Why 112+122+132+=π26\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = \frac{\pi^2}{6}

In the previous note that I've written I said sinx=x1!x33!+x55!x77!+\sin x = \frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots This can come in handy today! \newline We know the function image of sinx\sin x is: First we can say sinx(π+x)x(πx)\sin x \approx (\pi + x)x(\pi -x) because when x=πx = -\pi or 00 or π\pi the (π+x)x(πx)=0(\pi +x)x(\pi -x) = 0 and sinx=0\sin x = 0. But let we see the function image (the blue line): This fits too badly! (sinx)=cosx(\sin x)' = \cos x and when x=0x = 0 then (sinx)=1(\sin x)' = 1. Can we let [C1×(π+x)x(πx)]=1[C_1 \times (\pi +x)x(\pi -x)]' = 1 when x=0x = 0? \newline Let [C1×(π+x)x(πx)]=1[C_1 \times (\pi +x)x(\pi -x)]' = 1 which C1C_1 is a constant. So C1×[(π+x)x(πx)]=1C_1 \times [(\pi +x)x(\pi -x)]' = 1 when x=0x = 0. After expansion, derivation, only "π2\pi ^2" is left, and the others are all 0 because they contain x (ie 0). So C1×π2=1C_1 \times \pi ^2 = 1 C1=1π2C_1 = \frac{1}{\pi ^2} 1π2(π+x)x(πx)=(1+xπ)x(1xπ) \frac{1}{\pi ^2} (\pi + x)x(\pi -x) = (1+\frac{x}{\pi})x(1-\frac{x}{\pi}) Then it will become like this: It's great! \newlineThe same, sinx(2π+x)(1+xπ)x(1xπ)(2πx)\sin x \approx (2\pi +x)(1+\frac{x}{\pi})x(1-\frac{x}{\pi})(2\pi -x) C2×[(2π+x)(1+xπ)x(1xπ)(2πx)]=1C_2 \times [(2\pi +x)(1+\frac{x}{\pi})x(1-\frac{x}{\pi})(2\pi -x)]' = 1 only "(2π)2(2\pi)^2" is left, and the others are all 0 because they contain x (ie 0). So C2=1(2π)2C_2 = \frac{1}{(2\pi)^2} 1(2π)2(2π+x)(1+xπ)x(1xπ)(2πx)=(1+x2π)(1+xπ)x(1xπ)(1x(2π)2)\frac{1}{(2\pi)^2}(2\pi +x)(1+\frac{x}{\pi})x(1-\frac{x}{\pi})(2\pi -x) = (1 +\frac{x}{2\pi})(1+\frac{x}{\pi})x(1-\frac{x}{\pi})(1 -\frac{x}{(2\pi)^2}) And so on: C3=1(3π)2,C4=1(4π)2,Cn=1(nπ)2C_3 = \frac{1}{(3\pi)^2}, C_4=\frac{1}{(4\pi)^2}, C_n=\frac{1}{(n\pi)^2 }\cdots Therefore sinx=(1+x2π)(1+xπ)x(1xπ)(1x(2π)2)\sin x = \cdots (1 +\frac{x}{2\pi})(1+\frac{x}{\pi})x(1-\frac{x}{\pi})(1 -\frac{x}{(2\pi)^2}) \cdots We also know (a+b)(ab)=a2b2(a+b)(a-b)=a^2-b^2. Then sinx=x(1x2π2)(1x2(2π)2)(1x2(3π)2)\sin x = x(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{(2\pi)^2})(1-\frac{x^2}{(3\pi)^2}) \cdots and sinx\sin x also is x1!x33!+x55!x77!+\frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots So x1!x33!+x55!x77!+=x(1x2π2)(1x2(2π)2)(1x2(3π)2)\frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \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 x3x^3 items on the left and right are equal. In the left of the formula, it has x313!-x^3 \frac{1}{3!} and in the right of the formula -- how many x3x^3 does it have? We must have x3x^3, remove the first xx, it takes a x2\frac{x^2}{\cdots} to form x3x^3. So we have: x3×(13!)=x(x2π2x2(2π)2x2(3π)2)x^3 \times (-\frac{1}{3!})=x(-\frac{x^2}{\pi ^2} - \frac{x^2}{(2\pi)^2} - \frac{x^2}{(3\pi)^2}-\cdots) 13!=1π2+1(2π)2+1(3π)2+\frac{1}{3!} = \frac{1}{\pi^2} + \frac{1}{(2\pi)^2} + \frac{1}{(3\pi)^2} + \cdots 16=1π2(112+122+1(3π)2+)\frac{1}{6} = \frac{1}{\pi^2}(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{(3\pi)^2} + \cdots) π26=112+122+132+\frac{\pi^2}{6} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots Proven!

Note by Raymond Fang
5 months, 3 weeks ago

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Great! in this way you can also find the sum like ζ(4)=n=11n4=π490\zeta(4)=\sum_{n=1}^∞ \dfrac{1}{n^4}= \dfrac{π^4}{90} ζ(6)=n=11n6=π6945\zeta(6)=\sum_{n=1}^∞ \dfrac{1}{n^6}= \dfrac{π^6}{945} ζ(8)=n=11n8=π89450\zeta(8)=\sum_{n=1}^∞ \dfrac{1}{n^8}= \dfrac{π^8}{9450} ζ(10)=n=11n10=π1093555\zeta(10)=\sum_{n=1}^∞ \dfrac{1}{n^{10}}= \dfrac{π^{10}}{93555} For General ζ(2n)=(1)n+1β2n(2π)2n2(2n)!\zeta(2n)= \dfrac{(-1)^{n+1} \beta_{2n} (2π)^{2n}}{2(2n)!} Where βn\beta_n = nthn^{th} Bernoulli Number

Dwaipayan Shikari - 5 months, 3 weeks ago

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Great!

Raymond Fang - 5 months, 3 weeks ago

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