Question Number 123763 by mnjuly1970 last updated on 28/Nov/20
$$\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:\:{calculus}\:… \\ $$$$\:\:\:\:\:\mathscr{E}{valuate}\:::\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:=??? \\ $$
Answered by Snail last updated on 28/Nov/20
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}={tan}^{−\mathrm{1}} \left({x}\right) \\ $$$$ \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}={tan}^{−\mathrm{1}} \left(\mathrm{1}\right)=\pi/\mathrm{4} \\ $$
Commented by mnjuly1970 last updated on 28/Nov/20
$${thank}\:{you}\:{so}\:{much} \\ $$
Answered by Dwaipayan Shikari last updated on 28/Nov/20
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}+… \\ $$$${Another}\:{way} \\ $$$${cotx}=\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\pi−{x}}+\frac{\mathrm{1}}{\pi+{x}}−\frac{\mathrm{1}}{\mathrm{2}\pi−{x}}+\frac{\mathrm{1}}{\mathrm{2}\pi+{x}}+… \\ $$$${x}=\frac{\pi}{\mathrm{4}}\:\:\:\:\:\:\Rightarrow\mathrm{1}=\frac{\mathrm{4}}{\pi}−\frac{\mathrm{4}}{\mathrm{3}\pi}+\frac{\mathrm{4}}{\mathrm{5}\pi}−\frac{\mathrm{4}}{\mathrm{7}\pi}+… \\ $$$$\frac{\pi}{\mathrm{4}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}+… \\ $$
Commented by mnjuly1970 last updated on 28/Nov/20
$${very}\:{nice}\:{bravo}\:{sir}\:{dwaipayan}.. \\ $$