Question Number 135757 by Dwaipayan Shikari last updated on 15/Mar/21
$$\frac{\mathrm{1}.\mathrm{1}}{\mathrm{2}}+\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right)\mathrm{2}}{\mathrm{2}^{\mathrm{3}} }+\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right)\mathrm{3}}{\mathrm{2}^{\mathrm{4}} }+\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}\right)\mathrm{5}}{\mathrm{2}^{\mathrm{5}} }+… \\ $$
Commented by Dwaipayan Shikari last updated on 15/Mar/21
$${My}\:{way}\:{was} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{{n}} {x}^{{n}} ={S} \\ $$$${S}={H}_{\mathrm{1}} {x}+{H}_{\mathrm{2}} {x}^{\mathrm{2}} +{H}_{\mathrm{3}} {x}^{\mathrm{3}} +…\:\:\:\:\:\:\left({H}_{{n}} −{H}_{{n}−\mathrm{1}} =\frac{\mathrm{1}}{{n}}\right) \\ $$$${S}\left(\mathrm{1}−{x}\right)={H}_{\mathrm{1}} {x}+\left({H}_{\mathrm{2}} −{H}_{\mathrm{1}} \right){x}^{\mathrm{2}} +\left({H}_{\mathrm{3}} −{H}_{\mathrm{2}} \right){x}^{\mathrm{3}} +… \\ $$$${S}\left(\mathrm{1}−{x}\right)={x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+…\Rightarrow{S}=\frac{−{log}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}} \\ $$$${So} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{{n}} {F}_{{n}} {x}^{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{{n}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{x}\right)^{{n}} −{H}_{{n}} \left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}{x}\right)^{{n}} \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}.\frac{{log}\left(\mathrm{1}−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{x}\right)}{\mathrm{1}−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{x}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}.\frac{{log}\left(\mathrm{1}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}{x}\right)}{\mathrm{1}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}{x}} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Dwaipayan Shikari last updated on 15/Mar/21
$${Generalised} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} {F}_{{n}} }{\mathrm{2}^{{n}} }\:=\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}}{log}\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{3}−\sqrt{\mathrm{5}}}\right)\:\:\left({H}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:\:\:{and}\:{F}_{{n}−\mathrm{1}} +{F}_{{n}−\mathrm{2}} ={F}_{{n}} \right) \\ $$
Commented by Olaf last updated on 15/Mar/21
$$\mathrm{Sorry}\:\mathrm{sir},\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{pleasure}\:\mathrm{for}\:\mathrm{a}\:\mathrm{bad} \\ $$$$\mathrm{mathematician}\:\mathrm{like}\:\mathrm{me}\:\mathrm{to}\:\mathrm{try}\:\mathrm{to}\:\mathrm{answer} \\ $$$$\mathrm{the}\:\mathrm{questions}. \\ $$$$\mathrm{But},\:\mathrm{if}\:\mathrm{each}\:\mathrm{time}\:\mathrm{you}\:\mathrm{give}\:\mathrm{the}\:\mathrm{answers} \\ $$$$\mathrm{to}\:\mathrm{your}\:\mathrm{own}\:\mathrm{questions},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{goal}\:? \\ $$
Commented by Dwaipayan Shikari last updated on 15/Mar/21
$${I}\:{am}\:{inexperienced}\:{in}\:{Mathematics}\:{sir}!.{Each}\:{time}\:{when} \\ $$$${i}\:{find}\:{some}\:{nice}\:{things},\:{i}\:{share}\:{with}\:{this}\:{Community}.\:{There} \\ $$$${are}\:{many}\:{Nice}\:{teachers}\:\left({like}\:{you}\:{sir}!\right)\:{to}\:{interact}\:{with}\:{me}. \\ $$$${But}\:{sorry}\:{for}\:{that}\: \\ $$😔
Answered by Olaf last updated on 15/Mar/21
$$\mathrm{We}\:\mathrm{know}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:{s} \\ $$$$\mathrm{generates}\:\mathrm{the}\:\mathrm{serie}\:\underset{{n}\in\mathbb{N}} {\sum}{F}_{{n}} {z}^{{n}} \:: \\ $$$${s}\left({z}\right)\:=\:\frac{{z}}{\mathrm{1}−{z}−{z}^{\mathrm{2}} }\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{F}_{{n}} {z}^{{n}} ,\:\mid{z}\mid\:<\:\frac{\mathrm{1}}{\varphi} \\ $$$$\mathrm{for}\:{z}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\::\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{F}_{{n}} }{\mathrm{2}^{{n}} }\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\:=\:\mathrm{2} \\ $$$$\mathrm{Let}\:\Psi\left({z}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{H}_{{n}} {F}_{{n}} {z}^{{n}} \\ $$$$<\mathrm{work}\:\mathrm{in}\:\mathrm{progress}\:/> \\ $$