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Question Number 153103 by amin96 last updated on 04/Sep/21

(1/(2∙3))+(1/(4∙5∙6))+(1/(7∙8∙9))+…

$$\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}\centerdot\mathrm{5}\centerdot\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{7}\centerdot\mathrm{8}\centerdot\mathrm{9}}+\ldots \\ $$

Answered by puissant last updated on 04/Sep/21

S=Σ_(n=0) ^∞ (1/((3n+1)(3n+2)(3n+3))) =Σ_(n=0) ^∞ ((1/((3n+1)(3n+2)))−(1/((3n+1)(3n+3)))) =(1/9)Σ_(n=0) ^∞ ((1/((n+(1/3))(n+(2/3))))−(1/((n+(1/3))(n+1)))) =(1/9)Σ_(n=0) ^∞ (((ψ((2/3))−ψ((1/3)))/((2/3)−(1/3)))−((ψ(1)−ψ((1/3)))/(1−(1/3)))) =(1/9)(((π/( (√3)))/(1/3))−((−γ+(π/(2(√3)))+(3/2)ln3+γ)/(2/3))) =(1/9)(π(√3)−((π(√3))/4)−(9/4)ln3) ∴∵ S=(π/(4(√3)))−(1/4)ln3.

$${S}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)\left(\mathrm{3}{n}+\mathrm{3}\right)} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)}−\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{3}\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right)}−\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left({n}+\mathrm{1}\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}{\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\psi\left(\mathrm{1}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left(\frac{\frac{\pi}{\:\sqrt{\mathrm{3}}}}{\frac{\mathrm{1}}{\mathrm{3}}}−\frac{−\gamma+\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}+\frac{\mathrm{3}}{\mathrm{2}}{ln}\mathrm{3}+\gamma}{\frac{\mathrm{2}}{\mathrm{3}}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left(\pi\sqrt{\mathrm{3}}−\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{4}}−\frac{\mathrm{9}}{\mathrm{4}}{ln}\mathrm{3}\right) \\ $$$$ \\ $$$$\therefore\because\:\:\:{S}=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{\mathrm{4}}{ln}\mathrm{3}. \\ $$

Commented by amin96 last updated on 04/Sep/21

Thanks sir

$${Thanks}\:{sir} \\ $$

Commented by Tawa11 last updated on 05/Sep/21

weldone sir

$$\mathrm{weldone}\:\mathrm{sir} \\ $$

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