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Question Number 163430 by amin96 last updated on 06/Jan/22

Answered by qaz last updated on 07/Jan/22

$$\frac{\mathrm{A}}{\mathrm{B}}=\frac{\underset{\mathrm{n}=1}{\sum ^{999}}\frac{1}{(2\mathrm{n}-1)\left(2\mathrm{n}\right)}}{\underset{\mathrm{n}=1}{\sum ^{999}}\frac{1}{(999+\mathrm{n})(1999-\mathrm{n})}}$$$$=\frac{\underset{\mathrm{n}=1}{\sum ^{999}}(\frac{1}{2\mathrm{n}-1}-\frac{1}{2\mathrm{n}})={\mathrm{H}}_{1998}-\frac{1}{2}{\mathrm{H}}_{999}-\frac{1}{2}{\mathrm{H}}_{999}={\mathrm{H}}_{1998}-{\mathrm{H}}_{999}}{\frac{1}{2998}\underset{\mathrm{n}=1}{\sum ^{999}}(\frac{1}{999+\mathrm{n}}+\frac{1}{1999-\mathrm{n}})=\frac{1}{1499}\underset{\mathrm{n}=1000}{\sum ^{1998}}\frac{1}{\mathrm{n}}=\frac{1}{1499}({\mathrm{H}}_{1998}-{\mathrm{H}}_{999})}$$$$=1499$$

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