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Question Number 129084 by mnjuly1970 last updated on 12/Jan/21

       ...nice  calculus...     prove that::        (2/3)((1/4)−(1/(20))+(1/(56))−(1/(120))+(1/(220))−...)         =^(???) π −3

$$\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{20}}+\frac{\mathrm{1}}{\mathrm{56}}−\frac{\mathrm{1}}{\mathrm{120}}+\frac{\mathrm{1}}{\mathrm{220}}−...\right) \\ $$$$\:\:\:\:\:\:\:\overset{???} {=}\pi\:−\mathrm{3}\:\: \\ $$

Answered by Dwaipayan Shikari last updated on 12/Jan/21

(1/6)(1−(1/5)+(1/(14))−(1/(30))+(1/(55))−..)  =(1/6)(1−(1/(1^2 +2^2 ))+(1/(1^2 +2^2 +3^2 ))−(1/(1^2 +2^2 +3^2 +4^2 ))−...)  =Σ_(n=1) ^∞ (((−1)^(n+1) )/(n(n+1)(2n+1)))=π−3

$$\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{14}}−\frac{\mathrm{1}}{\mathrm{30}}+\frac{\mathrm{1}}{\mathrm{55}}−..\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }−...\right) \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}=\pi−\mathrm{3} \\ $$

Commented by mnjuly1970 last updated on 12/Jan/21

excellent  mr payan .  thanks alot...

$${excellent}\:\:{mr}\:{payan}\:. \\ $$$${thanks}\:{alot}... \\ $$

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