13
$\begingroup$

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 :)

$\endgroup$
3
  • 5
    $\begingroup$ Maybe you will like this: $$\sum_{k=1}^{\infty}\frac{k}{2^k}=2$$ $\endgroup$ – ClassicStyle Nov 1 '14 at 0:32
  • 1
    $\begingroup$ @TylerHG this is nice! $\endgroup$ – Guerlando OCs Nov 1 '14 at 0:40
  • $\begingroup$ @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. $\endgroup$ – Taladris May 30 '15 at 9:39
11
$\begingroup$

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

$\endgroup$
7
  • 1
    $\begingroup$ 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) $\endgroup$ – Milo Brandt Nov 1 '14 at 0:15
  • $\begingroup$ Shouldn't the rigth side summation to infinity diverge? $\endgroup$ – Guerlando OCs Nov 1 '14 at 0:15
  • $\begingroup$ No: we get a telescoping series $\endgroup$ – Ben Grossmann Nov 1 '14 at 0:16
  • 1
    $\begingroup$ Can you think of a same argument to show that $$\left(1+\frac{1}{n}\right)^n < 3$$ when $n\to\infty$? $\endgroup$ – Guerlando OCs Nov 1 '14 at 0:35
  • 3
    $\begingroup$ 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. $\endgroup$ – user21467 Nov 1 '14 at 1:46
17
$\begingroup$

Hint:

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

$\endgroup$
4
$\begingroup$

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

$\endgroup$
10
  • $\begingroup$ Good approach! :) $\endgroup$ – Guerlando OCs Nov 1 '14 at 0:35
  • $\begingroup$ Maybe I am stupid but do you get the strict inequality as well? $\endgroup$ – Peter Apr 20 '15 at 14:44
  • $\begingroup$ @Peter When $n=1$, then you get the strict inequality :). $\endgroup$ – Sherlock Holmes Apr 22 '15 at 1:47
  • $\begingroup$ It seems that I am really stupid but when $n=1$ the equation tells me that $1\leq 1$. $\endgroup$ – Peter Apr 22 '15 at 11:58
  • $\begingroup$ @Peter that is exactly right - $\leq$ means less than OR equal to. $\endgroup$ – Sherlock Holmes Apr 22 '15 at 18:52
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.