Proving that ${n+3\choose 3} =\frac{n+2}{2}\sum_{k=1}^{n+1}\csc^2\frac{k\pi}{n+2}$

Fancy physics predicts the equality $${n+3\choose 3} =\frac{n+2}{2}\;\sum_{k=1}^{n+1}\csc^2\frac{k\pi}{n+2}$$ which I can check (numerically and symbolically) for small $$n$$, but cannot prove for every $$n$$.

Does someone see an elegant way of doing this? Expressing $$\csc$$ in terms of exponentials allows one to see this as a sum involving roots of unity, but I don't see how to proceed.

• Why not write it as $${n+1\choose 3} =\frac{n}{2}\sum_{i=1}^{n-1}\frac1{\sin^2(i\pi/n)}?$$ Surely it must come from looking at the coefficients of a Chebyshev polynomial. – Lord Shark the Unknown Jun 1 at 18:02
• Physicists like their context, Shark. :) – The Count Jun 1 at 18:04
• @LordSharktheUnknown, or $$\sum_{k=1}^{n-1}\frac{1}{\sin^2(k\pi/n)} = \frac{n^2-1}{3}$$, or indeed $$H\left(\left\{1 - \cos\left(\frac{2\pi k}n\right) : 1 \le k < n \right\}\right) = \frac{6}{n+1}$$ where $H$ denotes the harmonic mean. – Peter Taylor Jun 1 at 22:03

Using parts of this question and answers $$\sum_{k=1}^{n+1}\frac{1}{\sin^2\left(\frac{k\pi}{n+2}\right)}= \sum_{k=1}^{n+1}\left(1+\cot^2\left(\frac{k\pi}{n+2}\right)\right)=\\ n+1+\sum_{k=1}^{n+1}\left(\cot^2\left(\frac{k\pi}{n+2}\right)\right)=n+1+\frac{(n+1)n}{3}=\frac{(n+3)(n+1)}{3}$$

• Beautiful! Nice and short. – Herri herri Jun 3 at 10:28

Proof using linear algebra :

Initial expression

$${m+3\choose 3} =\frac{m+2}{2}\sum_{k=1}^{m+1}\sin\left(\tfrac{k\pi}{m+2}\right)^{-2},\tag{1}$$

can be transformed, by expressing $${m+3\choose 3}=\dfrac16(m+3)(m+2)(m+1)$$ into :

$$\dfrac13(m+3)(m+1) =\sum_{k=1}^{m+1}\sin\left(\tfrac{k\pi}{m+2}\right)^{-2},\tag{1'}$$

Let us separate cases $$m$$ even and $$m$$ odd.

First case : $$m:=2n$$.

In this case, expression (1') is equivalent (due to the property $$\sin(\pi-x)=\sin(x)$$) to :

$$\dfrac13(2n+3)(2n+1) =1+2\sum_{k=1}^{n}\sin\left(\tfrac{k\pi}{2(n+1)}\right)^{-2},\tag{1''}$$

which is itself equivalent to :

$$\sum_{k=1}^n \dfrac{1}{4 \sin\left(\tfrac{k\pi}{2(n+1)}\right)^2}= \dfrac{n(n+2)}{6}\tag{2}$$

Consider the $$n \times n$$ tridiagonal matrix

$$D_n=\begin{pmatrix}2&-1&0&0&0&0&\cdots\\ -1&2&-1&0&0&0&\cdots\\ 0&-1&2&-1&0&0&\cdots\\ 0&0&-1&2&-1&0&\cdots\\ 0&0&0&-1&2&-1&\cdots\\ 0&0&0&0&-1&2&\cdots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots& \end{pmatrix}\tag{3}$$

The eigenvalues of $$D_n$$ are (see Eigenvalues of tridiagonal symmetric matrix with diagonal entries 2 and subdiagonal entries 1) :

$$\lambda_k:=4 \sin\left(\tfrac{k\pi}{2(n+1)}\right)^2, \ \ k=1,2,\cdots n\tag{4}$$

Remark : matrix $$D_n$$ is the classical discretized version of the second derivative (with a minus sign) : see the "Neumann case" in https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors_of_the_second_derivative.

Let us now consider $$E_n:=D_n^{-1}.$$

For example :

$$E_6:=D_6^{-1}=\dfrac17\begin{pmatrix} 6&5&4&3&2&1\\ 5&10&8&6&4&2\\ 4&8&12&9&6&3\\ 3&6&9&12&8&4\\ 2&4&6&8&10&5\\ 1&2&3&4&5&6 \end{pmatrix}\tag{5}$$

In these matrices, we just need the general fact that the diagonal entries of $$D_n$$ obey the following formula :

$$E_{n,kk}=\tfrac{1}{n+1}k(n+1-k)\tag{6}.$$

(see Remark 1 below for a proof of (6))

Therefore :

$$trace(E)=\tfrac{1}{n+1}(\sum_{k=1}^n k(n+1-k)) \tag{7}$$

$$=\tfrac{1}{n+1}((n+1)\tfrac{n(n+1)}{2}-\tfrac{n(n+1)(2n+1)}{6})=\tfrac{n(n+2)}{6} \tag{8}$$

i.e., the RHS of (2).

Besides,

$$trace(E)=\sum \dfrac{1}{\lambda_k}$$ gives the LHS of (2), establishing this relationship.

Remarks :

1) Proof of formula (6). We need a first result which is rather easy to establish by recurrence : $$\det(D_n)=n+1$$. The cofactor associated with $$E_{kk}$$ is obtained by cancelling the $$k$$th row and $$k$$th column of $$A_n$$, giving a diagonal-by-block determinant with blocks $$A_{k-1}$$ and $$A_{n-k)}$$, whose determinants are resp. $$k$$ and $$n-k+1$$. This minor is to be divided by the determinant of $$A_n,$$ i.e., $$n+1.$$

2) In (8), we have used the classical formulas for the sum of the $$n$$ first integers, resp. the squares of the $$n$$ first integers.

3) A slightly different version of matrix $$D$$ and its remarkable inverse can be found as second example in the following question : Looking for examples of Discrete / Continuous complementary approaches This is a typical case of an ill-conditionned matrix where a small perturbation (on the bottom right coefficient) gives a very different inverse.

4) Matrix $$D_n$$ is connected to Chebyshev polynomials of the 2nd kind $$U_n$$ ; see the answer https://math.stackexchange.com/q/1770607

Second case : $$m=2n-1$$

Relationship (1') (to be established) is :

$$\dfrac16 n(n+1) =\sum_{k=1}^{n}\left(4\sin\left(\tfrac{k\pi}{2n+1}\right)\right)^{-2},\tag{9}$$

Let us consider now the $$n \times n$$ tridiagonal matrix :

$$\Delta_n:=\begin{pmatrix}2&1&&&&\\1&2&1&&&\\&1&2&1&&\\&&\ddots&\ddots&\ddots\\&&&1&2&1\\&&&&1&3\end{pmatrix}$$

(please note the bottom right exceptional entry $$3$$).

It is not difficult to prove that $$\Delta_n$$ has the following eigenvalues :

$$\mu_k=4 \sin^2\left(\frac{k\pi}{2n+1}\right),\qquad k=1\dots n$$

One can prove as well that $$\det \Delta_n=2n+1$$.

The inverse $$\Gamma_n:=\Delta_n^{-1}$$ is a structured matrix whose diagonal elements are :

$$\Gamma_{n,kk}=k\dfrac{1}{2n+1}k(2n+1-2k)=k-2\dfrac{1}{2n+1}k^2\tag{10}$$

(proof using the same arguments as those used in Remark 1 above.)

Using (10), the trace of $$\Gamma$$ is

$$\sum_{k=1}^n k-2\dfrac{1}{2n+1}\sum_{k=1}^n k^2=\dfrac{n(n+1)}{6}$$

which is the LHS of (9), completing the proof.

• Interesting! Thanks for your answer. – Herri herri Jun 3 at 10:28
• I have now a proof covering the two cases (odd and even $m$). – Jean Marie Jun 3 at 21:14