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


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


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


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

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  • $\begingroup$ I haven't checked the arithmetic at the end, but otherwise, it looks right. This is calculus, though, not discrete math. $\endgroup$ – saulspatz Nov 11 '18 at 14:20
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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:$${1\over n}- {1\over n+1}={1\over n\cdot(n+1)} \leq {1\over n^2} $$

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