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Question Number 127682 by Dwaipayan Shikari last updated on 31/Dec/20

$$\frac{{\pi }^2}{1+\frac{{\pi }^2}{3-{\pi }^2+\frac{9{\pi }^2}{5-3{\pi }^2+\frac{25{\pi }^2}{7-5{\pi }^2+\frac{49{\pi }^2}{9-7{\pi }^2+\frac{81{\pi }^2}{11-9{\pi }^2+\frac{121{\pi }^2}{.....}}}}}}}$$

Commented by Dwaipayan Shikari last updated on 01/Jan/21

$$\pi {tan}^{-1}\left(\pi \right)$$

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

$${tan}^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+....$$$$=x+x(-\frac{x^2}{3})+x(-\frac{x^2}{3})(-\frac{3x^2}{5})+x(-\frac{x^2}{3})(-\frac{3x^2}{5})(-\frac{5x^2}{7})+..$$$$=\frac{x}{1+\frac{\frac{x^2}{3}}{1-\frac{x^2}{3}+\frac{\frac{x^2}{5}}{1-\frac{x^2}{5}+\frac{\frac{x^2}{7}}{1-\frac{x^2}{7}+...}}}}=\frac{x}{1+\frac{x^2}{3-x^2+\frac{3^2{x}^2}{5-3x^2+\frac{5^2{x}^2}{7-5x^2+..}}}}$$$$Takex=\pi$$

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

$$$$$$\mathrm{Thanks}\mathrm{sir}.\mathrm{My}\mathrm{respect}\mathrm{for}\mathrm{you}!$$

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