$\sum_{n=q_1}^{q_2-1}\frac{1}{n^2}<\frac{1}{q_1-1}$ inequality

Question

How can We prove that $$\sum_{n=q_1}^{q_2-1}\frac{1}{n^2}<\frac{1}{q_1-1}$$ for integer $q_1,q_2$ : $1\le q_1 \le q_2 $. Obvious esimation gives $$\sum_{n=q_1}^{q_2-1}\frac{1}{n^2}<\frac{q_2-q_1}{q_1^2}$$, and that doesn`t show anything. Thanks for your help.

Answer 1

For $1<q_1<q_2$ we obtain: $$\sum_{n=q_1}^{g_2-1}\frac{1}{n^2}<\sum_{n=q_1}^{g_2-1}\frac{1}{n(n-1)}=\sum_{n=q_1}^{g_2-1}\left(\frac{1}{n-1}-\frac{1}{n}\right)=\frac{1}{q_1-1}-\frac{1}{q_2-1}<\frac{1}{q_1-1}.$$

Answer 2

$\frac1{n^2}<\frac1{(n-1)n}=\frac1{n-1}-\frac1{n}\Rightarrow \sum\limits_{n=q_1}^{q_2-1}\left(\frac1{n-1}-\frac1{n}\right)=\frac1{q_1-1}-\frac1{q_2-1}<\frac1{q_1-1}$