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.

  • 5
    $\begingroup$ Starting a math club...+1 $\endgroup$
    – imranfat
    Commented Aug 25, 2016 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. $\endgroup$ Commented Apr 25, 2017 at 21:59
  • $\begingroup$ Also, you may want to look at power towers with your math club, like towers of the form $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}$$ $\endgroup$ Commented Apr 25, 2017 at 22:03

2 Answers 2


Let the middle expression $$\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}$$ be $S_0$. We can then compare the middle expression with two different sums, with 100 terms in each: $$S_1=\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=100\cdot \frac{1}{100}=1$$ and $$S_2=\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}=100\cdot \frac{1}{200}=\frac{1}{2}.$$ Since the denominators of the terms in $S_0$ is increasing, the values are getting smaller and smaller (Example: $\frac{1}{150}<\frac{1}{100}$), so $S_0<S_1\rightarrow S_0<1$.

Since all but one of the terms in $S_0$ is greater than $\frac{1}{200}$, (Example: $\frac{1}{150}>\frac{1}{200}$), $S_0>S_2\rightarrow S_0>\frac{1}{2}$.


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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