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Question Number 184144 by paul2222 last updated on 03/Jan/23

𝚺_(n=1) ^∞ 𝚺_(m=1) ^∞ (((βˆ’1)^(n+m) )/(nm(n+m)))

$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\underset{\boldsymbol{\mathrm{m}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\left(βˆ’\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}+\boldsymbol{\mathrm{m}}} }{\boldsymbol{\mathrm{nm}}\left(\boldsymbol{\mathrm{n}}+\boldsymbol{\mathrm{m}}\right)} \\ $$

Answered by SEKRET last updated on 04/Jan/23

Ξ£_(n=1) ^∞ (1/n)βˆ™Ξ£_(m=1) ^∞ (((βˆ’1)^(n+m) )/(mβˆ™(n+m))) f(x)_(x=βˆ’1) = Ξ£_(m=1) ^∞ (x^(n+m) /(mβˆ™(n+m))) f β€² (x) = Ξ£_(m=1) ^∞ (x^(n+mβˆ’1) /m)=x^(nβˆ’1) βˆ™Ξ£_(m=1) ^∞ (x^m /m) f β€² (x) = x^(nβˆ’1) βˆ™ ln((1/(1βˆ’x))) C=0 Ξ£_(n=1) ^∞ (((βˆ’1)/n)βˆ™(∫x^(nβˆ’1) βˆ™ln(1βˆ’x) dx))_(x=βˆ’1) ^( C=0) Ξ£_(n=1) ^∞ (βˆ’ ((x^n βˆ™ln(1βˆ’x) +B_x (n+1,0))/n^2 ) ) Ξ£_(n=1) ^∞ (((βˆ’(βˆ’1)^n βˆ™ln(2))/n^2 ) βˆ’ ((B_(βˆ’1) (n+1 ; 0))/n^2 ))= =βˆ’1βˆ™ln(2)βˆ™Ξ£_(n=1) ^∞ (((βˆ’1)^n )/n^2 ) + Ξ£_(n=1) ^∞ ((βˆ’B_(βˆ’1) (n+1,0))/n^2 )= = ((𝛑^2 βˆ™ln(2))/(12)) + (Ξ£_(n=1) ^∞ (((βˆ’1)^(n+1) 𝚿^0 ((n/2)+(1/2)))/(2n^2 )) +Ξ£_(n=0) ^∞ ((βˆ’πšΏ^0 ((n/2)+1))/(2n^2 ))) = ((𝛑^2 βˆ™ln(2))/(12)) +( βˆ’0.28159 βˆ’0.337683) β‰ˆ 0.0491823 ABDULAZIZ ABDUVALIYEV

$$\:\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\centerdot\underset{\boldsymbol{\mathrm{m}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}+\boldsymbol{\mathrm{m}}} }{\boldsymbol{\mathrm{m}}\centerdot\left(\boldsymbol{\mathrm{n}}+\boldsymbol{\mathrm{m}}\right)} \\ $$$$\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)_{\boldsymbol{\mathrm{x}}=βˆ’\mathrm{1}} \:\:=\:\underset{\boldsymbol{\mathrm{m}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}+\boldsymbol{\mathrm{m}}} }{\boldsymbol{\mathrm{m}}\centerdot\left(\boldsymbol{\mathrm{n}}+\boldsymbol{\mathrm{m}}\right)} \\ $$$$\:\:\:\boldsymbol{\mathrm{f}}\:'\:\left(\boldsymbol{\mathrm{x}}\right)\:=\:\underset{\boldsymbol{\mathrm{m}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}+\boldsymbol{\mathrm{m}}βˆ’\mathrm{1}} }{\boldsymbol{\mathrm{m}}}=\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}βˆ’\mathrm{1}} \centerdot\underset{\boldsymbol{\mathrm{m}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{m}}} }{\boldsymbol{\mathrm{m}}} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{f}}\:'\:\left(\boldsymbol{\mathrm{x}}\right)\:=\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}βˆ’\mathrm{1}} \:\centerdot\:\boldsymbol{\mathrm{ln}}\left(\frac{\mathrm{1}}{\mathrm{1}βˆ’\boldsymbol{\mathrm{x}}}\right)\:\:\:\:\:\:\boldsymbol{\mathrm{C}}=\mathrm{0} \\ $$$$\:\:\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\frac{βˆ’\mathrm{1}}{\boldsymbol{\mathrm{n}}}\centerdot\left(\int\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}βˆ’\mathrm{1}} \centerdot\boldsymbol{\mathrm{ln}}\left(\mathrm{1}βˆ’\boldsymbol{\mathrm{x}}\right)\:\boldsymbol{\mathrm{dx}}\right)\right)_{\boldsymbol{\mathrm{x}}=βˆ’\mathrm{1}} ^{\:\:\:\boldsymbol{\mathrm{C}}=\mathrm{0}} \\ $$$$\:\:\:\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(βˆ’\:\frac{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}} \centerdot\boldsymbol{\mathrm{ln}}\left(\mathrm{1}βˆ’\boldsymbol{\mathrm{x}}\right)\:+\boldsymbol{\mathrm{B}}_{\boldsymbol{\mathrm{x}}} \left(\boldsymbol{\mathrm{n}}+\mathrm{1},\mathrm{0}\right)}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }\:\right) \\ $$$$\:\:\:\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\frac{βˆ’\left(βˆ’\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} \centerdot\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }\:βˆ’\:\frac{\boldsymbol{\mathrm{B}}_{βˆ’\mathrm{1}} \left(\boldsymbol{\mathrm{n}}+\mathrm{1}\:;\:\mathrm{0}\right)}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }\right)= \\ $$$$\:\:=βˆ’\mathrm{1}\centerdot\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)\centerdot\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }\:+\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{βˆ’\boldsymbol{\mathrm{B}}_{βˆ’\mathrm{1}} \left(\boldsymbol{\mathrm{n}}+\mathrm{1},\mathrm{0}\right)}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }= \\ $$$$=\:\frac{\boldsymbol{\pi}^{\mathrm{2}} \centerdot\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)}{\mathrm{12}}\:+\:\left(\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \boldsymbol{\Psi}^{\mathrm{0}} \left(\frac{\boldsymbol{\mathrm{n}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\boldsymbol{\mathrm{n}}^{\mathrm{2}} }\:+\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{βˆ’\boldsymbol{\Psi}^{\mathrm{0}} \left(\frac{\boldsymbol{\mathrm{n}}}{\mathrm{2}}+\mathrm{1}\right)}{\mathrm{2}\boldsymbol{\mathrm{n}}^{\mathrm{2}} }\right) \\ $$$$=\:\frac{\boldsymbol{\pi}^{\mathrm{2}} \centerdot\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)}{\mathrm{12}}\:+\left(\:βˆ’\mathrm{0}.\mathrm{28159}\:βˆ’\mathrm{0}.\mathrm{337683}\right) \\ $$$$\:\:\approx\:\mathrm{0}.\mathrm{0491823} \\ $$$$\boldsymbol{{ABDULAZIZ}}\:\:\boldsymbol{{ABDUVALIYEV}} \\ $$$$\:\:\:\:\:\: \\ $$

Commented by paul2222 last updated on 11/Jan/23

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Answered by witcher3 last updated on 05/Jan/23

∫_0 ^1 ((ln^2 (1+x))/x)dx=βˆ’li_2 (βˆ’x)ln(1+x)+∫_0 ^1 ((li_2 (βˆ’x))/(1+x))dx....(E) Li_2 (βˆ’x)=ΞΆ(2)βˆ’ln(1+x)ln(βˆ’x)βˆ’li_2 (1+x) =li_2 (βˆ’x)ln(1+x)βˆ’Li_3 (1+x)+ΞΆ(2)ln(1+x) βˆ’βˆ«_0 ^1 ((ln(1+x))/(1+x))ln(βˆ’x) =βˆ’(1/2)ln^2 (1+x)ln(βˆ’x)+(1/2)∫((ln^2 (1+x))/x)dx =βˆ’2li_2 (βˆ’x)ln(1+x)βˆ’2Li_3 (1+x)+2ΞΆ(2)ln(1+x)βˆ’ln^2 (1+x)ln(βˆ’x) li_2 (βˆ’x)=ΞΆ(2)βˆ’ln(βˆ’x)ln(1+x)βˆ’li_2 (1+x) ∫((ln^2 (1+x))/x)dx=βˆ’2li_3 (1+x)+2li_2 (1+x)ln(1+x)+ln^2 (1+x)ln(βˆ’x)+c Let f(x)=Ξ£_(nβ‰₯1) Ξ£_(mβ‰₯1) (((βˆ’1)^(n+m) x^(n+m) )/(nm(n+m))),βˆ€x∈[βˆ’1,1] fβ€²(x)=ΣΣ(((βˆ’1)^(n+m) )/(nm))x^(n+mβˆ’1) =(1/x)ΣΣ(((βˆ’x)^n (βˆ’x)^m )/(n.m)) =(1/x){Ξ£(((βˆ’x)^n )/n)}{Ξ£(((βˆ’x)^m )/m)} Ξ£_(nβ‰₯1) (((βˆ’1)^(nβˆ’1) x^n )/n)=ln(1+x) f^β€² (x)=((ln^(2() (1+x))/x) f(x)=∫((ln^2 (1+x))/x)dx...(E) apply f(x)=βˆ’2Li_3 (1+x)+2Li_2 (1+x)ln(1+x)+ln(βˆ’x)ln^2 (1+x)+c f(0)=0 lim_(xβ†’0) {βˆ’2Li_3 (1+x)+2Li_2 (1+x)ln(1+x)+ln(βˆ’x)ln^2 (1+x)+c}=0 ⇔cβˆ’2ΞΆ(3)=0 c=2ΞΆ(3) f(x)=βˆ’2Li_3 (1+x)+2Li_2 (1+x)ln(1+x)+ln(βˆ’x)ln^2 (1+x)+2ΞΆ(3) our Sum=f(1)

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}{{x}}{dx}=βˆ’{li}_{\mathrm{2}} \left(βˆ’{x}\right){ln}\left(\mathrm{1}+{x}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{li}_{\mathrm{2}} \left(βˆ’{x}\right)}{\mathrm{1}+{x}}{dx}....\left({E}\right) \\ $$$${Li}_{\mathrm{2}} \left(βˆ’{x}\right)=\zeta\left(\mathrm{2}\right)βˆ’{ln}\left(\mathrm{1}+{x}\right){ln}\left(βˆ’{x}\right)βˆ’{li}_{\mathrm{2}} \left(\mathrm{1}+{x}\right) \\ $$$$={li}_{\mathrm{2}} \left(βˆ’{x}\right){ln}\left(\mathrm{1}+{x}\right)βˆ’{Li}_{\mathrm{3}} \left(\mathrm{1}+{x}\right)+\zeta\left(\mathrm{2}\right){ln}\left(\mathrm{1}+{x}\right) \\ $$$$βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{ln}\left(βˆ’{x}\right) \\ $$$$=βˆ’\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right){ln}\left(βˆ’{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}{{x}}{dx} \\ $$$$=βˆ’\mathrm{2}{li}_{\mathrm{2}} \left(βˆ’{x}\right){ln}\left(\mathrm{1}+{x}\right)βˆ’\mathrm{2}{Li}_{\mathrm{3}} \left(\mathrm{1}+{x}\right)+\mathrm{2}\zeta\left(\mathrm{2}\right){ln}\left(\mathrm{1}+{x}\right)βˆ’{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right){ln}\left(βˆ’{x}\right) \\ $$$${li}_{\mathrm{2}} \left(βˆ’{x}\right)=\zeta\left(\mathrm{2}\right)βˆ’{ln}\left(βˆ’{x}\right){ln}\left(\mathrm{1}+{x}\right)βˆ’{li}_{\mathrm{2}} \left(\mathrm{1}+{x}\right) \\ $$$$\int\frac{{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}{{x}}{dx}=βˆ’\mathrm{2}{li}_{\mathrm{3}} \left(\mathrm{1}+{x}\right)+\mathrm{2}{li}_{\mathrm{2}} \left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}+{x}\right)+{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right){ln}\left(βˆ’{x}\right)+{c} \\ $$$${Let}\:{f}\left({x}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{\left(βˆ’\mathrm{1}\right)^{{n}+{m}} {x}^{{n}+{m}} }{{nm}\left({n}+{m}\right)},\forall{x}\in\left[βˆ’\mathrm{1},\mathrm{1}\right] \\ $$$${f}'\left({x}\right)=\Sigma\Sigma\frac{\left(βˆ’\mathrm{1}\right)^{{n}+{m}} }{{nm}}{x}^{{n}+{m}βˆ’\mathrm{1}} =\frac{\mathrm{1}}{{x}}\Sigma\Sigma\frac{\left(βˆ’{x}\right)^{{n}} \left(βˆ’{x}\right)^{{m}} }{{n}.{m}} \\ $$$$=\frac{\mathrm{1}}{{x}}\left\{\Sigma\frac{\left(βˆ’{x}\right)^{{n}} }{{n}}\right\}\left\{\Sigma\frac{\left(βˆ’{x}\right)^{{m}} }{{m}}\right\} \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(βˆ’\mathrm{1}\right)^{{n}βˆ’\mathrm{1}} {x}^{{n}} }{{n}}={ln}\left(\mathrm{1}+{x}\right) \\ $$$${f}^{'} \left({x}\right)=\frac{{ln}^{\mathrm{2}\left(\right.} \left(\mathrm{1}+{x}\right)}{{x}} \\ $$$${f}\left({x}\right)=\int\frac{{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}{{x}}{dx}...\left({E}\right)\:{apply}\: \\ $$$${f}\left({x}\right)=βˆ’\mathrm{2}{Li}_{\mathrm{3}} \left(\mathrm{1}+{x}\right)+\mathrm{2}{Li}_{\mathrm{2}} \left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}+{x}\right)+{ln}\left(βˆ’{x}\right){ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)+{c} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{βˆ’\mathrm{2}{Li}_{\mathrm{3}} \left(\mathrm{1}+{x}\right)+\mathrm{2}{Li}_{\mathrm{2}} \left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}+{x}\right)+{ln}\left(βˆ’{x}\right){ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)+{c}\right\}=\mathrm{0} \\ $$$$\Leftrightarrow{c}βˆ’\mathrm{2}\zeta\left(\mathrm{3}\right)=\mathrm{0} \\ $$$${c}=\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$$ \\ $$$${f}\left({x}\right)=βˆ’\mathrm{2}{Li}_{\mathrm{3}} \left(\mathrm{1}+{x}\right)+\mathrm{2}{Li}_{\mathrm{2}} \left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}+{x}\right)+{ln}\left(βˆ’{x}\right){ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)+\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$${our}\:{Sum}={f}\left(\mathrm{1}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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