New year Infinite continued fractions

Calculus Level 3

π 2 1 + π 2 3 − π 2 + 9 π 2 5 − 3 π 2 + 25 π 2 7 − 5 π 2 + ⋯ = ? \Large \frac{π^2}{1+\frac{π^2}{3-π^2 +\frac{9π^2}{5-3π^2 +\frac{25π^2 }{7-5π^2 +\cdots}}}} = \ ?

References:


The answer is 3.966.

Dwaipayan Shikari by Dwaipayan Shikari , 17, India

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1 solution

tan ⁡ − 1 ( x ) = x − x 3 3 + x 5 5 − x 7 7 + ⋯ = x + x ( − x 2 3 ) + x ( − x 2 3 ) ( − 3 x 2 5 ) + x ( − x 2 3 ) ( − 3 x 2 5 ) ( − 5 x 2 7 ) + ⋯ \tan^{-1}(x) = x-\frac{x^3}{3} +\frac{x^5}{5} -\frac{x^7}{7} +\cdots= 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})+\cdots

= x 1 + x 2 3 1 − ( x 2 3 ) + 3 x 2 5 1 − ( x 2 5 ) + ⋯ = \huge\frac{x}{1+\frac{\frac{x^2}{3}}{1-(\frac{x^2}{3})+\frac{\frac{3x^2}{5}}{1-(\frac{x^2}{5})+\cdots}}}

= x 1 + x 2 3 − x 2 + 9 x 2 5 − 3 x 2 + 25 x 2 5 − 7 x 2 + ⋯ =\huge \frac{x}{1+\frac{x^2}{3-x^2 +\frac{9x^2}{5-3x^2 +\frac{25x^2}{5-7x^2 +\cdots}}}}

Take x = π x=π then , π 2 1 + π 2 3 − π 2 + 9 π 2 5 − 3 π 2 + 25 π 2 7 − 5 π 2 + ⋯ = π tan ⁡ − 1 ( π ) = ≈ 3.966 \large\dfrac{π^2}{1+\frac{π^2}{3-π^2 +\frac{9π^2}{5-3π^2 +\frac{25π^2 }{7-5π^2 +\cdots}}}} =π \tan^{-1}(π) =\color{#E81990}\boxed{≈3.966} Relevant wiki Euler continued fractions And Continued fractions

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