Prove $\frac{b-a+1}{ab} \leq \sum_{i=a}^b \frac{1}{i^2} \leq \frac{b-a+1}{(a-1)b}$

Question

Given $b>a>1; \ a,b \in Z^+$. Prove $\frac{b-a+1}{ab} \leq \sum_{i=a}^b \frac{1}{i^2} \leq \frac{b-a+1}{(a-1)b}. \ $ I mostly care about the second inequality. Thank you!

Answer 1

For the second inequality just observe that $$ \frac{1}{i^2} \leq \frac{1}{i(i-1)} = \frac{1}{i-1} - \frac{1}{i}, $$ hence $$ \sum_{i=a}^b \frac{1}{i^2} \leq \sum_{i=a}^b\left(\frac{1}{i-1} - \frac{1}{i}\right) = \frac{1}{a-1} - \frac{1}{b} = \frac{b-a+1}{(a-1)b}\,. $$

The first inequality can be proved by induction on $b$.

Let $a\in\mathbb{Z}^+$ be fixed, and let us prove that $$ S(b) := \sum_{i=a}^{b} \frac{1}{i^2} - \frac{b-a+1}{ab}\geq 0 \qquad \forall b\geq a. $$ If $b = a$ then $S(b) = 0$. Assume now that the inequality holds for some $b\geq a$, and let us prove that it holds for $b+1$.

Using the induction assumption $S(b) \geq 0$, we have that: $$ \begin{split} S(b+1) & = S(b) + \frac{1}{(b+1)^2} -\frac{b-a+2}{a(b+1)} + \frac{b-a+1}{ab} \\ & \geq \frac{1}{(b+1)^2} -\frac{b-a+2}{a(b+1)} + \frac{b-a+1}{ab} \\ & = \frac{b-a+1}{ab(b+1)^2} > 0. \end{split} $$