Question Number 84982 by M±th+et£s last updated on 18/Mar/20
$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} \:{H}_{{n}+\mathrm{1}} }{{n}^{\mathrm{3}} −{n}} \\ $$
Answered by Kamel Kamel last updated on 18/Mar/20
$$=\frac{\mathrm{5}}{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}−\zeta\left(\mathrm{3}\right) \\ $$$${See}\:{my}\:{detailed}\:{solution}\:{in}\:{RMM}\:{magazine}. \\ $$
Answered by mind is power last updated on 19/Mar/20
$${H}_{{n}} {H}_{{n}+\mathrm{1}} ={H}_{{n}} \left({H}_{{n}} +\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$${Page}\:\mathrm{1}\: \\ $$$$={H}_{{n}} ^{\mathrm{2}} +\frac{{H}_{{n}} }{{n}+\mathrm{1}} \\ $$$$\underset{{n}\geqslant\mathrm{2}} {\sum}.\frac{{H}_{{n}} ^{\mathrm{2}} +\frac{{H}_{{n}} }{{n}+\mathrm{1}}}{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)} \\ $$$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{H}_{{n}} ^{\mathrm{2}} }{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}+\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{H}_{{n}} }{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${let}\:{S}=\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{H}_{{n}} }{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{4}\left({n}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{4}\left({n}+\mathrm{1}\right)} \\ $$$${S}=\underset{{n}\geqslant\mathrm{2}} {\sum}\left(\frac{−{H}_{{n}} }{{n}}+\frac{{H}_{{n}} }{\mathrm{4}\left({n}−\mathrm{1}\right)}+\frac{{H}_{{n}} }{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{3}{H}_{{n}} }{\mathrm{4}\left({n}+\mathrm{1}\right)}\right) \\ $$$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\left(\frac{−{H}_{{n}} }{{n}}+\frac{{H}_{{n}−\mathrm{1}} +\frac{\mathrm{1}}{{n}}}{\mathrm{4}\left({n}−\mathrm{1}\right)}+\frac{{H}_{{n}+\mathrm{1}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{3}\left\{{H}_{{n}+\mathrm{1}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}\right\}}{\mathrm{4}\left({n}+\mathrm{1}\right)}\right) \\ $$$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\left(−\frac{{H}_{{n}} }{{n}}+\frac{{H}_{{n}−\mathrm{1}} }{\mathrm{4}\left({n}−\mathrm{1}\right)}+\frac{\mathrm{3}{H}_{{n}+\mathrm{1}} }{\mathrm{4}\left({n}+\mathrm{1}\right)}\right)+\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{\mathrm{4}{n}\left({n}−\mathrm{1}\right)}−\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\underset{{n}\geqslant\mathrm{2}} {\sum}−\frac{\mathrm{3}}{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{H}_{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${the}\:{hardest}\:{one}\:{is}\:{Just} \\ $$$$\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{H}_{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }={A} \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{H}_{{n}} }{{n}^{\mathrm{2}} }={T} \\ $$$$−\frac{{H}_{{n}} }{{n}}=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}−\mathrm{1}} {log}\left(\mathrm{1}−{t}\right){dt} \\ $$$$\Rightarrow−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{H}_{{n}} }{{n}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{n}−\mathrm{1}} }{{n}}{log}\left(\mathrm{1}−{t}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{t}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{t}^{{n}} }{{n}}{log}\left(\mathrm{1}−{t}\right)\:\:{dt} \\ $$$$\Sigma\frac{{H}_{{n}} }{{n}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)}{{t}}{dt}= \\ $$$${log}\left(\mathrm{1}−{t}\right)=−{y}\Rightarrow{t}=\mathrm{1}−{e}^{−{y}} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{y}^{\mathrm{2}} {e}^{−{y}} }{\mathrm{1}−{e}^{−{y}} }{dy}={T} \\ $$$$\Gamma\left(\mathrm{3}\right)\zeta\left(\mathrm{3}\right)=\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$${A}={T}−\frac{{H}_{\mathrm{1}} }{\mathrm{1}}−\frac{{H}_{\mathrm{2}} }{\mathrm{4}}=\mathrm{2}\zeta\left(\mathrm{3}\right)−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}.\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{2}\zeta\left(\mathrm{3}\right)−\frac{\mathrm{11}}{\mathrm{8}} \\ $$$${S}=\underset{{n}\geqslant\mathrm{2}} {\sum}\left(−\frac{{H}_{{n}} }{{n}}+\frac{{H}_{{n}−\mathrm{1}} }{\mathrm{4}\left({n}−\mathrm{1}\right)}+\frac{\mathrm{3}{H}_{{n}+\mathrm{1}} }{\mathrm{4}\left({n}+\mathrm{1}\right)}\underset{={D}} {\right)}+\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}\geqslant\mathrm{2}} {\sum}\left(\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{1}}{{n}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\zeta\left(\mathrm{3}\right)−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\right) \\ $$$$−\frac{\mathrm{3}}{\mathrm{4}}\left(\zeta\left(\mathrm{2}\right)−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{2}\zeta\left(\mathrm{3}\right)−\frac{\mathrm{11}}{\mathrm{8}} \\ $$$${D}=\left(−\frac{{H}_{\mathrm{2}} }{\mathrm{2}}+\frac{{H}_{\mathrm{1}} }{\mathrm{2}}\:+\frac{{H}_{\mathrm{2}} }{\mathrm{4}.\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{9}}{\mathrm{16}}=−\frac{\mathrm{1}}{\mathrm{16}}\:\:\:\bigstar\underset{{n}\geqslant\mathrm{2}} {\sum}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}−\mathrm{1}}\right)=−\mathrm{1} \\ $$$${S}=−\frac{\mathrm{1}}{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{16}}−\frac{\mathrm{3}\zeta\left(\mathrm{2}\right)}{\mathrm{4}}+\frac{\mathrm{15}}{\mathrm{16}}+\mathrm{2}\zeta\left(\mathrm{3}\right)−\frac{\mathrm{11}}{\mathrm{8}} \\ $$$$\frac{\mathrm{3}\zeta\left(\mathrm{3}\right)}{\mathrm{2}}−\frac{\mathrm{3}\zeta\left(\mathrm{2}\right)}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{16}}={S} \\ $$$$ \\ $$
Answered by mind is power last updated on 19/Mar/20
$${F}=\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{H}_{{n}} ^{\mathrm{2}} }{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)} \\ $$$${Page}\:\mathrm{2} \\ $$$$\frac{{H}_{{n}} ^{\mathrm{2}} +{H}_{{n}} ^{\left(\mathrm{2}\right)} }{{n}}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} {log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right){dx} \\ $$$$\Rightarrow\Sigma\frac{{H}_{{n}} ^{\mathrm{2}} +{H}_{{n}} ^{\left(\mathrm{2}\right)} }{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{x}^{{n}−\mathrm{1}} {log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}{dx}={R} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right).\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{x}^{{n}−\mathrm{1}} }{\mathrm{2}}\left(\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right){dx}.\left(−\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{1}−{x}\right)−\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{x}^{{n}−\mathrm{1}} }{\mathrm{2}\left({n}+\mathrm{1}\right)}\right){dx} \\ $$$$=−\int\frac{{log}^{\mathrm{3}} \left(\mathrm{1}−{x}\right){dx}}{\mathrm{2}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right).\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }.\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{dx}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{+\infty} {u}^{\mathrm{3}} {e}^{−{u}} {du}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right).\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(−{log}\left(\mathrm{1}−{x}\right)−{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\right){dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}.\Gamma\left(\mathrm{4}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{3}} \left(\mathrm{1}−{x}\right)}{{x}^{\mathrm{2}} }{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{\mathrm{4}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{u}^{\mathrm{2}} {e}^{−{u}} }{\mathrm{1}−{e}^{−{u}} }=\Gamma\left(\mathrm{3}\right)\zeta\left(\mathrm{3}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{3}} \left(\mathrm{1}−{x}\right)}{{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{+\infty} \frac{{u}^{\mathrm{3}} {e}^{−{u}} }{\left(\mathrm{1}−{e}^{−{u}} \right)^{\mathrm{2}} }{du}={z} \\ $$$$\frac{{t}}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{1}−{t}}+\frac{\mathrm{1}}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }=−\underset{{k}\geqslant\mathrm{0}} {\sum}{t}^{{k}} +\Sigma\left({k}+\mathrm{1}\right){t}^{{k}} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} {y}^{\mathrm{3}} .\left(\underset{{k}=\mathrm{0}} {\overset{+\infty} {\sum}}{ke}^{−{ky}} \right){du} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\int_{\mathrm{0}} ^{+\infty} \frac{{u}^{\mathrm{3}} }{{k}^{\mathrm{3}} }{e}^{−{u}} {du} \\ $$$$=\Gamma\left(\mathrm{4}\right)\zeta\left(\mathrm{3}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}{dx}=\Gamma\left(\mathrm{3}\right).\zeta\left(\mathrm{3}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right){dx}=\Gamma\left(\mathrm{3}\right) \\ $$$${R}=−\frac{\Gamma\left(\mathrm{4}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\mathrm{4}\right)\zeta\left(\mathrm{3}\right)+\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\mathrm{3}\right)\zeta\left(\mathrm{3}\right)+\frac{\Gamma\left(\mathrm{3}\right)}{\mathrm{4}} \\ $$$$=−\mathrm{3}+\mathrm{4}\zeta\left(\mathrm{3}\right)+\frac{\mathrm{3}}{\mathrm{2}}=−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{4}\zeta\left(\mathrm{3}\right) \\ $$$$ \\ $$
Answered by mind is power last updated on 20/Mar/20
$${Page}\:\mathrm{3} \\ $$$$\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{{H}_{{n}} ^{\left(\mathrm{2}\right)} }{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)} \\ $$$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\left(−\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\right){H}_{{n}} ^{\left(\mathrm{2}\right)} \\ $$$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\left(−\frac{{H}_{{n}} ^{\left(\mathrm{2}\right)} }{{n}}+\frac{{H}_{{n}} ^{\left(\mathrm{2}\right)} }{\mathrm{2}\left({n}−\mathrm{1}\right)}+\frac{{H}_{{n}} ^{\left(\mathrm{2}\right)} }{\mathrm{2}\left({n}+\mathrm{1}\right)}\right) \\ $$$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\left(−\frac{{H}_{{n}} ^{\left(\mathrm{2}\right)} }{{n}}+\frac{{H}_{{n}−\mathrm{1}} ^{\left(\mathrm{2}\right)} }{\mathrm{2}\left({n}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)^{\mathrm{3}} }+\frac{{H}_{{n}+\mathrm{1}} ^{\left(\mathrm{2}\right)} }{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\right) \\ $$$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\left(−\frac{{H}_{{n}} ^{\left(\mathrm{2}\right)} }{{n}}+\frac{{H}_{{n}−\mathrm{1}} ^{\left(\mathrm{2}\right)} }{\mathrm{2}\left({n}−\mathrm{1}\right)}+\frac{{H}_{{n}+\mathrm{1}} ^{\left(\mathrm{2}\right)} }{\mathrm{2}\left({n}+\mathrm{1}\right)}\right)+\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)^{\mathrm{3}} }−\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\left(−\frac{{H}_{\mathrm{2}} ^{\left(\mathrm{2}\right)} }{\mathrm{2}}+\frac{{H}_{\mathrm{1}} ^{\left(\mathrm{2}\right)} }{\mathrm{2}}+\frac{{H}_{\mathrm{2}} ^{\left(\mathrm{2}\right)} }{\mathrm{4}}\right)+\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\zeta\left(\mathrm{3}\right)−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)+\frac{\mathrm{9}}{\mathrm{16}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${i}\:{Will}\:{finish}\:{Later}\: \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 19/Mar/20
$${thank}\:{you}\:{sir}\:{nice}\:{work} \\ $$
Commented by mind is power last updated on 20/Mar/20
$${withe}\:{pleasur}\:{have}\:{You}\:{the}\:{value}\:\:{of}\:{this}\:{sum}\:? \\ $$$${so}\:{long}\:{may}\:{bee}\:\:{i}\:{did}\:{mistack} \\ $$
Commented by M±th+et£s last updated on 20/Mar/20
$${sir}\:{kamal}\:{value}\:{is}\:{right}\:{sir} \\ $$
Answered by Kamel Kamel last updated on 20/Mar/20
Commented by M±th+et£s last updated on 20/Mar/20
$${great}\:{solution}\:{sir}\:{thank}\:{you} \\ $$