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Question Number 8958 by Sopheak last updated on 07/Nov/16
Prove that 1+(1/2)+(1/3)+....+(1/(2009))=2009−((1/2)+(2/3)+(3/4)+...+((2008)/(2009)))
$${Prove}\:{that}\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+….+\frac{\mathrm{1}}{\mathrm{2009}}=\mathrm{2009}−\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{4}}+…+\frac{\mathrm{2008}}{\mathrm{2009}}\right) \\ $$
Answered by sou1618 last updated on 07/Nov/16
1+(1/2)+(1/3)+(1/4)...+(1/(2009))  =(1−(0/1))+(1−(1/2))+(1−(2/3))+(1−(3/4))...+(1−((2008)/(2009)))  =2009−((0/1)+(1/2)+(2/3)+(3/4)...+((2008)/(2009)))  =2009−((1/2)+(2/3)+(3/4)...+((2008)/(2009)))
$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}…+\frac{\mathrm{1}}{\mathrm{2009}} \\ $$$$=\left(\mathrm{1}−\frac{\mathrm{0}}{\mathrm{1}}\right)+\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}\right)+\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\right)…+\left(\mathrm{1}−\frac{\mathrm{2008}}{\mathrm{2009}}\right) \\ $$$$=\mathrm{2009}−\left(\frac{\mathrm{0}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{4}}…+\frac{\mathrm{2008}}{\mathrm{2009}}\right) \\ $$$$=\mathrm{2009}−\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{4}}…+\frac{\mathrm{2008}}{\mathrm{2009}}\right) \\ $$$$ \\ $$

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