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Question Number 216914 by hardmath last updated on 24/Feb/25

$$\mathrm{Prove}\mathrm{that}:$$$$\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{2000}>\frac{5}{8}$$

Answered by MrGaster last updated on 24/Feb/25

$$\underset{k=1001}{\sum ^{2000}}\frac{1}{k}>\frac{5}{8}$$$${\int }_{1001}^{2001}\frac{1}{x}dx>\frac{5}{8}$$$$\ln \left(20001\right)-\ln \left(1001\right)>\frac{5}{8}$$$$\ln \left(\frac{2001}{1001}\right)>\frac{5}{8}$$$$\ln (2.000999)>\frac{5}{8}$$$$0.693147>0.626$$$$[\mathrm{Q}.\mathrm{E}.\mathrm{D}]$$

Commented by mehdee7396 last updated on 25/Feb/25

$$why\mathrm{\Sigma }\frac{1}{k}>\frac{5}{8}\Rightarrow \int \frac{1}{x}dx>\frac{5}{8}?$$$$$$

Answered by mehdee7396 last updated on 25/Feb/25

$$s>\frac{0.1999}{2000}+\frac{0.1999}{2000}+...+\frac{0.1999}{2000}$$$$=\frac{1999}{2000}>\frac{5}{8}$$

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