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Question Number 128181 by Algoritm last updated on 05/Jan/21

Commented by Algoritm last updated on 05/Jan/21

Prove that

$$\mathrm{Prove}\:\mathrm{that} \\ $$

Answered by Dwaipayan Shikari last updated on 05/Jan/21

1+(1/2)+(1/3)+(1/4)+...+(1/n)  =1+1((1/2))+1((1/2))((2/3))+1((1/2))((2/3))((3/4))+...+1((1/2))...((((n−1))/(2n−1)))  =(1/(1−((1/2)/(1+(1/2)−((2/3)/(1+(2/3)−((3/4)/(1+(3/4)−...(((n−1))/(2n−1))))))))))=(1/(1−(1^2 /(3−(2^2 /(5−(3^2 /(7−(4^2 /(...−(((n−1)^2 )/(2n−1))))))))))))

$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+...+\frac{\mathrm{1}}{{n}} \\ $$$$=\mathrm{1}+\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{2}}{\mathrm{3}}\right)+\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\left(\frac{\mathrm{3}}{\mathrm{4}}\right)+...+\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)...\left(\frac{\left({n}−\mathrm{1}\right)}{\mathrm{2}{n}−\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}−\frac{\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}}−...\frac{\left({n}−\mathrm{1}\right)}{\mathrm{2}{n}−\mathrm{1}}}}}}=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{3}−\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{5}−\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{7}−\frac{\mathrm{4}^{\mathrm{2}} }{...−\frac{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}{n}−\mathrm{1}}}}}}} \\ $$

Commented by Dwaipayan Shikari last updated on 05/Jan/21

see Euler continued fractions https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula

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