What is the value of ζ(4)\zeta (4)?

We know ζ(n)=11n+12n+13n+\zeta (n) = \frac{1}{1^n}+\frac{1}{2^n}+\frac{1}{3^n} + \cdots so ζ(4)=114+124+134+\zeta (4) = \frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4} + \cdots At last note I said x1!x33!+x55!=x(1x2π2)(1x2(2π)2)(1x2(3π)2)\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 x5x^5 items on the left and right are equal. In the left of the formula, it has x515!x^5 \frac{1}{5!} and in the right of the formula -- how many x5x^5 does it have? We must have x5x^5, remove the first xx, it takes two x2\frac{x^2}{\cdots} to form x5x^5. So we have: 15!=1π2(1(2π)2+1(3π)2+)+1(2π)2(1(3π)2+1(4π)2+)+\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 π45!=112(122+132+)+122(132+142+)+\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 S1=112+122+132+=π26S_1=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots = \frac{\pi^2}{6} then π4120=112(S1112)+122(S1112122)+\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 π4120=S1(112+122+)(114+124+)112(122+132+)122(132+142+)\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 π4120=S12(114+124+)[112(S1112)+122(S1112122)+]\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] π4120=π436ζ(4)π4120\frac{\pi^4}{120}=\frac{\pi^4}{36}-\zeta(4)-\frac{\pi^4}{120} ζ(4)=π436(π4120+π4120)\zeta(4)=\frac{\pi^4}{36} - (\frac{\pi^4}{120} + \frac{\pi^4}{120}) ζ(4)=π436π460\zeta(4)=\frac{\pi^4}{36}-\frac{\pi^4}{60} ζ(4)=π490\zeta(4)=\frac{\pi^4}{90}

Note by Raymond Fang
3 weeks, 2 days ago

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

Yajat Shamji - 3 weeks, 1 day ago

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:-)

Raymond Fang - 3 weeks, 1 day ago

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Great ! For more higher terms use ζ(2n)=(1)n+1β2n(2π)2n2(2n)!\zeta(2n)= \dfrac{(-1)^{n+1} \beta_{2n}(2π)^{2n}}{2(2n)!} Or use your great way

Dwaipayan Shikari - 3 weeks ago

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:-)

Raymond Fang - 3 weeks ago

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

Raymond Fang - 3 weeks ago

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One thing for sure - it's irrational!

Yajat Shamji - 2 weeks, 6 days ago

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ζ(3)=1+123+133+...1.20205690315959428540\zeta(3) = 1 + \frac{1}{2^3} + \frac{1}{3^3} + ... \approx 1.20205690315959428540

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

Yajat Shamji - 2 weeks, 6 days ago

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@Yajat Shamji Thanks!

Raymond Fang - 2 weeks, 6 days ago

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Generally n=1f(n)=1f(x)dx+f(z)+f(1)2+k=1β2k(2k)!(limz(f2k+1(z)f2k+1(1)))\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

Dwaipayan Shikari - 2 weeks, 6 days ago

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