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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 :)

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    $\begingroup$ Maybe you will like this: $$\sum_{k=1}^{\infty}\frac{k}{2^k}=2$$ $\endgroup$ – ClassicStyle Nov 1 '14 at 0:32
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    $\begingroup$ @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
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Hint: for $n \geq 2$, $$ \frac 1{n^2} \leq \frac{1}{n(n-1)} = \frac1{n-1} - \frac 1n $$

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    $\begingroup$ 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 $\endgroup$ – Omnomnomnom Nov 1 '14 at 0:16
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    $\begingroup$ 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
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    $\begingroup$ 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
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Hint:

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

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

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

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

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