# Why $\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = \frac{\pi^2}{6}$

In the previous note that I've written I said $\sin x = \frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$ This can come in handy today! $\newline$ We know the function image of $\sin x$ is: First we can say $\sin x \approx (\pi + x)x(\pi -x)$ because when $x = -\pi$ or $0$ or $\pi$ the $(\pi +x)x(\pi -x) = 0$ and $\sin x = 0$. But let we see the function image (the blue line): This fits too badly! $(\sin x)' = \cos x$ and when $x = 0$ then $(\sin x)' = 1$. Can we let $[C_1 \times (\pi +x)x(\pi -x)]' = 1$ when $x = 0$? $\newline$ Let $[C_1 \times (\pi +x)x(\pi -x)]' = 1$ which $C_1$ is a constant. So $C_1 \times [(\pi +x)x(\pi -x)]' = 1$ when $x = 0$. After expansion, derivation, only "$\pi ^2$" is left, and the others are all 0 because they contain x (ie 0). So $C_1 \times \pi ^2 = 1$ $C_1 = \frac{1}{\pi ^2}$ $\frac{1}{\pi ^2} (\pi + x)x(\pi -x) = (1+\frac{x}{\pi})x(1-\frac{x}{\pi})$ Then it will become like this: It's great! $\newline$The same, $\sin x \approx (2\pi +x)(1+\frac{x}{\pi})x(1-\frac{x}{\pi})(2\pi -x)$ $C_2 \times [(2\pi +x)(1+\frac{x}{\pi})x(1-\frac{x}{\pi})(2\pi -x)]' = 1$ only "$(2\pi)^2$" is left, and the others are all 0 because they contain x (ie 0). So $C_2 = \frac{1}{(2\pi)^2}$ $\frac{1}{(2\pi)^2}(2\pi +x)(1+\frac{x}{\pi})x(1-\frac{x}{\pi})(2\pi -x) = (1 +\frac{x}{2\pi})(1+\frac{x}{\pi})x(1-\frac{x}{\pi})(1 -\frac{x}{(2\pi)^2})$ And so on: $C_3 = \frac{1}{(3\pi)^2}, C_4=\frac{1}{(4\pi)^2}, C_n=\frac{1}{(n\pi)^2 }\cdots$ Therefore $\sin x = \cdots (1 +\frac{x}{2\pi})(1+\frac{x}{\pi})x(1-\frac{x}{\pi})(1 -\frac{x}{(2\pi)^2}) \cdots$ We also know $(a+b)(a-b)=a^2-b^2$. Then $\sin x = x(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{(2\pi)^2})(1-\frac{x^2}{(3\pi)^2}) \cdots$ and $\sin x$ also is $\frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$ So $\frac{x}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = x(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{(2\pi)^2})(1-\frac{x^2}{(3\pi)^2}) \cdots$ Therefore, the split items with $x^3$ items on the left and right are equal. In the left of the formula, it has $-x^3 \frac{1}{3!}$ and in the right of the formula -- how many $x^3$ does it have? We must have $x^3$, remove the first $x$, it takes a $\frac{x^2}{\cdots}$ to form $x^3$. So we have: $x^3 \times (-\frac{1}{3!})=x(-\frac{x^2}{\pi ^2} - \frac{x^2}{(2\pi)^2} - \frac{x^2}{(3\pi)^2}-\cdots)$ $\frac{1}{3!} = \frac{1}{\pi^2} + \frac{1}{(2\pi)^2} + \frac{1}{(3\pi)^2} + \cdots$ $\frac{1}{6} = \frac{1}{\pi^2}(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{(3\pi)^2} + \cdots)$ $\frac{\pi^2}{6} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots$ Proven! Note by Raymond Fang
4 months, 2 weeks ago

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Great! in this way you can also find the sum like $\zeta(4)=\sum_{n=1}^∞ \dfrac{1}{n^4}= \dfrac{π^4}{90}$ $\zeta(6)=\sum_{n=1}^∞ \dfrac{1}{n^6}= \dfrac{π^6}{945}$ $\zeta(8)=\sum_{n=1}^∞ \dfrac{1}{n^8}= \dfrac{π^8}{9450}$ $\zeta(10)=\sum_{n=1}^∞ \dfrac{1}{n^{10}}= \dfrac{π^{10}}{93555}$ For General $\zeta(2n)= \dfrac{(-1)^{n+1} \beta_{2n} (2π)^{2n}}{2(2n)!}$ Where $\beta_n$= $n^{th}$ Bernoulli Number

- 4 months, 2 weeks ago

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Great!

- 4 months, 2 weeks ago

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