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$1<\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{3001}<1\frac{1}{3}$
The first part is trivial with $AM-HM $ inequality. Having problem with the second part.

Soham
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  • Tatan - see edit to your photo post - here - if acceptable you can resubmit or ask for it to be undeleted. Max size was exceeded - RM – Russell McMahon Nov 17 '15 at 18:14
  • @RussellMcMahon-How can I ask for it to be undeleted?The post was deleted by a moderator and I cannot vote to open it or comment it to address John Cavan,requesting him to undelete it......By the way please consider commenting in Photography Stack Exchange. – Soham Nov 18 '15 at 14:51
  • I've asked them if it can be undeleted. The "flag"button under the photo is still active - for me anyway. – Russell McMahon Nov 19 '15 at 07:02
  • @RussellMcMahon-It has been undeleted.Thanks for your help:-)!!! – Soham Nov 19 '15 at 15:53

3 Answers3

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For $0 \leqslant k \leqslant 1000$

$$\frac{1}{2001-k} + \frac{1}{2001+k} = \frac{4002}{2001^2 - k^2} \leqslant \frac{4002}{2001^2 - 1000^2} = \frac{4002}{1001 \cdot 3001} < \frac{1}{1000 \frac{1}{2}} \cdot \frac{4}{3}.$$

Adding up the inequalities for $k = 1, 2, \ldots, 1000$ and the inequality divided by $2$ for $k = 0$, we obtain

$$\frac{1}{1001} + \frac{1}{1002} + \ldots + \frac{1}{3001} < \frac{1000 \frac{1}{2}}{1000 \frac{1}{2}} \cdot \frac{4}{3} = \frac{4}{3}$$

Adayah
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3

$\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{3001}<\int_{1000}^{3001}\frac{1}{x}=\ln(\frac{3001}{1000})\approx 1.1$

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    Maybe use a short Taylor series with remainder to replace that last $\approx$ with a rigorous upper bound. –  Aug 14 '15 at 12:01
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One can add up. To do it in one's head, forget temporarily about the last term. Then note that the first quarter of the rest have sum less than $1/2$, the next quarter have sum less than $1/3$, and so on, for a total less than $77/60$. Adding the last term does not change things appreciably.

André Nicolas
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