Beyond Basel

Note that $\begin{aligned} \sin(2n+1)x & = \; \mathrm{Im}\left((\cos x + i \sin x)^{2n+1}\right) \; = \; \sum_{m=0}^n (-1)^m {2n+1 \choose 2m+1} \cos^{2n-2m}x \sin^{2m+1}x \\ & = \; \sin^{2n+1}x F_n(\cot^2x) \end{aligned}$ where $F_n(X)$ is the degree $n$ polynomial $F_n(X) \; = \; \sum_{m=0}^n (-1)^m{2n+1 \choose 2m+1}X^{n-m} \; = \; {2n+1 \choose 1} X^n - {2n+1 \choose 3}X^{n-1} + {2n+1 \choose 5}X^{n-2} + \cdots$ Note that the zeros of $F_n(X)$ are the values of $\cot^2x$ where $\sin(2n+1)x = 0$ but $\sin x \neq 0$ , and so $F_n(X)$ has $n$ distinct real zeros, namely $u_p \; = \; \cot^2 \left(\tfrac{p\pi}{2n+1}\right) \hspace{2cm} p \; = \; 1,2,...,n$ Thus we know that $\begin{aligned} e_1 \; = \; \sum_{p=1}^n u_p & = \; {2n+1 \choose 1}^{-1}{2n+1 \choose 3} \; = \; \tfrac13n(2n-1) \\ e_2 \; = \; \sum_{1 \le p < q \le n} u_p u_q & = \; {2n+1 \choose 1}^{-1}{2n+1 \choose 5} \; = \; \tfrac{1}{30}n(n-1)(2n-1)(2n-3)\\ e_3 \; = \; \sum_{1 \le p < q < r \le n} u_p u_q u_r & = \; {2n+1 \choose 1}^{-1}{2n+1 \choose 7} \; = \; \tfrac{1}{630}n(n-1)(n-2)(2n-1)(2n-3)(2n-5) \end{aligned}$ and hence $\begin{aligned} \sum_{p=1}^n \cot^2\big(\tfrac{p \pi}{2n+1}\big) & = \; e_1 \; = \; \tfrac13n(2n-1) \\ \sum_{p=1}^n \cot^4\big(\tfrac{p \pi}{2n+1}\big) & = \; e_1^2 - 2e_2 \; = \; \tfrac{1}{45}n(2n-1)(4n^2 + 10n - 9)\\ \sum_{p=1}^n \cot^6\big(\tfrac{p \pi}{2n+1}\big) & = \; e_1^3 - 3e_1e_2 + 3e_3 \; = \; \tfrac{1}{945}n(2n-1)(32n^4 + 112n^3+ 8n^2 - 252n + 135) \end{aligned}$ which makes the answer $32 + 112 + 8 + 252 + 135 + 945 = \boxed{1484}$ .

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If $0 < x < \tfrac12\pi$ then $\sin x < x < \tan x$ , and hence $\cot x < x^{-1} < \mathrm{cosec}\, x$ , so that $\cot^6x \; < \; x^{-6} \;< \; \mathrm{cosec}^6 x \; = \; (\cot^2 x + 1)^3 \; = \; \cot^6x + 3\cot^4x + 3\cot^2x + 1$ and hence $\sum_{p=1}^n \cot^6\big(\tfrac{p \pi}{2n+1}\big) \; < \; \tfrac{(2n+1)^6}{\pi^6}\sum_{p=1}^n \frac{1}{p^6} \; < \; \sum_{p=1}^n \big[\cot^2\big(\tfrac{p \pi}{2n+1}\big) + 1\big]^3$ Since $\sum_{p=1}^n \cot^2\big(\tfrac{p \pi}{2n+1}\big) \; \sim \; \tfrac23n^2 \hspace{1cm} \sum_{p=1}^n \cot^4\big(\tfrac{p \pi}{2n+1}\big) \; \sim \; \tfrac{8}{45}n^4 \hspace{1cm} \sum_{p=1}^n \cot^6\big(\tfrac{p \pi}{2n+1}\big) \; \sim \; \tfrac{64}{945}n^6$ as $n \to \infty$ , we easily deduce that $\lim_{n \to \infty} \tfrac{1}{(2n+1)^6}\sum_{p=1}^n \cot^6\big(\tfrac{p\pi}{2n+1}\big) \; = \; \lim_{n \to \infty} \tfrac{1}{(2n+1)^6}\sum_{p=1}^n \big[\cot^2\big(\tfrac{p\pi}{2n+1}\big) + 1\big]^3 \; = \; \tfrac{1}{945}$ and hence $\sum_{p=1}^\infty \tfrac{1}{p^6} \; = \; \tfrac{1}{945}\pi^6$

My skills do not run this deep. This problem was so far beyond me that I knew I would have to approach it more like a detective than a mathematician.

First I directly evaluated the first 6 terms on my calculator: $\frac{1}{27}, \frac{34}{5}, \frac{563}{7}, \frac{11692}{27}, 1575, 4510$

The common denominator is 945. That's just like the bonus. Let's assume the continued terms are either integers or don't require any other factors in their denominators. OK then $F=945$

Now to tackle the polynomial. I have 6 terms but I only need 5 for a fourth-degree polynomial. Good. I can check my result with it.

$\frac{1}{27}=\frac{1}{945}(1)(1)(35)$

$\frac{34}{5}=\frac{1}{945}(2)(3)(1071)$

$\frac{563}{7}=\frac{1}{945}(3)(5)(5067)$

$\frac{11692}{27}=\frac{1}{945}(4)(7)(14615)$

$1575=\frac{1}{945}(5)(9)(33075)$

$4510=\frac{1}{945}(6)(11)(64575)$

I used my calculator to fit a polynomial to the first 5 terms and it nicely fit the 6th. Looking good.

The polynomial is $32x^{4}+112x^{3}+8x^{2}-252x+135$

The $x$ term is the only negative! Another clue that I am on the right track.

Then the moment of truth: $32+112+8+252+135+945=1484$

Try it and... Advanced Problem #5, you are MINE!