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The Basel Problem.md

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