Question Number 99005 by bobhans last updated on 18/Jun/20
$$\underset{\mathrm{m}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{mn}\left(\mathrm{m}+\mathrm{n}\right)}\:?\: \\ $$
Answered by maths mind last updated on 18/Jun/20
$$\underset{{m}\geqslant\mathrm{1}} {\sum}.\frac{\mathrm{1}}{{m}^{\mathrm{2}} }\underset{{n}\geqslant\mathrm{1}} {\sum}.\frac{\left({n}+{m}\right)−{n}}{{n}\left({n}+{m}\right)} \\ $$$$=\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{m}^{\mathrm{2}} }.\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+{m}}\right) \\ $$$$=\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{m}^{\mathrm{2}} }.{H}_{{m}} \\ $$$$=\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{{H}_{{m}} }{{m}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right){dx}=−\frac{{H}_{{n}} }{{n}} \\ $$$$\Rightarrow\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{{H}_{{m}} }{{m}^{\mathrm{2}} }=−\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{m}\geqslant\mathrm{1}} {\sum}\frac{{x}^{{m}−\mathrm{1}} }{{m}}{ln}\left(\mathrm{1}−{x}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{{x}}\left(\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{{x}^{{m}} }{{m}}\right){ln}\left(\mathrm{1}−{x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({t}\right)}{\mathrm{1}−{t}}{dt}\:\:\:\:,{ln}\left({t}\right)=−{r}\Rightarrow \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{2}} {e}^{−{t}} {dt}}{\mathrm{1}−{e}^{−{t}} }=\int_{\mathrm{0}} ^{+\infty} \frac{{t}^{\mathrm{2}} {dt}}{{e}^{{t}} −\mathrm{1}}=\Gamma\left(\mathrm{3}\right)\zeta\left(\mathrm{3}\right) \\ $$$${i}\:{used}\:\int_{\mathrm{0}} ^{+\infty} \frac{{t}^{{a}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}=\Gamma\left({a}\right)\zeta\left({a}\right)\:{for}\:{Re}\left({a}\right)>\mathrm{1} \\ $$$${we}\:{get}\Gamma\left(\mathrm{3}\right)\zeta\left(\mathrm{3}\right)=\mathrm{2}\zeta\left(\mathrm{3}\right)=\underset{{m}\geqslant\mathrm{1}} {\sum}.\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{mn}\left({n}+{m}\right)} \\ $$
Answered by bemath last updated on 18/Jun/20
![∫_0 ^1 ∫_0 ^1 ∫_0 ^1 ((z dxdydz)/((1−xz)(1−yz))) = ∫_0 ^1 z {∫_0 ^1 (dx/(1−xz)) ∫_0 ^1 (dy/(1−yz)) } dz = ∫_0 ^1 z {[ ((−ln(1−xz))/z) ]_0 ^1 × [((−ln(1−yz))/z)]_0 ^1 } dz = ∫_0 ^1 ((( ln(1−z))^2 )/z) dz , set 1−z =t = ∫_0 ^1 ((( ln(t))^2 )/(1−t)) dt = ∫_0 ^1 (ln (t))^2 Σ_(n=0) ^∞ t^n = Σ_(n=0) ^∞ ∫_0 ^1 t^n (ln (t))^2 dt = 2 Σ_(n=0) ^∞ (1/((n+1)^3 )) = 2 Σ_(n=0) ^∞ (1/n^3 ) = 2 ζ(3) ■](Q99019.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{z}\:\mathrm{dxdydz}}{\left(\mathrm{1}−\mathrm{xz}\right)\left(\mathrm{1}−\mathrm{yz}\right)}\:= \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{z}\:\left\{\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\mathrm{1}−\mathrm{xz}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dy}}{\mathrm{1}−\mathrm{yz}}\:\right\}\:\mathrm{dz}\:= \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{z}\:\left\{\left[\:\frac{−\mathrm{ln}\left(\mathrm{1}−\mathrm{xz}\right)}{\mathrm{z}}\:\right]_{\mathrm{0}} ^{\mathrm{1}} \:×\:\left[\frac{−\mathrm{ln}\left(\mathrm{1}−\mathrm{yz}\right)}{\mathrm{z}}\right]_{\mathrm{0}} ^{\mathrm{1}} \right\}\:\mathrm{dz} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(\:\mathrm{ln}\left(\mathrm{1}−\mathrm{z}\right)\right)^{\mathrm{2}} }{\mathrm{z}}\:\mathrm{dz}\:,\:\mathrm{set}\:\mathrm{1}−\mathrm{z}\:=\mathrm{t} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(\:\mathrm{ln}\left(\mathrm{t}\right)\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{t}}\:\mathrm{dt}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{ln}\:\left(\mathrm{t}\right)\right)^{\mathrm{2}} \:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{t}^{\mathrm{n}} \\ $$$$=\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{t}^{\mathrm{n}} \:\left(\mathrm{ln}\:\left(\mathrm{t}\right)\right)^{\mathrm{2}} \:\mathrm{dt}\: \\ $$$$=\:\mathrm{2}\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} }\:=\:\mathrm{2}\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\: \\ $$$$=\:\mathrm{2}\:\zeta\left(\mathrm{3}\right)\:\blacksquare \\ $$
Commented by maths mind last updated on 18/Jun/20
$${nice} \\ $$$$\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right).\left({m}+\mathrm{1}\right)\left({n}+{m}+\mathrm{2}\right)} \\ $$$${just}\:{a}\:{littel}\:{justifaction} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {dx}.\int_{\mathrm{0}} ^{\mathrm{1}} {y}^{{m}} {dy}.\int_{\mathrm{0}} ^{\mathrm{1}} {z}^{{n}+{m}+\mathrm{1}} {dz} \\ $$$$=\frac{\mathrm{1}}{{n}+\mathrm{1}}.\frac{\mathrm{1}}{{m}+\mathrm{1}}.\frac{\mathrm{1}}{\left({n}+{m}+\mathrm{2}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {z}\int_{\mathrm{0}} ^{\mathrm{1}} \left({xz}\right)^{{n}} {dx}.\int_{\mathrm{0}} ^{\mathrm{1}} \left({zy}\right)^{{m}} {dy}.{dz} \\ $$$$\underset{{m}} {\sum}.\underset{{n}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} {z}\int_{\mathrm{0}} ^{\mathrm{1}} \left({xz}\right)^{{n}} {dx}.\int_{\mathrm{0}} ^{\mathrm{1}} \left({yz}\right)^{{m}} {dy}.{dz} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{zdxdydz}}{\left(\mathrm{1}−{xz}\right)\left(\mathrm{1}−{yz}\right)}=\Sigma\Sigma\frac{\mathrm{1}}{{nm}\left({n}+{m}\right)} \\ $$
Commented by bemath last updated on 18/Jun/20
$$\mathrm{great}\:\mathrm{sir} \\ $$