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Question Number 84982 by M±th+et£s last updated on 18/Mar/20

Σ_(n=2) ^∞ ((H_n H_(n+1) )/(n^3 −n))

$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} \:{H}_{{n}+\mathrm{1}} }{{n}^{\mathrm{3}} −{n}} \\ $$

Answered by Kamel Kamel last updated on 18/Mar/20

=(5/2)−(π^2 /(24))−ζ(3) See my detailed solution in RMM magazine.

$$=\frac{\mathrm{5}}{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}−\zeta\left(\mathrm{3}\right) \\ $$$${See}\:{my}\:{detailed}\:{solution}\:{in}\:{RMM}\:{magazine}. \\ $$

Answered by mind is power last updated on 19/Mar/20

H_n H_(n+1) =H_n (H_n +(1/(n+1))) Page 1 =H_n ^2 +(H_n /(n+1)) Σ_(n≥2) .((H_n ^2 +(H_n /(n+1)))/(n(n−1)(n+1))) =Σ_(n≥2) (H_n ^2 /(n(n−1)(n+1)))+Σ_(n≥2) (H_n /(n(n−1)(n+1)^2 )) let S=Σ_(n≥2) (H_n /(n(n−1)(n+1)^2 )) (1/(n(n−1)(n+1)^2 ))=−(1/n)+(1/(4(n−1)))+(1/(2(n+1)^2 ))+(3/(4(n+1))) S=Σ_(n≥2) (((−H_n )/n)+(H_n /(4(n−1)))+(H_n /(2(n+1)^2 ))+((3H_n )/(4(n+1)))) =Σ_(n≥2) (((−H_n )/n)+((H_(n−1) +(1/n))/(4(n−1)))+((H_(n+1) −(1/(n+1)))/(2(n+1)^2 ))+((3{H_(n+1) −(1/(n+1))})/(4(n+1)))) =Σ_(n≥2) (−(H_n /n)+(H_(n−1) /(4(n−1)))+((3H_(n+1) )/(4(n+1))))+Σ_(n≥2) (1/(4n(n−1)))−Σ_(n≥2) (1/(2(n+1)^3 )) Σ_(n≥2) −(3/(4(n+1)^2 ))+Σ_(n≥2) (H_(n+1) /((n+1)^2 )) the hardest one is Just Σ_(n≥2) (H_(n+1) /((n+1)^2 ))=A Σ_(n≥1) (H_n /n^2 )=T −(H_n /n)=∫_0 ^1 t^(n−1) log(1−t)dt ⇒−Σ_(n≥1) (H_n /n^2 )=∫_0 ^1 (t^(n−1) /n)log(1−t)dt =∫_0 ^1 (1/t)Σ_(n≥1) (t^n /n)log(1−t) dt Σ(H_n /n^2 )=∫_0 ^1 ((log^2 (1−t))/t)dt= log(1−t)=−y⇒t=1−e^(−y) =∫_0 ^(+∞) ((y^2 e^(−y) )/(1−e^(−y) ))dy=T Γ(3)ζ(3)=2ζ(3) A=T−(H_1 /1)−(H_2 /4)=2ζ(3)−1−(1/4).(3/2)=2ζ(3)−((11)/8) S=Σ_(n≥2) (−(H_n /n)+(H_(n−1) /(4(n−1)))+((3H_(n+1) )/(4(n+1))))_(=D) +(1/4)Σ_(n≥2) ((1/(n−1))−(1/n))−(1/2)(ζ(3)−1−(1/8)) −(3/4)(ζ(2)−1−(1/4))+2ζ(3)−((11)/8) D=(−(H_2 /2)+(H_1 /2) +(H_2 /(4.2))) =(1/2)−(9/(16))=−(1/(16)) ★Σ_(n≥2) ((1/n)−(1/(n−1)))=−1 S=−(1/(16))−(1/4)−((ζ(3))/2)+(9/(16))−((3ζ(2))/4)+((15)/(16))+2ζ(3)−((11)/8) ((3ζ(3))/2)−((3ζ(2))/4)+(1/(16))=S

$${H}_{{n}} {H}_{{n}+\mathrm{1}} ={H}_{{n}} \left({H}_{{n}} +\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$${Page}\:\mathrm{1}\: \\ $$$$={H}_{{n}} ^{\mathrm{2}} +\frac{{H}_{{n}} }{{n}+\mathrm{1}} \\ $$$$\underset{{n}\geqslant\mathrm{2}} {\sum}.\frac{{H}_{{n}} ^{\mathrm{2}} +\frac{{H}_{{n}} }{{n}+\mathrm{1}}}{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)} \\ $$$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{H}_{{n}} ^{\mathrm{2}} }{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}+\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{H}_{{n}} }{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${let}\:{S}=\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{H}_{{n}} }{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{4}\left({n}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{4}\left({n}+\mathrm{1}\right)} \\ $$$${S}=\underset{{n}\geqslant\mathrm{2}} {\sum}\left(\frac{−{H}_{{n}} }{{n}}+\frac{{H}_{{n}} }{\mathrm{4}\left({n}−\mathrm{1}\right)}+\frac{{H}_{{n}} }{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{3}{H}_{{n}} }{\mathrm{4}\left({n}+\mathrm{1}\right)}\right) \\ $$$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\left(\frac{−{H}_{{n}} }{{n}}+\frac{{H}_{{n}−\mathrm{1}} +\frac{\mathrm{1}}{{n}}}{\mathrm{4}\left({n}−\mathrm{1}\right)}+\frac{{H}_{{n}+\mathrm{1}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{3}\left\{{H}_{{n}+\mathrm{1}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}\right\}}{\mathrm{4}\left({n}+\mathrm{1}\right)}\right) \\ $$$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\left(−\frac{{H}_{{n}} }{{n}}+\frac{{H}_{{n}−\mathrm{1}} }{\mathrm{4}\left({n}−\mathrm{1}\right)}+\frac{\mathrm{3}{H}_{{n}+\mathrm{1}} }{\mathrm{4}\left({n}+\mathrm{1}\right)}\right)+\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{\mathrm{4}{n}\left({n}−\mathrm{1}\right)}−\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\underset{{n}\geqslant\mathrm{2}} {\sum}−\frac{\mathrm{3}}{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{H}_{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${the}\:{hardest}\:{one}\:{is}\:{Just} \\ $$$$\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{H}_{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }={A} \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{H}_{{n}} }{{n}^{\mathrm{2}} }={T} \\ $$$$−\frac{{H}_{{n}} }{{n}}=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}−\mathrm{1}} {log}\left(\mathrm{1}−{t}\right){dt} \\ $$$$\Rightarrow−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{H}_{{n}} }{{n}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{n}−\mathrm{1}} }{{n}}{log}\left(\mathrm{1}−{t}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{t}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{t}^{{n}} }{{n}}{log}\left(\mathrm{1}−{t}\right)\:\:{dt} \\ $$$$\Sigma\frac{{H}_{{n}} }{{n}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)}{{t}}{dt}= \\ $$$${log}\left(\mathrm{1}−{t}\right)=−{y}\Rightarrow{t}=\mathrm{1}−{e}^{−{y}} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{y}^{\mathrm{2}} {e}^{−{y}} }{\mathrm{1}−{e}^{−{y}} }{dy}={T} \\ $$$$\Gamma\left(\mathrm{3}\right)\zeta\left(\mathrm{3}\right)=\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$${A}={T}−\frac{{H}_{\mathrm{1}} }{\mathrm{1}}−\frac{{H}_{\mathrm{2}} }{\mathrm{4}}=\mathrm{2}\zeta\left(\mathrm{3}\right)−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}.\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{2}\zeta\left(\mathrm{3}\right)−\frac{\mathrm{11}}{\mathrm{8}} \\ $$$${S}=\underset{{n}\geqslant\mathrm{2}} {\sum}\left(−\frac{{H}_{{n}} }{{n}}+\frac{{H}_{{n}−\mathrm{1}} }{\mathrm{4}\left({n}−\mathrm{1}\right)}+\frac{\mathrm{3}{H}_{{n}+\mathrm{1}} }{\mathrm{4}\left({n}+\mathrm{1}\right)}\underset{={D}} {\right)}+\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}\geqslant\mathrm{2}} {\sum}\left(\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{1}}{{n}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\zeta\left(\mathrm{3}\right)−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\right) \\ $$$$−\frac{\mathrm{3}}{\mathrm{4}}\left(\zeta\left(\mathrm{2}\right)−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{2}\zeta\left(\mathrm{3}\right)−\frac{\mathrm{11}}{\mathrm{8}} \\ $$$${D}=\left(−\frac{{H}_{\mathrm{2}} }{\mathrm{2}}+\frac{{H}_{\mathrm{1}} }{\mathrm{2}}\:+\frac{{H}_{\mathrm{2}} }{\mathrm{4}.\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{9}}{\mathrm{16}}=−\frac{\mathrm{1}}{\mathrm{16}}\:\:\:\bigstar\underset{{n}\geqslant\mathrm{2}} {\sum}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}−\mathrm{1}}\right)=−\mathrm{1} \\ $$$${S}=−\frac{\mathrm{1}}{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{16}}−\frac{\mathrm{3}\zeta\left(\mathrm{2}\right)}{\mathrm{4}}+\frac{\mathrm{15}}{\mathrm{16}}+\mathrm{2}\zeta\left(\mathrm{3}\right)−\frac{\mathrm{11}}{\mathrm{8}} \\ $$$$\frac{\mathrm{3}\zeta\left(\mathrm{3}\right)}{\mathrm{2}}−\frac{\mathrm{3}\zeta\left(\mathrm{2}\right)}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{16}}={S} \\ $$$$ \\ $$

Answered by mind is power last updated on 19/Mar/20

F=Σ_(n≥2) (H_n ^2 /(n(n−1)(n+1))) Page 2 ((H_n ^2 +H_n ^((2)) )/n)=∫_0 ^1 x^(n−1) log^2 (1−x)dx ⇒Σ((H_n ^2 +H_n ^((2)) )/(n(n−1)(n+1)))=∫_0 ^1 Σ_(n≥2) ((x^(n−1) log^2 (1−x))/((n−1)(n+1)))dx=R =∫_0 ^1 log^2 (1−x).Σ_(n≥2) (x^(n−1) /2)((1/(n−1))−(1/(n+1)))dx =∫_0 ^1 log^2 (1−x)dx.(−(1/2)log(1−x)−Σ_(n≥2) (x^(n−1) /(2(n+1))))dx =−∫((log^3 (1−x)dx)/2)dx−(1/2)∫_0 ^1 log^2 (1−x).((1/x^2 ).Σ_(k≥2) (x^(n+1) /(n+1))dx) =−(1/2)∫_0 ^(+∞) u^3 e^(−u) du−(1/2)∫_0 ^1 log^2 (1−x).((1/x^2 )(−log(1−x)−x−(x^2 /2)))dx =−(1/2).Γ(4)+(1/2)∫_0 ^1 ((log^3 (1−x))/x^2 )dx+(1/2)∫_0 ^1 ((log^2 (1−x))/x)+∫_0 ^1 ((log^2 (1−x))/4)dx ∫_0 ^1 ((log^2 (1−x))/x)dx =∫_0 ^(+∞) ((u^2 e^(−u) )/(1−e^(−u) ))=Γ(3)ζ(3) ∫_0 ^1 ((log^3 (1−x))/x^2 )dx=∫_0 ^(+∞) ((u^3 e^(−u) )/((1−e^(−u) )^2 ))du=z (t/((1−t)^2 ))=−(1/(1−t))+(1/((1−t)^2 ))=−Σ_(k≥0) t^k +Σ(k+1)t^k =∫_0 ^(+∞) y^3 .(Σ_(k=0) ^(+∞) ke^(−ky) )du =Σ_(k=0) ^(+∞) ∫_0 ^(+∞) (u^3 /k^3 )e^(−u) du =Γ(4)ζ(3) ∫_0 ^1 ((log^2 (1−x))/x)dx=Γ(3).ζ(3) ∫_0 ^1 log^2 (1−x)dx=Γ(3) R=−((Γ(4))/2)+(1/2)Γ(4)ζ(3)+(1/2)Γ(3)ζ(3)+((Γ(3))/4) =−3+4ζ(3)+(3/2)=−(3/2)+4ζ(3)

$${F}=\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{H}_{{n}} ^{\mathrm{2}} }{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)} \\ $$$${Page}\:\mathrm{2} \\ $$$$\frac{{H}_{{n}} ^{\mathrm{2}} +{H}_{{n}} ^{\left(\mathrm{2}\right)} }{{n}}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} {log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right){dx} \\ $$$$\Rightarrow\Sigma\frac{{H}_{{n}} ^{\mathrm{2}} +{H}_{{n}} ^{\left(\mathrm{2}\right)} }{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{x}^{{n}−\mathrm{1}} {log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}{dx}={R} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right).\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{x}^{{n}−\mathrm{1}} }{\mathrm{2}}\left(\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right){dx}.\left(−\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{1}−{x}\right)−\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{x}^{{n}−\mathrm{1}} }{\mathrm{2}\left({n}+\mathrm{1}\right)}\right){dx} \\ $$$$=−\int\frac{{log}^{\mathrm{3}} \left(\mathrm{1}−{x}\right){dx}}{\mathrm{2}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right).\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }.\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{dx}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{+\infty} {u}^{\mathrm{3}} {e}^{−{u}} {du}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right).\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(−{log}\left(\mathrm{1}−{x}\right)−{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\right){dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}.\Gamma\left(\mathrm{4}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{3}} \left(\mathrm{1}−{x}\right)}{{x}^{\mathrm{2}} }{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{\mathrm{4}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{u}^{\mathrm{2}} {e}^{−{u}} }{\mathrm{1}−{e}^{−{u}} }=\Gamma\left(\mathrm{3}\right)\zeta\left(\mathrm{3}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{3}} \left(\mathrm{1}−{x}\right)}{{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{+\infty} \frac{{u}^{\mathrm{3}} {e}^{−{u}} }{\left(\mathrm{1}−{e}^{−{u}} \right)^{\mathrm{2}} }{du}={z} \\ $$$$\frac{{t}}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{1}−{t}}+\frac{\mathrm{1}}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }=−\underset{{k}\geqslant\mathrm{0}} {\sum}{t}^{{k}} +\Sigma\left({k}+\mathrm{1}\right){t}^{{k}} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} {y}^{\mathrm{3}} .\left(\underset{{k}=\mathrm{0}} {\overset{+\infty} {\sum}}{ke}^{−{ky}} \right){du} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\int_{\mathrm{0}} ^{+\infty} \frac{{u}^{\mathrm{3}} }{{k}^{\mathrm{3}} }{e}^{−{u}} {du} \\ $$$$=\Gamma\left(\mathrm{4}\right)\zeta\left(\mathrm{3}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}{dx}=\Gamma\left(\mathrm{3}\right).\zeta\left(\mathrm{3}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right){dx}=\Gamma\left(\mathrm{3}\right) \\ $$$${R}=−\frac{\Gamma\left(\mathrm{4}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\mathrm{4}\right)\zeta\left(\mathrm{3}\right)+\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\mathrm{3}\right)\zeta\left(\mathrm{3}\right)+\frac{\Gamma\left(\mathrm{3}\right)}{\mathrm{4}} \\ $$$$=−\mathrm{3}+\mathrm{4}\zeta\left(\mathrm{3}\right)+\frac{\mathrm{3}}{\mathrm{2}}=−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{4}\zeta\left(\mathrm{3}\right) \\ $$$$ \\ $$

Answered by mind is power last updated on 20/Mar/20

Page 3 Σ_(n≥2) (H_n ^((2)) /(n(n−1)(n+1))) =Σ_(n≥2) (−(1/n)+(1/(2(n−1)))+(1/(2(n+1))))H_n ^((2)) =Σ_(n≥2) (−(H_n ^((2)) /n)+(H_n ^((2)) /(2(n−1)))+(H_n ^((2)) /(2(n+1)))) =Σ_(n≥2) (−(H_n ^((2)) /n)+(H_(n−1) ^((2)) /(2(n−1)))+(1/(2(n−1)^3 ))+(H_(n+1) ^((2)) /(2(n+1)^3 ))−(1/(2(n+1)^3 ))) =Σ_(n≥2) (−(H_n ^((2)) /n)+(H_(n−1) ^((2)) /(2(n−1)))+(H_(n+1) ^((2)) /(2(n+1))))+Σ_(n≥2) (1/(2(n−1)^3 ))−Σ_(n≥2) (1/(2(n+1)^3 )) =(−(H_2 ^((2)) /2)+(H_1 ^((2)) /2)+(H_2 ^((2)) /4))+((ζ(3))/2)−(1/2)(ζ(3)−1−(1/8)) =(1/2)−(1/4)(1+(1/4))+(9/(16))=(3/4) i Will finish Later

$${Page}\:\mathrm{3} \\ $$$$\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{H}_{{n}} ^{\left(\mathrm{2}\right)} }{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)} \\ $$$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\left(−\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\right){H}_{{n}} ^{\left(\mathrm{2}\right)} \\ $$$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\left(−\frac{{H}_{{n}} ^{\left(\mathrm{2}\right)} }{{n}}+\frac{{H}_{{n}} ^{\left(\mathrm{2}\right)} }{\mathrm{2}\left({n}−\mathrm{1}\right)}+\frac{{H}_{{n}} ^{\left(\mathrm{2}\right)} }{\mathrm{2}\left({n}+\mathrm{1}\right)}\right) \\ $$$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\left(−\frac{{H}_{{n}} ^{\left(\mathrm{2}\right)} }{{n}}+\frac{{H}_{{n}−\mathrm{1}} ^{\left(\mathrm{2}\right)} }{\mathrm{2}\left({n}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)^{\mathrm{3}} }+\frac{{H}_{{n}+\mathrm{1}} ^{\left(\mathrm{2}\right)} }{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\right) \\ $$$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\left(−\frac{{H}_{{n}} ^{\left(\mathrm{2}\right)} }{{n}}+\frac{{H}_{{n}−\mathrm{1}} ^{\left(\mathrm{2}\right)} }{\mathrm{2}\left({n}−\mathrm{1}\right)}+\frac{{H}_{{n}+\mathrm{1}} ^{\left(\mathrm{2}\right)} }{\mathrm{2}\left({n}+\mathrm{1}\right)}\right)+\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)^{\mathrm{3}} }−\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\left(−\frac{{H}_{\mathrm{2}} ^{\left(\mathrm{2}\right)} }{\mathrm{2}}+\frac{{H}_{\mathrm{1}} ^{\left(\mathrm{2}\right)} }{\mathrm{2}}+\frac{{H}_{\mathrm{2}} ^{\left(\mathrm{2}\right)} }{\mathrm{4}}\right)+\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\zeta\left(\mathrm{3}\right)−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)+\frac{\mathrm{9}}{\mathrm{16}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${i}\:{Will}\:{finish}\:{Later}\: \\ $$$$ \\ $$

Commented by M±th+et£s last updated on 19/Mar/20

thank you sir nice work

$${thank}\:{you}\:{sir}\:{nice}\:{work} \\ $$

Commented by mind is power last updated on 20/Mar/20

withe pleasur have You the value of this sum ? so long may bee i did mistack

$${withe}\:{pleasur}\:{have}\:{You}\:{the}\:{value}\:\:{of}\:{this}\:{sum}\:? \\ $$$${so}\:{long}\:{may}\:{bee}\:\:{i}\:{did}\:{mistack} \\ $$

Commented by M±th+et£s last updated on 20/Mar/20

sir kamal value is right sir

$${sir}\:{kamal}\:{value}\:{is}\:{right}\:{sir} \\ $$

Answered by Kamel Kamel last updated on 20/Mar/20

Commented by M±th+et£s last updated on 20/Mar/20

great solution sir thank you

$${great}\:{solution}\:{sir}\:{thank}\:{you} \\ $$

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