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

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 :)

• Maybe you will like this: $$\sum_{k=1}^{\infty}\frac{k}{2^k}=2$$ Nov 1, 2014 at 0:32
• @TylerHG this is nice! Nov 1, 2014 at 0:40
• @TylerHG: could you expand it to an answer? The usual (but not unique) way to compare two series $\sum_{n \geqslant 1} a_n$ and $\sum_{n \geqslant 1} b_n$ is to show that $a_n<b_n$ for all ${n \geqslant 1}$. But here, $\frac{n}{2^n}<\frac{1}{n^2}$ for $n$ (not so) large enough. So I am curious about the details. Thanks. May 30, 2015 at 9:39

Hint:

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

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

• Moreover, this can prove that $\frac{1}{i}+\sum_{n=1}^{i}\frac{1}{n^2}$ is an upper bound to the sequence for any $i$ (which yields arbitrarily tight bounds) Nov 1, 2014 at 0:15
• Shouldn't the rigth side summation to infinity diverge? Nov 1, 2014 at 0:15
• No: we get a telescoping series Nov 1, 2014 at 0:16
• Can you think of a same argument to show that $$\left(1+\frac{1}{n}\right)^n < 3$$ when $n\to\infty$? Nov 1, 2014 at 0:35
• If $x>0$ then $e^x\le(1+\frac xn)^{n+1}$; take $x=2$ and $n=1$ to get $e^2\le 9$; so $(1+\frac1n)^n < e \le 3$. But this should really be a separate question.
– user21467
Nov 1, 2014 at 1:46

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 $\sum_{n=1}^\infty \frac{1}{n^2} \leq 2$.

• Good approach! :) Nov 1, 2014 at 0:35
• Maybe I am stupid but do you get the strict inequality as well? Apr 20, 2015 at 14:44
• @Peter When $n=1$, then you get the strict inequality :). Apr 22, 2015 at 1:47
• It seems that I am really stupid but when $n=1$ the equation tells me that $1\leq 1$. Apr 22, 2015 at 11:58
• @Peter that is exactly right - $\leq$ means less than OR equal to. Apr 22, 2015 at 18:52

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

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