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Question Number 220395 by hardmath last updated on 12/May/25
Find: Ω =Σ_(x=1) ^∞ Σ_(y=1) ^∞ (1/(x^2 y^3 (x^2 + 1)(y + 2))) = ?
$$\mathrm{Find}:\:\:\:\Omega\:=\underset{\boldsymbol{\mathrm{x}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\underset{\boldsymbol{\mathrm{y}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:\mathrm{y}^{\mathrm{3}} \:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left(\mathrm{y}\:+\:\mathrm{2}\right)}\:=\:? \\ $$
Answered by MathematicalUser2357 last updated on 12/May/25
(1/(288))(24ζ(3)+9−2π^2 )(3+π^2 −3πcoth π)
$$\frac{\mathrm{1}}{\mathrm{288}}\left(\mathrm{24}\zeta\left(\mathrm{3}\right)+\mathrm{9}−\mathrm{2}\pi^{\mathrm{2}} \right)\left(\mathrm{3}+\pi^{\mathrm{2}} −\mathrm{3}\pi\mathrm{coth}\:\pi\right) \\ $$
Commented by hardmath last updated on 12/May/25
solution
$$\mathrm{solution} \\ $$
Answered by MrGaster last updated on 12/May/25
(1):Ω=(Σ_(x=1) ^∞ (1/(x^2 (x^2 +1))))(Σ_(y=1) ^∞ (1/(y^3 (y+2)))) Σ_(x=1) ^∞ (1/(x^2 (x^2 +1)))=Σ_(x=1) ^∞ ((1/x^2 )−(1/(x^2 +1)))=(π^2 /6)−(((π cosh π−1)/2)) Σ_(y=1) ^∞ (1/(y^3 (y+2)))=(1/8)Σ_(y=1) ^∞ (1/y)−(1/4)Σ_(y=1) ^∞ (1/y^2 )+(1/2)Σ_(y=1) ^∞ (1/y^3 )−(1/8)Σ_(y=1) ^∞ (1/(y+2)) =(3/(16))−(π^2 /(24))+((ζ(3))/2) Ω=(((π^2 +3−3π cosh π)/6))(((24ζ(3)+9−2π^2 )/(48))) Ω=(1/(288))(24ζ(3)+9−2π^2 )(3+π^2 −3π cosh π) (2):Σ_(x=1) ^∞ Σ_(y=1) ^∞ (1/(x^2 y^3 (x^2 + 1)(y + 2))) =(Σ_(x=1) ^∞ (1/(x^2 (x^2 +1))))(Σ_(y=1) ^∞ (1/(y^3 (y+2)))) Σ_(x=1) ^∞ (1/(x^2 (x^2 +1)))=Σ_(x=1) ^∞ ((1/x^2 )−(1/(x^2 −1))) =ζ(2)−((π/2)cosh π−(1/2)) =(π^2 /6)−(π/2)cosh π+(1/2) =((π^2 +3−3π cosh π)/6) Σ_(y=1) ^∞ (1/(y^3 (y+2)))=(1/8)Σ_(y=1) ^∞ ((1/y)−(2/y^2 )+(4/y^3 )−(1/(y+2))) =(1/8)(ζ(1)−2ζ(2)+4ζ(3)−Σ_(y=3) ^∞ (1/y)) =(1/8)(γ+(3/2)−2∙(π^2 /6)+4ζ(3)) =((24ζ(3)+9−2π^2 )/(48)) Ω=(((π^2 +3−3π cosh π)/6))(((24ζ(3)+9−2π^2 )/(48))) Ω=(1/(288))(24ζ(3)+9−2π^2 )(3+π^2 −3π cosh π)
$$\left(\mathrm{1}\right):\Omega=\left(\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}\right)\left(\underset{{y}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{y}^{\mathrm{3}} \left({y}+\mathrm{2}\right)}\right) \\ $$$$\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}=\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\left(\frac{\pi\:\mathrm{cosh}\:\pi−\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\underset{{y}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{y}^{\mathrm{3}} \left({y}+\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{8}}\underset{{y}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{y}}−\frac{\mathrm{1}}{\mathrm{4}}\underset{{y}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{y}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\underset{{y}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{y}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{8}}\underset{{y}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{y}+\mathrm{2}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{16}}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}+\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{2}} \\ $$$$\Omega=\left(\frac{\pi^{\mathrm{2}} +\mathrm{3}−\mathrm{3}\pi\:\mathrm{cosh}\:\pi}{\mathrm{6}}\right)\left(\frac{\mathrm{24}\zeta\left(\mathrm{3}\right)+\mathrm{9}−\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{48}}\right) \\ $$$$\Omega=\frac{\mathrm{1}}{\mathrm{288}}\left(\mathrm{24}\zeta\left(\mathrm{3}\right)+\mathrm{9}−\mathrm{2}\pi^{\mathrm{2}} \right)\left(\mathrm{3}+\pi^{\mathrm{2}} −\mathrm{3}\pi\:\mathrm{cosh}\:\pi\right) \\ $$$$\left(\mathrm{2}\right):\underset{\boldsymbol{\mathrm{x}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\underset{\boldsymbol{\mathrm{y}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:\mathrm{y}^{\mathrm{3}} \:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left(\mathrm{y}\:+\:\mathrm{2}\right)}\: \\ $$$$=\left(\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}\right)\left(\underset{{y}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{y}^{\mathrm{3}} \left({y}+\mathrm{2}\right)}\right) \\ $$$$\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}=\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$$=\zeta\left(\mathrm{2}\right)−\left(\frac{\pi}{\mathrm{2}}\mathrm{cosh}\:\pi−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\pi}{\mathrm{2}}\mathrm{cosh}\:\pi+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\pi^{\mathrm{2}} +\mathrm{3}−\mathrm{3}\pi\:\mathrm{cosh}\:\pi}{\mathrm{6}} \\ $$$$\underset{{y}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{y}^{\mathrm{3}} \left({y}+\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{8}}\underset{{y}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{y}}−\frac{\mathrm{2}}{{y}^{\mathrm{2}} }+\frac{\mathrm{4}}{{y}^{\mathrm{3}} }−\frac{\mathrm{1}}{{y}+\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\zeta\left(\mathrm{1}\right)−\mathrm{2}\zeta\left(\mathrm{2}\right)+\mathrm{4}\zeta\left(\mathrm{3}\right)−\underset{{y}=\mathrm{3}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{y}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\gamma+\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{2}\centerdot\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\mathrm{4}\zeta\left(\mathrm{3}\right)\right) \\ $$$$=\frac{\mathrm{24}\zeta\left(\mathrm{3}\right)+\mathrm{9}−\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$$\Omega=\left(\frac{\pi^{\mathrm{2}} +\mathrm{3}−\mathrm{3}\pi\:\mathrm{cosh}\:\pi}{\mathrm{6}}\right)\left(\frac{\mathrm{24}\zeta\left(\mathrm{3}\right)+\mathrm{9}−\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{48}}\right) \\ $$$$\Omega=\frac{\mathrm{1}}{\mathrm{288}}\left(\mathrm{24}\zeta\left(\mathrm{3}\right)+\mathrm{9}−\mathrm{2}\pi^{\mathrm{2}} \right)\left(\mathrm{3}+\pi^{\mathrm{2}} −\mathrm{3}\pi\:\mathrm{cosh}\:\pi\right) \\ $$
Commented by hardmath last updated on 12/May/25
Amazing solution as always, thank you very much, my valuable professor...
$$ \\ $$Amazing solution as always, thank you very much, my valuable professor…
Commented by mr W last updated on 12/May/25
maybe you meant coth π instead of cosh π.
$${maybe}\:{you}\:{meant}\:\mathrm{coth}\:\pi\:{instead}\:{of} \\ $$$$\mathrm{cosh}\:\pi. \\ $$

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