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Prove the following inequality

$$\zeta (4)\le 1.1$$

I saw on the site some proofs for $\zeta(4)$ that use Fourier or Euler's way for computing its precise value, and that's fine and I can use it. Still, I wonder if there is a simpler way around for proving
this inequality. Thanks!

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3 Answers 3

up vote 8 down vote accepted

$$\zeta(4) < \sum_{n=1}^{6} \frac{1}{n^{4}} + \int_{6}^{\infty} \frac{dx}{x^4} < 1.1$$

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nice! Thanks! (+1) –  Chris's sis Sep 23 '12 at 14:31
1  
(a) I suspect it should be $\int_{6}^{\infty}$ rather than $\int_{7}^{\infty}$ to justify the first $\lt$ –  Henry Sep 23 '12 at 15:04
2  
(b) $$\zeta(4) < \sum_{n=1}^{3} \frac{1}{n^{4}} + \int_{3}^{\infty} \frac{dx}{x^4} < 1.0872$$ seems to work –  Henry Sep 23 '12 at 15:05
    
@ Henry: I just checked that and you're perfectly right! (+1) Thank you. The second version looks better. –  Chris's sis Sep 23 '12 at 15:18

$$ \begin{eqnarray} \zeta(4)&<&\sum_{n=1}^{9}\frac{1}{n^4} +\left(\sum_{n=10}^{\infty}\frac{1}{n^2}\right)^2 \\ &=& \sum_{n=1}^{9}\frac{1}{n^4} + \left(\frac{\pi^2}{6}-\sum_{n=1}^{9}\frac{1}{n^2}\right)^2 \\ &=& 1.0929965... \end{eqnarray} $$

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it seems a bit harder :-) (+1) –  Chris's sis Sep 23 '12 at 15:13

One has $${1\over (n-{1\over2})^3}-{1\over (n+{1\over2})^3}={3n^2+{1\over4}\over (n^2-{1\over4})^3}>{3\over n^4}\ .$$ Therefore one obtains the telescopic sum $$\eqalign{\sum_{n=1}^\infty{1\over n^4}&=1+\sum_{n=2}^\infty{1\over n^4}< 1+{1\over3}\sum_{n=2}^\infty\Bigl({1\over (n-{1\over2})^3}-{1\over (n+{1\over2})^3}\Bigr)\cr &=1+{1\over3}{1\over(2-{1\over2})^3}=1+{8\over81}<1.1\ .\cr}$$

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thank you for your solution. –  Chris's sis Sep 23 '12 at 18:06

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