New Basel Problem Proof

This sum is equal to $\frac 23 4^n-\frac 23$ . We can show this by induction. The base case of n=1 is straightforward. By the double angle identity for sin(x), $\large \frac 23 4^n-\frac 23= \sum_{j=1}^{2^{n}-1} \frac{1}{\sin^2 \left(\frac{\pi j}{2^{n+1}}\right)}=\frac 14 \sum_{j=1}^{2^{n}-1} \frac{1}{\sin^2 \left(\frac{\pi j}{2^{n+2}}\right)\cos^2 \left(\frac{\pi j}{2^{n+2}}\right)}$ We can use the identity $1=\cos^2 x +\sin^2 x$ on the numerator inside the sum to get $\large \frac 14 \sum_{j=1}^{2^{n}-1} \frac{1}{\sin^2 \left(\frac{\pi j}{2^{n+2}}\right)}+\frac{1}{\cos^2 \left(\frac{\pi j}{2^{n+2}}\right)}=\frac 14 \sum_{j=1}^{2^{n}-1} \frac{1}{\sin^2 \left(\frac{\pi j}{2^{n+2}}\right)}+\frac 14 \sum_{j=1}^{2^{n}-1} \frac{1}{\sin^2 \left(\frac{2^{n+1} \pi-\pi j}{2^{n+2}}\right)}=\frac 14 \bigg( \sum_{j=1}^{2^{n+1}-1} \bigg( \frac{1}{\sin^2 \left(\frac{\pi j}{2^{n+2}}\right)}\bigg) - \frac{1}{\sin^2 \left(\frac{\pi}{4}\right)}\bigg)$ Rearranging where we started and where we ended up gives us $\large \sum_{j=1}^{2^{n+1}-1} \frac{1}{\sin^2 \left(\frac{\pi j}{2^{n+2}}\right)}=\frac 23 4^{n+1}-\frac 23$ which completes the proof.

Bonus: Because $\frac{1}{\sin^2(x)}=1+\frac{1}{\tan^2(x)}$ , we have $\large \sum_{j=1}^{2^{n}-1} \frac{1}{\tan^2 \left(\frac{\pi j}{2^{n+1}}\right)}=\frac 23 4^n-2^n+\frac 13$ Also, $\sin(x) \leq x \leq \tan(x) \implies \frac{1}{\tan^2(x)} \leq \frac{1}{x^2} \leq \frac{1}{\sin^2(x)}$ for $0 \leq x \leq \frac{\pi}{2}$ so $\large \frac 23 4^n-2^n+\frac 13 \leq \sum_{j=1}^{2^{n}-1} \frac{1}{\left(\frac{\pi j}{2^{n+1}}\right)^2} \leq \frac 23 4^n-\frac 23 \implies \frac 23 - \frac{1}{2^n} + \frac{1}{3*4^n} \leq \large \sum_{j=1}^{2^n-1} \frac{1}{\frac{\pi^2}{4} j^2} \leq \frac 23 - \frac{2}{3*4^n}$ Taking the limit as n goes to infinity gives $\large \frac 23 \leq \large \sum_{j=1}^{\infty} \frac{1}{\frac{\pi^2}{4} j^2} \leq \frac 23 \implies \sum_{j=1}^{\infty} \frac{1}{j^2}=\frac {\pi^2}{6}$

Note that $\sin 2Nx \; = \; \sum_{j=0}^{N-1}{2N \choose 2j+1}(-1)^j \cos^{2N-2j-1}x \sin^{2j+1}x \; = \; \sin^{2N-1}x \,\cos x\, F_N(\cot^2x)$ where $F_N(X) \; = \; \sum_{j=0}^{N-1}(-1)^j {2N \choose 2j+1}X^{N-j-1} \; = \; {2N \choose 1}X^{N-1} - {2N \choose 3}X^{N-2} + \cdots$ It is clear that the roots of $F_N(X)$ are $\cot^2\tfrac{j\pi}{2N}$ for $1 \le j \le N-1$ , and hence we deduce that $\sum_{j=1}^{N-1} \cot^2 \tfrac{j\pi}{2N} \; = \; {2N \choose 1}^{-1} {2N \choose 3} \; = \; \tfrac13(N-1)(2N-1)$ and so $\sum_{j=1}^{N-1} \mathrm{cosec}^2 \tfrac{j\pi}{2N} \; = \; N-1 + \tfrac13(N-1)(2N-1) \; = \; \tfrac23(N^2-1)$ Putting $N = 2^n$ gives the sum $\tfrac23(4^n - 1)$ for this question, making the answer $2+3+4+2+3 = \boxed{14}$ .

We can prove the formula for $\zeta(2)$ without using the $2^n$ version of this result. Since $\sin x < x < \tan x$ for $0 < x < \tfrac12\pi$ , we obtain $\sum_{j=1}^{n-1} \cot^2 \tfrac{j\pi}{2n} \; < \; \tfrac{4n^2}{\pi^2} \sum_{j=1}^{n-1} j^{-2} \; < \; \sum_{j=1}^{n-1} \mathrm{cosec}^2 \tfrac{j\pi}{2n}$ and hence $\tfrac13(n-1)(2n-1) \; < \; \tfrac{4n^2}{\pi^2}\sum_{j=1}^{n-1} j^{-2} \; < \; \tfrac23(n^2-1)$ which gives $\tfrac{(n-1)(2n-1)}{12n^2}\pi^2 \; < \; \sum_{j=1}^{n-1}j^{-2} \; < \; \tfrac{n^2-1}{6n^2}\pi^2$ and the result is immediate, since the two outside terms both converge to $\tfrac16\pi^2$ as $n \to \infty$ .