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Question Number 127644 by Dwaipayan Shikari last updated on 31/Dec/20

$$\frac{1}{1-\frac{1}{2-\frac{\frac{1}{2}}{\frac{3}{2}-\frac{\frac{1}{3}}{\frac{4}{3}-\frac{\frac{1}{4}}{\frac{5}{4}-\frac{\frac{1}{5}}{\frac{6}{5}-...}}}}}}$$

Commented by Dwaipayan Shikari last updated on 31/Dec/20

$$a_0+a_0{a}_1+a_0{a}_2{a}_1+a_0{a}_1{a}_2{a}_3+.....=\frac{a_0}{1-\frac{a_1}{1+a_1-\frac{a_2}{1+a_2-\frac{a_3}{1+a_3-\frac{a_4}{1+a_4..}}}}}$$$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+...$$$$=1+1.x+x\left(\frac{x}{2}\right)+x\left(\frac{x}{2}\right)\left(\frac{x}{3}\right)+x.\frac{x}{2}.\frac{x}{3}.\frac{x}{4}+..$$$$=\frac{1}{1-\frac{x}{1+x-\frac{\frac{x}{2}}{1+\frac{x}{2}-\frac{\frac{x}{3}}{1+\frac{x}{3}-\frac{\frac{x}{4}}{1+\frac{x}{4}-..}}}}}$$$$e=\frac{1}{1-\frac{1}{2-\frac{\frac{1}{2}}{\frac{3}{2}-\frac{\frac{1}{3}}{\frac{4}{3}-\frac{\frac{1}{4}}{\frac{5}{4}..}}}}}$$

Commented by MathSh last updated on 31/Dec/20

$$e$$

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