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1-1-4-1-8-1-10-1-14-1-16-




Question Number 123650 by Dwaipayan Shikari last updated on 27/Nov/20
1−(1/4)+(1/8)−(1/(10))+(1/(14))−(1/(16))+...
$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{14}}−\frac{\mathrm{1}}{\mathrm{16}}+… \\ $$
Commented by MJS_new last updated on 27/Nov/20
sorry I misread it...
$$\mathrm{sorry}\:\mathrm{I}\:\mathrm{misread}\:\mathrm{it}… \\ $$
Commented by Dwaipayan Shikari last updated on 27/Nov/20
How do you find it sir?
$${How}\:{do}\:{you}\:{find}\:{it}\:{sir}? \\ $$
Commented by MJS_new last updated on 27/Nov/20
=1−Σ(1/(4+6n))+Σ(1/(8+6n))=  =1−(1/9)Σ(1/(n^2 +2n+8/9))=1−((1/2)−((√3)/(18))π)=  =((√3)/(18))π+(1/2)
$$=\mathrm{1}−\Sigma\frac{\mathrm{1}}{\mathrm{4}+\mathrm{6}{n}}+\Sigma\frac{\mathrm{1}}{\mathrm{8}+\mathrm{6}{n}}= \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}\Sigma\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{8}/\mathrm{9}}=\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{18}}\pi\right)= \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{18}}\pi+\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Dwaipayan Shikari last updated on 27/Nov/20
Yes sir! i have also find (1/2)+(π/(6(√3)))
$${Yes}\:{sir}!\:{i}\:{have}\:{also}\:{find}\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{6}\sqrt{\mathrm{3}}} \\ $$

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