Question Number 127682 by Dwaipayan Shikari last updated on 31/Dec/20
$$\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{3}−\pi^{\mathrm{2}} +\frac{\mathrm{9}\pi^{\mathrm{2}} }{\mathrm{5}−\mathrm{3}\pi^{\mathrm{2}} +\frac{\mathrm{25}\pi^{\mathrm{2}} }{\mathrm{7}−\mathrm{5}\pi^{\mathrm{2}} +\frac{\mathrm{49}\pi^{\:\mathrm{2}} }{\mathrm{9}−\mathrm{7}\pi^{\mathrm{2}} +\frac{\mathrm{81}\pi^{\mathrm{2}} }{\mathrm{11}−\mathrm{9}\pi^{\mathrm{2}} +\frac{\mathrm{121}\pi^{\mathrm{2}} }{…..}}}}}}} \\ $$
Commented by Dwaipayan Shikari last updated on 01/Jan/21
$$\pi{tan}^{−\mathrm{1}} \left(\pi\right) \\ $$
Commented by A8;15: last updated on 03/Jan/21
$$\mathrm{It}\:\mathrm{will}\:\mathrm{be}\:\mathrm{nice}\:\mathrm{if}\:\mathrm{you}\:\mathrm{show}\:\mathrm{us}\:\mathrm{a}\:\mathrm{proof},\mathrm{sir}! \\ $$
Answered by Dwaipayan Shikari last updated on 03/Jan/21
$${tan}^{−\mathrm{1}} {x}={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}}+…. \\ $$$$\:\:\:\:\:\:\:\:\:=\:\:{x}+{x}\left(−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\right)+{x}\left(−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\right)\left(−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{5}}\right)+{x}\left(−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\right)\left(−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{5}}\right)\left(−\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{7}}\right)+.. \\ $$$$\:\:\:\:\:\:\:=\:\:\frac{{x}}{\mathrm{1}+\frac{\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}+\frac{\frac{{x}^{\mathrm{2}} }{\mathrm{5}}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{5}}+\frac{\frac{{x}^{\mathrm{2}} }{\mathrm{7}}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{7}}+…}}}}=\frac{{x}}{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}−{x}^{\mathrm{2}} +\frac{\mathrm{3}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{5}−\mathrm{3}{x}^{\mathrm{2}} +\frac{\mathrm{5}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{7}−\mathrm{5}{x}^{\mathrm{2}} +..}}}} \\ $$$${Take}\:{x}=\pi \\ $$
Commented by Dwaipayan Shikari last updated on 03/Jan/21
https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula
Commented by A8;15: last updated on 03/Jan/21
$$ \\ $$$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{My}\:\mathrm{respect}\:\mathrm{for}\:\mathrm{you}! \\ $$