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

Commented by Algoritm last updated on 05/Jan/21

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

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

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

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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