The Basel Problem

Math NT

Basel Problem Solution

Base Sum

$$\begin{gathered} \frac{{\pi }^2}{4} \\ =\frac{{\pi }^2}{4}{\csc }^2(\frac{\pi }{2}) \\ =\frac{{\pi }^2}{4^2}({\csc }^2(\frac{\pi }{4})+{\csc }^2(\frac{\pi }{4}+\pi )) \end{gathered}$$

Do this operation $a$ times, with the above equation being the second time:

$$\begin{gathered} =\frac{{\pi }^2}{4^{a+1}}\sum _{n=1}^{2^a}{\csc }^2(\frac{\pi }{2^{a+1}}+\frac{\pi }{2^a}) \\ =\sum _{n=1}^{2^a}\frac{{\pi }^2}{4^{a+1}}{\csc }^2(\frac{\pi }{2^{a+1}}+\frac{\pi }{2^a}) \\ =\sum _{n=1}^{2^a}\frac{{\pi }^2}{4^{a+1}}{\csc }^2(\frac{\pi }{2^{a+1}}+\frac{\pi }{2^a}) \\ =\sum _{n=1}^{2^a}(\frac{2^{a+1}}{\pi }\sin (\frac{\pi }{2^{a+1}}+\frac{\pi }{2^a}){)}^{-2} \end{gathered}$$

As $a$ approaches $\infty$:

$$=2\sum _{n=1}^{\infty }(2n-1{)}^{-2}$$

Therefore:

$$\sum _{n=1}^{\infty }(2n-1{)}^{-2}=\frac{{\pi }^2}{8}$$

Manipulating this Sum

$$\begin{gathered} \sum _{n=1}^{\infty }(2n{)}^{-2}=\frac{1}{4}\sum _{n=1}^{\infty }{n}^{-2} \\ \sum _{n=1}^{\infty }(2n-1{)}^{-2}=\frac{3}{4}\sum _{n=1}^{\infty }{n}^{-2} \\ \frac{{\pi }^2}{8}=\frac{3}{4}\sum _{n=1}^{\infty }{n}^{-2} \\ \frac{{\pi }^2}{6}=\sum _{n=1}^{\infty }{n}^{-2} \end{gathered}$$

Therefore

$$\frac{{\pi }^2}{6}=\sum _{n=1}^{\infty }{n}^{-2}$$