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 :)

Maybe you will like this: $$\sum_{k=1}^{\infty}\frac{k}{2^k}=2$$ — ClassicStyle, Nov 01 '14 at 00:32
@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. — Taladris, May 30 '15 at 09:39

score 18 · Answer 1 · answered Nov 01 '14 at 00:12

18

Hint:

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

answered Nov 01 '14 at 00:12

Aaron Maroja

17,869

score 12 · Accepted Answer · answered Nov 01 '14 at 00:12

12

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

answered Nov 01 '14 at 00:12

Ben Grossmann

234,171
12
184
355

1

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) – Milo Brandt Nov 01 '14 at 00:15
Shouldn't the rigth side summation to infinity diverge? – Poperton Nov 01 '14 at 00:15
No: we get a telescoping series – Ben Grossmann Nov 01 '14 at 00:16
1

Can you think of a same argument to show that $$\left(1+\frac{1}{n}\right)^n < 3$$ when $n\to\infty$? – Poperton Nov 01 '14 at 00:35
@GuerlandoOCs OCs What about $2 \leq e = 1 + 1 + \frac{1}{2!} + \ldots + \frac{1}{n!} \leq 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \ldots + \frac{1}{2^{n-1}} < 3$. – Aaron Maroja Nov 01 '14 at 01:25
3

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. – Nov 01 '14 at 01:46
It should definitely be a seperate question. – Aaron Maroja Nov 01 '14 at 01:59

score 4 · Answer 3 · edited May 30 '15 at 08:31

4

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

edited May 30 '15 at 08:31

Peter

3,441

answered Nov 01 '14 at 00:13

Sherlock Holmes

1,928

Good approach! :) – Poperton Nov 01 '14 at 00:35
Maybe I am stupid but do you get the strict inequality as well? – Peter Apr 20 '15 at 14:44
@Peter When $n=1$, then you get the strict inequality :). – Sherlock Holmes Apr 22 '15 at 01:47
It seems that I am really stupid but when $n=1$ the equation tells me that $1\leq 1$. – Peter Apr 22 '15 at 11:58
@Peter that is exactly right - $\leq$ means less than OR equal to. – Sherlock Holmes Apr 22 '15 at 18:52
Put it differently: You have your statement, how to you conclude that $\sum_{n=1}^\infty \frac{1}{n^2} < 2$ ? – Peter Apr 22 '15 at 20:35
Still not answered.... – Peter Apr 30 '15 at 23:03
@Peter $2 - \frac{1}{n}<2$ – Sherlock Holmes May 30 '15 at 00:43
That does not answer my question. Note that $\frac{1}{n} > 0$ for all $n\in \mathbb{N}$ but obviously $\lim_{n\to\infty} \frac{1}{n} = 0 $. – Peter May 30 '15 at 08:07
@SherlockHolmes I fixed the problem and edited your post. Now it's fine but this proves the weaker statement I meant. – Peter May 30 '15 at 10:21

score 3 · Answer 4 · answered Apr 20 '15 at 14:40

3

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

answered Apr 20 '15 at 14:40

Samrat Mukhopadhyay

16,568

score 2 · Answer 5 · answered Nov 01 '14 at 01:56

2

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

answered Nov 01 '14 at 01:56

Jack D'Aurizio

361,689

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

5 Answers5

Linked

Related