Reference Request for inequalities without variables

Question

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.

Answer 1

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: $\frac{1}{150}>\frac{1}{200}$), $S_0>S_2\rightarrow S_0>\frac{1}{2}$.

Answer 2

The right inequality.

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