# 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?
$$\lim\limits_{n \to \infty} \int_2^n (2x+3)^{-2} \,dx \le 7^{-2} + 9^{-2} + 11^{-2} + ... \le \lim\limits_{n \to \infty} \int_2^n (2x+1)^{-2} \,dx$$

$$\lim\limits_{n \to \infty} \int_2^n (2x+3)^{-2} \,dx = \lim\limits_{n \to \infty}\frac{-1}2(2n+3)^{-1}+\frac12(7)^{-1} = \frac1{14}$$

$$\lim\limits_{n \to \infty} \int_2^n (2x+1)^{-2} \,dx = \lim\limits_{n \to \infty}\frac{-1}2(2n+1)^{-1}+\frac12(5)^{-1} = \frac1{10}$$

$$\therefore \frac1{14} \le 7^{-2} + 9^{-2} + 11^{-2} + ... \le \frac1{10}$$
$$\therefore 1.22253... \le 1 + 3^{-2} + 5^{-2} + ... \le 1.25111...$$
$$\therefore 1.222 \le 1 + 3^{-2} + 5^{-2} + ... \le 1.252$$

• I haven't checked the arithmetic at the end, but otherwise, it looks right. This is calculus, though, not discrete math. – saulspatz Nov 11 '18 at 14:20

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