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{x}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}.\frac{{log}\left(\mathrm{1}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}{x}\right)}{\mathrm{1}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}{x}} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by Dwaipayan Shikari last updated on 15/Mar/21

Generalised Σ_(n=1) ^∞ ((H_n F_n )/2^n ) =((3+(√5))/( (√5)))log(((3+(√5))/(3−(√5)))) (H_n =Σ_(k=1) ^n (1/k) and F_(n−1) +F_(n−2) =F_n )

$${Generalised} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} {F}_{{n}} }{\mathrm{2}^{{n}} }\:=\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}}{log}\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{3}−\sqrt{\mathrm{5}}}\right)\:\:\left({H}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:\:\:{and}\:{F}_{{n}−\mathrm{1}} +{F}_{{n}−\mathrm{2}} ={F}_{{n}} \right) \\ $$

Commented by Olaf last updated on 15/Mar/21

Sorry sir, but it′s a pleasure for a bad mathematician like me to try to answer the questions. But, if each time you give the answers to your own questions, what is the goal ?

$$\mathrm{Sorry}\:\mathrm{sir},\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{pleasure}\:\mathrm{for}\:\mathrm{a}\:\mathrm{bad} \\ $$$$\mathrm{mathematician}\:\mathrm{like}\:\mathrm{me}\:\mathrm{to}\:\mathrm{try}\:\mathrm{to}\:\mathrm{answer} \\ $$$$\mathrm{the}\:\mathrm{questions}. \\ $$$$\mathrm{But},\:\mathrm{if}\:\mathrm{each}\:\mathrm{time}\:\mathrm{you}\:\mathrm{give}\:\mathrm{the}\:\mathrm{answers} \\ $$$$\mathrm{to}\:\mathrm{your}\:\mathrm{own}\:\mathrm{questions},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{goal}\:? \\ $$

Commented by Dwaipayan Shikari last updated on 15/Mar/21

I am inexperienced in Mathematics sir!.Each time when i find some nice things, i share with this Community. There are many Nice teachers (like you sir!) to interact with me. But sorry for that 😔

$${I}\:{am}\:{inexperienced}\:{in}\:{Mathematics}\:{sir}!.{Each}\:{time}\:{when} \\ $$$${i}\:{find}\:{some}\:{nice}\:{things},\:{i}\:{share}\:{with}\:{this}\:{Community}.\:{There} \\ $$$${are}\:{many}\:{Nice}\:{teachers}\:\left({like}\:{you}\:{sir}!\right)\:{to}\:{interact}\:{with}\:{me}. \\ $$$${But}\:{sorry}\:{for}\:{that}\: \\ $$😔

Answered by Olaf last updated on 15/Mar/21

We know that the function s generates the serie Σ_(n∈N) F_n z^n : s(z) = (z/(1−z−z^2 )) = Σ_(n=0) ^∞ F_n z^n , ∣z∣ < (1/ϕ) for z = (1/2) : Σ_(n=0) ^∞ (F_n /2^n ) = ((1/2)/(1−(1/2)−((1/2))^2 )) = 2 Let Ψ(z) = Σ_(n=0) ^∞ H_n F_n z^n <work in progress />

$$\mathrm{We}\:\mathrm{know}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:{s} \\ $$$$\mathrm{generates}\:\mathrm{the}\:\mathrm{serie}\:\underset{{n}\in\mathbb{N}} {\sum}{F}_{{n}} {z}^{{n}} \:: \\ $$$${s}\left({z}\right)\:=\:\frac{{z}}{\mathrm{1}−{z}−{z}^{\mathrm{2}} }\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{F}_{{n}} {z}^{{n}} ,\:\mid{z}\mid\:<\:\frac{\mathrm{1}}{\varphi} \\ $$$$\mathrm{for}\:{z}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\::\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{F}_{{n}} }{\mathrm{2}^{{n}} }\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\:=\:\mathrm{2} \\ $$$$\mathrm{Let}\:\Psi\left({z}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{H}_{{n}} {F}_{{n}} {z}^{{n}} \\ $$$$<\mathrm{work}\:\mathrm{in}\:\mathrm{progress}\:/> \\ $$

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