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Prove that $\dfrac{1}{2} < \dfrac{1}{101} + \dfrac{1}{102} + \dots + \dfrac {1}{200} < 1$

I started a math club at my school in the hopes of promoting math interest, and I want to interest my members with challenging problems, which aren't too challenging.

This is an example of an inequality problem that would be good for the club. I'd like a source of more inequalities like this which depend on clever groupings and observations in order to solve them.

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    $\begingroup$ Starting a math club...+1 $\endgroup$ – imranfat Aug 25 '16 at 14:24
  • $\begingroup$ If your club is interested in sums like that, you might want to try the problem of proving that the harmonic series diverges. Also, inequalities involving trigonometric functions can be hard to prove and often have to be proven geometrically. $\endgroup$ – Franklin Pezzuti Dyer Apr 25 '17 at 21:59
  • $\begingroup$ Also, you may want to look at power towers with your math club, like towers of the form $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}$$ $\endgroup$ – Franklin Pezzuti Dyer Apr 25 '17 at 22:03
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The right inequality.

Since $f(x)=\frac{1}{x}$ is a convex function, we have: $$\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}<50\left(\frac{1}{101}+\frac{1}{200}\right)<1$$

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