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Question Number 143031 by rs4089 last updated on 09/Jun/21

Answered by mnjuly1970 last updated on 09/Jun/21

(π^2 /6)+ln^2 (2)....

$$\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+{ln}^{\mathrm{2}} \left(\mathrm{2}\right).... \\ $$

Answered by Dwaipayan Shikari last updated on 09/Jun/21

S=Σ_(n=1) ^∞ H_n ^2 x^n =H_1 ^2 x+H_2 ^2 x^2 +H_3 x^3 +...  S(1−x)=x+(H_2 ^2 −H_1 ^2 )x^2 +(H_3 ^2 −H_2 ^2 )x^3 +...  S(1−x)=x+((H_2 +H_1 )/2)x^2 +((H_2 +H_3 )/3)x^3 +...    H_(n−1) =H_n −(1/n)  S(1−x)=Σ_(n=1) ^∞ ((H_n +H_(n−1) )/n)x^n =Σ_(n=1) ^∞ ((2H_n −(1/n))/n)x^n   ⇒S(1−x)=2Σ_(n=1) ^∞ (H_n /n)x^n −Σ_(n=1) ^∞ (x^n /n^2 )            Σ_(n=1) ^∞ H_n x^(n−1) =−((log(1−x))/(x(1−x)))  Σ_(n=1) ^∞ (H_n /(n2^n ))=−∫_0 ^(1/2) ((log(1−x))/x)dx−∫_0 ^(1/2) ((log(1−x))/(1−x))dx  =Li_2 ((1/2))+(1/2)log^2 (2)  Li_2 (1−x)+Li_2 (x)=(π^2 /6)−(1/2)log(x)log(1−x)  ⇒Li_2 ((1/2))=(π^2 /(12))−(1/2)log^2 (2)  S(1−(1/2))=2(Li_2 ((1/2))+log^2 (2))−Li_2 ((1/2))  S=2Li_2 ((1/2))+2log^2 (2)=(π^2 /6)+log^2 (2)

$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{{n}} ^{\mathrm{2}} {x}^{{n}} ={H}_{\mathrm{1}} ^{\mathrm{2}} {x}+{H}_{\mathrm{2}} ^{\mathrm{2}} {x}^{\mathrm{2}} +{H}_{\mathrm{3}} {x}^{\mathrm{3}} +... \\ $$$${S}\left(\mathrm{1}−{x}\right)={x}+\left({H}_{\mathrm{2}} ^{\mathrm{2}} −{H}_{\mathrm{1}} ^{\mathrm{2}} \right){x}^{\mathrm{2}} +\left({H}_{\mathrm{3}} ^{\mathrm{2}} −{H}_{\mathrm{2}} ^{\mathrm{2}} \right){x}^{\mathrm{3}} +... \\ $$$${S}\left(\mathrm{1}−{x}\right)={x}+\frac{{H}_{\mathrm{2}} +{H}_{\mathrm{1}} }{\mathrm{2}}{x}^{\mathrm{2}} +\frac{{H}_{\mathrm{2}} +{H}_{\mathrm{3}} }{\mathrm{3}}{x}^{\mathrm{3}} +...\:\:\:\:{H}_{{n}−\mathrm{1}} ={H}_{{n}} −\frac{\mathrm{1}}{{n}} \\ $$$${S}\left(\mathrm{1}−{x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} +{H}_{{n}−\mathrm{1}} }{{n}}{x}^{{n}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{H}_{{n}} −\frac{\mathrm{1}}{{n}}}{{n}}{x}^{{n}} \\ $$$$\Rightarrow{S}\left(\mathrm{1}−{x}\right)=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{{n}}{x}^{{n}} −\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{{n}} {x}^{{n}−\mathrm{1}} =−\frac{{log}\left(\mathrm{1}−{x}\right)}{{x}\left(\mathrm{1}−{x}\right)} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{{n}\mathrm{2}^{{n}} }=−\int_{\mathrm{0}} ^{\mathrm{1}/\mathrm{2}} \frac{{log}\left(\mathrm{1}−{x}\right)}{{x}}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}/\mathrm{2}} \frac{{log}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}}{dx} \\ $$$$={Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{log}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$${Li}_{\mathrm{2}} \left(\mathrm{1}−{x}\right)+{Li}_{\mathrm{2}} \left({x}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}{log}\left({x}\right){log}\left(\mathrm{1}−{x}\right) \\ $$$$\Rightarrow{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{2}}{log}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$${S}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}\left({Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+{log}^{\mathrm{2}} \left(\mathrm{2}\right)\right)−{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${S}=\mathrm{2}{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{2}{log}^{\mathrm{2}} \left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+{log}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 09/Jun/21

greate....

$${greate}.... \\ $$

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