Question Number 222478 by mnjuly1970 last updated on 28/Jun/25
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{{H}_{{n}} }{{n}^{\mathrm{2}} }\:=\:? \\ $$$$\:{note}:\:\:\:{H}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+…+\frac{\mathrm{1}}{{n}}\: \\ $$
Answered by MrGaster last updated on 28/Jun/25
![H_n =∫_0 ^1 ((1−t^n )/(1−t))dt S=Σ_(n=1) ^∞ (−1)^(n−1) (1/n^2 )∫_0 ^1 ((1−t^n )/(1−t))dt=∫_0 ^1 (1/(1−t))Σ_(n=1) ^∞ (−1)^(n−1) ((1−t^n )/n^2 )dt Σ_(n=1) ^∞ (−1)^(n−1) (1/n^2 )=η(2)=(π^2 /(12)) Σ_(n=1) ^∞ (−1)^(n−1) (t^m /n^2 )=∫_0 ^t ((ln(1+u))/u)du=−Li_2 (−t) S=∫_0 ^1 (1/(1−t))((π^2 /(12))+Li_2 (−t))dt Li_2 (−t)=−∫_0 ^1 ((ln(1+u))/u)du S=∫_0 ^1 (1/(1−t))((π^2 /(12))−(−∫_0 ^1 ((ln(1+u))/u)du))dt=∫_0 ^1 (1/(1−t)) (π^2 /(12))dt−∫_0 ^1 (1/(1−t))∫_0 ^1 ((ln(1+u))/u)dudt {(x,t)∣0<u<t<1}≡0<u<1,u<t<1 ∫_0 ^1 (1/(1−t))∫_0 ^1 ((ln(1+u))/u)dudt=∫_0 ^1 ((ln(1+u))/u)∫(1/(1−t))dtdu ∫_0 ^1 (1/(1−t))dt=[−ln(1−t)]_0 ^1 =ln(1/(1−u)) ∫_0 ^1 (1/(1−t))dt=(π^2 /(12))∫_0 ^1 ln(1/(1−u))du=(π^2 /(12))∫_0 ^1 (−ln(1+u)))du=(π^2 /(12))∫_0 ^1 Σ_(k=1) ^∞ (u^k /k)du=(π^2 /(12))Σ_(k=1) ^∞ (1/k) (1/(k+1))=(π^2 /(12))Σ_(k=1) ^∞ ((1/k)−(1/(k+1)))=(π^2 /(12))∙1 ∫_0 ^1 ((ln(1+u)ln(1−u))/u)du=−∫_0 ^1 (1/u)Σ_(m=1) ^∞ Σ_(k=1) ^∞ (−1)^(m−1) (1/m) (1/k)u^(m+k) du=−Σ_(m=1) ^∞ Σ_(k=1) ^∞ (−1)^(m+1) (1/(mk)) (1/(m+k)) Σ_(m=1) ^∞ Σ_(k=1) ^∞ (−1)^(m+1) (1/(mk(m+k)))=(5/8)ζ(3)](https://www.tinkutara.com/question/Q222486.png)
$$ \\ $$$${H}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{t}^{{n}} }{\mathrm{1}−{t}}{dt} \\ $$$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{\mathrm{1}}{{n}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{t}^{{n}} }{\mathrm{1}−{t}}{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−{t}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{\mathrm{1}−{t}^{{n}} }{{n}^{\mathrm{2}} }{dt} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\eta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{{t}^{{m}} }{{n}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{{t}} \frac{\mathrm{ln}\left(\mathrm{1}+{u}\right)}{{u}}{du}=−\mathrm{Li}_{\mathrm{2}} \left(−{t}\right) \\ $$$${S}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−{t}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\mathrm{Li}_{\mathrm{2}} \left(−{t}\right)\right){dt} \\ $$$$\mathrm{Li}_{\mathrm{2}} \left(−{t}\right)=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{u}\right)}{{u}}{du} \\ $$$${S}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−{t}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\left(−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{u}\right)}{{u}}{du}\right)\right){dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−{t}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{12}}{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−{t}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{u}\right)}{{u}}{dudt} \\ $$$$\left\{\left({x},{t}\right)\mid\mathrm{0}<{u}<{t}<\mathrm{1}\right\}\equiv\mathrm{0}<{u}<\mathrm{1},{u}<{t}<\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−{t}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{u}\right)}{{u}}{dudt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{u}\right)}{{u}}\int\frac{\mathrm{1}}{\mathrm{1}−{t}}{dtdu} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−{t}}{dt}=\left[−\mathrm{ln}\left(\mathrm{1}−{t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{ln}\frac{\mathrm{1}}{\mathrm{1}−{u}} \\ $$$$\left.\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−{t}}{dt}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\frac{\mathrm{1}}{\mathrm{1}−{u}}{du}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{ln}\left(\mathrm{1}+{u}\right)\right)\right){du}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{u}^{{k}} }{{k}}{du}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}}\:\frac{\mathrm{1}}{{k}+\mathrm{1}}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\centerdot\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{u}\right)\mathrm{ln}\left(\mathrm{1}−{u}\right)}{{u}}{du}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{u}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{m}−\mathrm{1}} \frac{\mathrm{1}}{{m}}\:\frac{\mathrm{1}}{{k}}{u}^{{m}+{k}} {du}=−\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{m}+\mathrm{1}} \frac{\mathrm{1}}{{mk}}\:\frac{\mathrm{1}}{{m}+{k}} \\ $$$$\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{m}+\mathrm{1}} \frac{\mathrm{1}}{{mk}\left({m}+{k}\right)}=\frac{\mathrm{5}}{\mathrm{8}}\zeta\left(\mathrm{3}\right) \\ $$
Commented by mnjuly1970 last updated on 08/Jul/25
$$\: \\ $$