Proof that $\sum_{1}^{\infty} \frac{1}{n^2} <2$

Question

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 :)

Answer 1

Hint: for $n \geq 2$, $$ \frac 1{n^2} \leq \frac{1}{n(n-1)} = \frac1{n-1} - \frac 1n $$

Answer 2

Hint:

$$\sum_{n=1}^{\infty} \frac{1}{n^2} < 1+ \int_{1}^{\infty} \frac{1}{x^2}dx $$

Answer 3

You can use induction to prove the inequality

$1+\frac{1}{2^2}+\cdots+\frac{1}{n^2} \leq 2-\frac{1}{n}$ for $n \geq 1$, i.e. $\sum_{i=1}^{n} \frac{1}{i^2} \leq 2 - \frac{1}{n}\to 2$ as $n\to\infty$.

This short proof, however, only proves the weaker statement $\sum_{n=1}^\infty \frac{1}{n^2} \leq 2$.

Answer 4

$$\zeta(2)=\frac{5}{4}+\sum_{n=3}^{+\infty}\frac{1}{n^2}\leq\frac{5}{4}+4\sum_{n=3}^{+\infty}\frac{1}{(2n-1)(2n+1)}=\frac{5}{4}+\frac{2}{5}=\frac{33}{20}.$$

Answer 5

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