# What is the value of $\zeta (4)$?

We know $\zeta (n) = \frac{1}{1^n}+\frac{1}{2^n}+\frac{1}{3^n} + \cdots$ so $\zeta (4) = \frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4} + \cdots$ At last note I said $\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots=x(1-\frac{x^2}{\pi ^2})(1-\frac{x^2}{(2\pi)^2})(1-\frac{x^2}{(3\pi)^2})\cdots$ Therefore, the split items with $x^5$ items on the left and right are equal. In the left of the formula, it has $x^5 \frac{1}{5!}$ and in the right of the formula -- how many $x^5$ does it have? We must have $x^5$, remove the first $x$, it takes two $\frac{x^2}{\cdots}$ to form $x^5$. So we have: $\frac{1}{5!}=\frac{1}{\pi^2}(\frac{1}{(2\pi)^2}+\frac{1}{(3\pi)^2}+\cdots)+\frac{1}{(2\pi)^2}(\frac{1}{(3\pi)^2}+\frac{1}{(4\pi)^2}+\cdots)+\cdots$ $\frac{\pi^4}{5!}=\frac{1}{1^2}(\frac{1}{2^2}+\frac{1}{3^2}+\cdots)+\frac{1}{2^2}(\frac{1}{3^2}+\frac{1}{4^2}+\cdots)+\cdots$ Let $S_1=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots = \frac{\pi^2}{6}$ then $\frac{\pi^4}{120}=\frac{1}{1^2}(S_1-\frac{1}{1^2})+\frac{1}{2^2}(S_1-\frac{1}{1^2}-\frac{1}{2^2})+\cdots$ $\frac{\pi^4}{120}=S_1(\frac{1}{1^2}+\frac{1}{2^2}+\cdots)-(\frac{1}{1^4}+\frac{1}{2^4}+\cdots)-\frac{1}{1^2}(\frac{1}{2^2}+\frac{1}{3^2}+\cdots)-\frac{1}{2^2}(\frac{1}{3^2}+\frac{1}{4^2}+\cdots)-\cdots$ $\frac{\pi^4}{120}=S_1^2-(\frac{1}{1^4}+\frac{1}{2^4}+\cdots)-[\frac{1}{1^2}(S_1-\frac{1}{1^2})+\frac{1}{2^2}(S_1-\frac{1}{1^2}-\frac{1}{2^2})+\cdots]$ $\frac{\pi^4}{120}=\frac{\pi^4}{36}-\zeta(4)-\frac{\pi^4}{120}$ $\zeta(4)=\frac{\pi^4}{36} - (\frac{\pi^4}{120} + \frac{\pi^4}{120})$ $\zeta(4)=\frac{\pi^4}{36}-\frac{\pi^4}{60}$ $\zeta(4)=\frac{\pi^4}{90}$ Note by Raymond Fang
6 months ago

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Really, really interesting! Do more - keep inviting me as well!

- 6 months ago

:-)

- 6 months ago

Great ! For more higher terms use $\zeta(2n)= \dfrac{(-1)^{n+1} \beta_{2n}(2π)^{2n}}{2(2n)!}$ Or use your great way

- 6 months ago

:-)

- 6 months ago

What is the value of $\zeta(3)$? (This is an unsolved problem so far.)

- 6 months ago

One thing for sure - it's irrational!

- 6 months ago

$\zeta(3) = 1 + \frac{1}{2^3} + \frac{1}{3^3} + ... \approx 1.20205690315959428540$

The value of $\zeta(3)$ is called Apéry's constant

- 6 months ago

Thanks!

- 6 months ago

Generally $\sum_{n=1}^∞ f(n)= \int_1^∞ f(x)dx +\dfrac{f(z)+f(1)}{2} +\sum_{k=1}^∞ \dfrac{\beta_{2k}}{(2k)!}(\lim_{z\rightarrow{∞}} (f^{2k+1}(z) -f^{2k+1}(1)))$

Using this we can find the approximation

- 6 months ago