Proving $\sum\limits_{l=1}^n \sum\limits _{k=1}^{n-1}\tan \frac {lk\pi }{2n+1}\tan \frac {l(k+1)\pi }{2n+1}=0$

Solution 1:

Micah already pointed the way in a comment: The identity

$$\tan a \tan b = \frac{\tan a - \tan b}{\tan(a-b)} - 1$$

causes the inner sum to telescope. To make full use of this, let's extend the inner sum to $k=2n$:

$$ \begin{align} \sum _{l=1}^{n}\sum _{k=1}^{2n}\tan \frac {lk\pi } {2n+1}\tan \frac {l(k+1) \pi } {2n+1} &= \sum _{l=1}^{n}\sum _{k=1}^{2n}\left(\frac{\tan \frac {l(k+1)\pi } {2n+1}-\tan \frac {lk \pi } {2n+1}}{\tan\frac {l\pi}{2n+1}}-1\right) \\ &= \sum _{l=1}^{n}\left(\frac{\tan \frac {l(2n+1)\pi } {2n+1}-\tan \frac {l\pi } {2n+1}}{\tan\frac {l\pi}{2n+1}}-2n\right) \\ &= \sum _{l=1}^{n}(0-1-2n) \\ &= -n(2n+1)\;. \end{align} $$

This sum contains each of the terms we want to sum twice, with mirror symmetry, and it contains one additional term in the middle for $k=n$. Thus, for our sum to vanish, we need

$$ \sum _{l=1}^{n}\tan \frac {ln\pi } {2n+1}\tan \frac {l(n+1) \pi } {2n+1}=-n(2n+1)\;. $$

The arguments of the two factors add to $l\pi$, so they're negatives of each other, so we're looking for

$$ -\sum _{l=1}^{n}\tan^2 \frac {ln\pi } {2n+1}\;. $$

We can again extend the sum to $2n$ to double it, since the arguments form pairs that add up to $n\pi$; then, since $n$ and $2n+1$ are coprime, we can replace $ln$ by $l$ while traversing the same arguments; and then we can set the upper limit back to $n$, since the arguments still add up to $\pi$ in pairs. Thus, what we need is

$$ \sum _{l=1}^{n}\tan^2 \frac {l\pi } {2n+1}=n(2n+1)\;. $$

How to find this sum is shown at Prove that $\sum\limits_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$. Adapting the argument in the accepted answer there for our odd denominator, we obtain

$$ \left(\cos\frac{k\pi}{2n+1}+\mathrm i\sin\frac{k\pi}{2n+1}\right)^{2n+1}=(-1)^k\;, $$

taking the imaginary part,

$$ \sum_{r=0}^n\binom{2n+1}{2r+1}\left(\cos\frac{k\pi}{2n+1}\right)^{2n-2r}\left(\mathrm i\sin\frac{k\pi}{2n+1}\right)^{2r+1}=0\;, $$

dividing by $\left(\cos\frac{k\pi}{2n+1}\right)^{2n+1}$,

$$ \sum_{r=0}^n\binom{2n+1}{2r+1}\left(\mathrm i\tan\frac{k\pi}{2n+1}\right)^{2r+1}=0\;, $$

and dividing by $\tan\frac{k\pi}{2n+1}$ and letting $x=\tan^2\frac{k\pi}{2n+1}$,

$$ \sum_{r=0}^n\binom{2n+1}{2r+1}(-x)^r=0\;. $$

Then Vieta's formula shows that the sum of the roots of this equation is

$$\frac{\binom{2n+1}2}{\binom{2n+1}0}=n(2n+1)\;,$$

as required.

[Update:]

This answer suggests an alternative, more elementary way to calculate the sum of the squares of the tangents: On the grid $\frac{l\pi}{2n+1}$, the tangent decomposes into $n$ mutually orthogonal sines, each of whose dot product with itself is $2n+1$, so the dot product of the tangent with itself is $n(2n+1)$.

Solution 2:

Starting with $$ \tan(x-y) = \frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}\tag{1} $$ we get $$ \tan(x)\tan(y)=\frac{\tan(x)-\tan(y)}{\tan(x-y)}-1\tag{2} $$ Thus, $$ \tan\left(\frac{l(k+1)\pi}{2n+1}\right)\tan\left(\frac{lk\pi}{2n+1}\right) =\frac{\tan\left(\frac{l(k+1)\pi}{2n+1}\right)-\tan\left(\frac{lk\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}-1\tag{3} $$ Therefore, because of the telescoping sum, $$ \begin{align} \sum_{k=1}^{n-1}\tan\left(\frac{l(k+1)\pi}{2n+1}\right)\tan\left(\frac{lk\pi}{2n+1}\right) &=\frac{\tan\left(\frac{ln\pi}{2n+1}\right)-\tan\left(\frac{l\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}-(n-1)\\ &=\frac{\tan\left(\frac{ln\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}-n\tag{4} \end{align} $$ Note that $$ \frac{\tan\left(\frac{(2n+1-l)n\pi}{2n+1}\right)}{\tan\left(\frac{(2n+1-l)\pi}{2n+1}\right)} =\frac{\tan\left(\frac{ln\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}\tag{5} $$ so that by replacing the odd $l$s with even $2n+1-l$s and using $\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}$, we get $$ \begin{align} \sum_{l=1}^n\frac{\tan\left(\frac{ln\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)} &=\sum_{l=1}^n\frac{\tan\left(\frac{2ln\pi}{2n+1}\right)}{\tan\left(\frac{2l\pi}{2n+1}\right)}\\ &=-\sum_{l=1}^n\frac{\tan\left(\frac{l\pi}{2n+1}\right)}{\tan\left(\frac{2l\pi}{2n+1}\right)}\\ &=\frac12\sum_{l=1}^n\left(\tan^2\left(\frac{l\pi}{2n+1}\right)-1\right)\tag{6} \end{align} $$

Using contour integration, we will compute $$ \sum_{l=1}^n\tan^2\left(\frac{\pi l}{2n+1}\right)=n(2n+1)\tag{7} $$ Note that $$ \frac{(2n+1)/z}{z^{2n+1}-1}\tag{8} $$ has simple poles. It has residue $1$ at $z=e^{\large\frac{2\pi li}{2n+1}}$ for each $l$ and residue $-(2n+1)$ at $z=0$.

Furthermore, at $z=e^{i\theta}$, $$ -\left(\dfrac{z-1}{z+1}\right)^2=\tan^2(\theta/2)\tag{9} $$ Because the total residue of $$ f(z)=\left(\frac{z-1}{z+1}\right)^2\frac{(2n+1)/z}{z^{2n+1}-1}\tag{10} $$ is $0$, we get that the sum of its residues at $z=0$ and $z=-1$ equals $$ \sum_{l=1}^{2n}\tan^2\left(\frac{\pi l}{2n+1}\right)=2\sum_{l=1}^n\tan^2\left(\frac{\pi l}{2n+1}\right)\tag{11} $$ First, $$ \mathrm{Res}_{z=0}f(z)=-(2n+1)\tag{12} $$ Next, $$ \begin{align} \mathrm{Res}_{z=-1}f(z) &=\mathrm{Res}_{z=0}f(z-1)\\ &=\mathrm{Res}_{z=0}(2n+1)\left(\frac{z-2}{z}\right)^2\frac1{1-z}\frac1{1+(1-z)^{2n+1}}\\ &=\mathrm{Res}_{z=0}(2n+1)\left(1-\frac4z+\frac4{z^2}\right)(1+z+\dots)\frac12\left(1+\frac{2n+1}{2}z+\dots\right)\\ &=\mathrm{Res}_{z=0}\frac{2n+1}{2}\left(1-\frac4z+\frac4{z^2}\right)\left(1+\frac{2n+3}{2}z+\dots\right)\\ &=\mathrm{Res}_{z=0}\frac{4n+2}{z^2}+\frac{(2n+1)^2}{z}+\dots\\ &=(2n+1)^2\tag{13} \end{align} $$ Combining $(11)$, $(12)$, and $(13)$, yields $(7)$.

Combining $(4)$, $(6)$, and $(7)$ yields $$ \sum_{k=1}^{n-1}\tan\left(\frac{l(k+1)\pi}{2n+1}\right)\tan\left(\frac{lk\pi}{2n+1}\right)=0\tag{14} $$

Solution 3:

This isn't an answer, just some numerical results that don't fit in the comments.

If you first perform the sum over $l$, the result is always an integer multiple of $2n+1$. I don't see a pattern in the multipliers, but I thought I'd post them in case someone else does. The numbers in the table are the multipliers $m$ in

$$ \sum _{l=1}^{n}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=m(2n+1)\;, $$

with $n$ increasing downward and $k$ to the right.

$$ \begin{array}{r|rr} &1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17\\\hline 2&0\\ 3&-1&1\\ 4&0&-1&1\\ 5&-1&-1&1&1\\ 6&0&0&-2&0&2\\ 7&-1&0&-2&1&0&2\\ 8&0&-2&2&-2&0&0&2\\ 9&-1&1&-1&-3&1&1&-1&3\\ 10&0&-1&-1&2&-3&0&-1&1&3\\ 11&-1&-1&-1&1&-3&1&1&-1&1&3\\ 12&0&0&0&-2&2&-4&0&0&0&0&4\\ 13&-1&0&0&-1&0&-4&3&-1&1&-1&0&4\\ 14&0&-2&0&-2&2&2&-4&0&0&-2&2&0&4\\ 15&-1&1&-3&3&-3&1&-5&3&-1&1&-1&-1&1&5\\ 16&0&-1&1&-2&-1&1&2&-5&-1&1&0&-1&0&1&5\\ 17&-1&-1&1&-1&-3&2&0&-5&3&-1&1&1&-2&0&1&5\\ 18&0&0&-2&0&2&-4&0&4&-6&2&-2&0&-2&2&0&0&6\\ \end{array} $$