TITLE: Proof that $\sum_{1}^{\infty} \frac{1}{n^2} <2$
QUESTION [13 upvotes]: I know how to prove that
$$\sum_1^{\infty} \frac{1}{n^2}<2$$ because
$$\sum_1^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}<2$$
But I wanted to prove it using only inequalities. Is there a way to do it? Can you think of an inequality such that you can calculate the limit of both sides, and the limit of the rigth side is $2$?
Is there a good book about inequalities that helps to prove that a sum is less than a given quantity?
This is not a homework problem, its a self posed problem that I was thinking about :)

REPLY [3 votes]: Still another proof: $$\sum_{n\ge 1}\frac{1}{n^2}\le \sum_{k\ge 0}\frac{2^{(k+1)}-2^k}{2^{2k}}=\sum_{k\ge 0}\frac{1}{2^k}=2$$