Show that $1.222 \le 1 + 3^{-2} + 5^{-2} + ... \le 1.252$ Show that $1.222 \le 1 + 3^{-2} + 5^{-2} + ... \le 1.252$ without calculate the value of $1 + 3^{-2} + 5^{-2} + ...$
 Can I do in this way?
 $$\lim\limits_{n \to \infty} \int_2^n (2x+3)^{-2} \,dx \le 7^{-2} + 9^{-2} + 11^{-2} + ... \le \lim\limits_{n \to \infty} \int_2^n (2x+1)^{-2} \,dx$$
 $$\lim\limits_{n \to \infty} \int_2^n (2x+3)^{-2} \,dx = \lim\limits_{n \to \infty}\frac{-1}2(2n+3)^{-1}+\frac12(7)^{-1} = \frac1{14}$$
$$\lim\limits_{n \to \infty} \int_2^n (2x+1)^{-2} \,dx = \lim\limits_{n \to \infty}\frac{-1}2(2n+1)^{-1}+\frac12(5)^{-1} = \frac1{10}$$
$$\therefore \frac1{14} \le 7^{-2} + 9^{-2} + 11^{-2} + ... \le \frac1{10}$$
$$\therefore 1.22253... \le 1 + 3^{-2} + 5^{-2} + ... \le 1.25111... $$
$$\therefore 1.222 \le 1 + 3^{-2} + 5^{-2} + ... \le 1.252 $$

Idea: Perhaps this will help:
$$ {1\over n^2}\leq {1\over n\cdot(n-1)} =  {1\over n-1}- {1\over n}$$
and vice versa:$${1\over n}- {1\over n+1}={1\over n\cdot(n+1)}  \leq {1\over n^2} $$
