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Question Number 159173 by HongKing last updated on 13/Nov/21

Answered by qaz last updated on 14/Nov/21

Ω=Σ_(k=1) ^∞ arctan ((1/(1+(k+(5/3))(k+(8/3))))) =Σ_(k=1) ^∞ arctan (((k+(8/3))−(k+(5/3)))/(1+(k+(5/3))(k+(8/3)))) =Σ_(k=1) ^∞ (arctan (k+(8/3))−arctan (k+(5/3))) =(π/2)−arctan (8/3)

$$\Omega=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{arctan}\:\left(\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{k}+\frac{\mathrm{5}}{\mathrm{3}}\right)\left(\mathrm{k}+\frac{\mathrm{8}}{\mathrm{3}}\right)}\right) \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{arctan}\:\frac{\left(\mathrm{k}+\frac{\mathrm{8}}{\mathrm{3}}\right)−\left(\mathrm{k}+\frac{\mathrm{5}}{\mathrm{3}}\right)}{\mathrm{1}+\left(\mathrm{k}+\frac{\mathrm{5}}{\mathrm{3}}\right)\left(\mathrm{k}+\frac{\mathrm{8}}{\mathrm{3}}\right)} \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{arctan}\:\left(\mathrm{k}+\frac{\mathrm{8}}{\mathrm{3}}\right)−\mathrm{arctan}\:\left(\mathrm{k}+\frac{\mathrm{5}}{\mathrm{3}}\right)\right) \\ $$$$=\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\:\frac{\mathrm{8}}{\mathrm{3}} \\ $$

Commented by HongKing last updated on 14/Nov/21

Cool my dear Ser, thank you so much

$$\mathrm{Cool}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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