public-notes/The Basel Problem.md

#Math #NT

Basel Problem Solution

Base Sum

\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))

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

= \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} \\

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

\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} \\

Therefore

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