<s>[INST] Prove $\frac{b-a+1}{ab} \leq \sum_{i=a}^b \frac{1}{i^2} \leq \frac{b-a+1}{(a-1)b}$ 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!
 [/INST] 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:
$$
\begin{split}
S(b+1) & = S(b) + \frac{1}{(b+1)^2} -\frac{b-a+2}{a(b+1)} + \frac{b-a+1}{ab}
\\ & \geq \frac{1}{(b+1)^2} -\frac{b-a+2}{a(b+1)} + \frac{b-a+1}{ab}
\\ & =
\frac{b-a+1}{ab(b+1)^2} > 0.
\end{split}
$$
</s>