Conjecture regarding the series $\sum_{k=0}^\infty\frac{\tan\left( \frac{2k+1}{2}\pi\sqrt{n}\right)}{(2k+1)^3}$

Question

In 2018 this question was posted on AoPS:

Prove that $$\sum_{k=0}^\infty\frac{\tan\left( \frac{2k+1}{2}\pi\color{red}{\sqrt{15}}\right)}{(2k+1)^3}=-\frac{\pi^3}{32\color{red}{\sqrt{15}}} $$

Before attempting to prove it, I noticed that $\sqrt{15}$ appears on both sides of the equation. This lead almost immediately to the question:

For what natural numbers $n$ the following holds? $$\sum_{k=0}^\infty\frac{\tan\left( \frac{2k+1}{2}\pi\color{red}{\sqrt{n}}\right)}{(2k+1)^3}=-\frac{\pi^3}{32\color{red}{\sqrt{n}}}$$

I tabled the values for $n$ ranging from $1$ to $200$ and found that the only values satisfying our equation in this range are $$n=3,15,35,63,99,143,195$$ A quick check on the OEIS revealed that these numbers are the sequence $a_m=4m^2-1$, or in other words the numbers that can be written as the product of two consecutive odd integers.

User @pisco linked a previous answer where it is shown that $\sum_{k=0}^\infty\frac{\tan\left( \frac{2k+1}{2}\pi\color{red}{\sqrt{n}}\right)}{(2k+1)^3}$ is of the form $\frac{\pi^3}{\sqrt{n}}\cdot q(n)$ where $q(n)\in\mathbb{Q}$ for every positive integer $n$. So the big question becomes:

What is the relation between $n$ and $q(n)$?

And the previous investigation leads to:

Conjecture 1: Is it true that $q(4m^2-1)=-\frac{1}{32}$?

Further investigation suggested another conjecture:

Conjecture 2 Is it true that $q(4m^2+1)=\frac{1}{32}$?

In the link provided, there is an algorithm to calculate $q(n)$ for each $n$ using the continued fraction of $\sqrt{n}$, but apart from realizing that \begin{align*} \sqrt{4m^2-1} &= [2m+1;\overline{1,4m+2}] \\ \sqrt{4m^2+1} &= [2m;\overline{4m}] \\ \end{align*} I couldn't elaborate further on it to understand what is going on, so I ask your advice.

In addition, I strongly believe there is another way to evaluate the sum, one that does not involve continued fractions. This claim is supported by the fact that if we plug $m=0$ in the sequence $4m^2-1$, resulting in $\sqrt{n}=\sqrt{-1}=i$, the result still holds, even though $i$ cannot be expressed as a real continued fraction.

I've been looking for this other way with no success for now, will update if I find something.

Answer 1

Not an answer, but it is easy to show the sum in question can always be evaluated nicely. Let $F(\alpha) = \sum_{n\geq 1} \frac{\cot(\pi n \alpha)}{n^3}$ defined here, then $$\sum_{k=0}^\infty\frac{\tan\left( \frac{2k+1}{2}\pi\sqrt{n}\right)}{(2k+1)^3}= \sum_{k=1}^\infty \frac{\tan\left( \frac{k}{2}\pi\sqrt{n}\right)}{k^3} - \frac{1}{8}\sum_{k=1}^\infty \frac{\tan\left(k\pi\sqrt{n}\right)}{k^3}$$ Using $\tan(x) = \cot(x) - 2\cot(2x)$, we see your sum is $$F(\frac{\sqrt{n}}{2}) - \frac{17}{8} F(\sqrt{n}) + \frac{1}{4}F(2\sqrt{n})$$ As shown in the link, $F(r\sqrt{n}) \in \frac{\pi^3\mathbb{Q}}{\sqrt{n}}$ for any non-square rational $n$, so is your sum.

Using the Mathematica code there, we obtain, for example: $$\sum_{k=0}^\infty\frac{\tan\left( \frac{2k+1}{2}\pi\sqrt{17}\right)}{(2k+1)^3} = \frac{\pi ^3}{32 \sqrt{17}}$$ $$\sum_{k=0}^\infty\frac{\tan\left( \frac{2k+1}{2}\pi\sqrt{61}\right)}{(2k+1)^3} = -\frac{2743 \pi ^3}{24352 \sqrt{61}}$$ $$\sum_{k=0}^\infty\frac{\tan\left( \frac{2k+1}{2}\pi\sqrt{211}\right)}{(2k+1)^3} = \frac{102530236189 \pi ^3}{204402190432 \sqrt{211}}$$

A generalization can be made by replacing cube in denominator to any odd power.

Answer 2

The already known resuls due to pisco from the answer to the MSE question 3019766 should lead to a good end. Since nobody did so far the work, i will drop an answer below.

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Let us use the functions: $$ \small \begin{aligned} F(a) &= \frac 1{1^3}\cot(\pi a\cdot 1) + \frac 1{2^3}\cot(\pi a\cdot 2) + \frac 1{3^3}\cot(\pi a\cdot 3) + \frac 1{4^3}\cot(\pi a\cdot 4) + \dots \\ &=\sum_{n\ge 1} \frac 1{n^3}\cot(\pi a\cdot n) \ , \\[2mm] \bbox[lightgreen]{G(a)} &= \frac 1{1^3}\tan\left(\frac{\pi a}2\cdot 1\right) + \frac 1{3^3}\tan\left(\frac{\pi a}2\cdot 3\right) + \frac 1{5^3}\tan\left(\frac{\pi a}2\cdot 5\right) + \frac 1{7^3}\tan\left(\frac{\pi a}2\cdot 7\right) + \dots \\ &\bbox[lightgreen]{=\sum_{\substack{n\ge 1\\n\text{ odd}}} \frac 1{n^3}\tan\left(\frac{\pi a}2\cdot n\right) } \ . \end{aligned} $$ We want the values of $G$ at specific values $a=\sqrt{4m^2\pm 1}$.

The function $F$ appears in loc. cit. and has the obvious properties $F(a+1)=F(a)$ (it is periodic with period one), $F(-a)=-F(a)$ (it is an odd function, since $\cot$ it is), and the functional equation, obtained by applying the residue theorem for $\cot(\pi z)\; \cot(\pi az)\;/\; z^3$: $$ \tag{*} \bbox[lightblue]{\qquad F(a) + a^2F\left(\frac 1a\right) = \underbrace{\frac{\pi^3}{90 a}(a^4 -5a^2+1)}_{=:R(a)}\ . \qquad} $$ The fact that for $a$ for instance equal to $\sqrt N$ we can use continued fractions, the periodicity, and the functional equation $(*)$ to get a linear system over $\Bbb Q(a,\pi)$ with one unknown being $F(a)$, leads to a procedure to compute such a value $F(\sqrt N)$. We use the same idea below.

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For an odd $n$ we have $$ \tan\left(\frac{\pi a}2\cdot n\right) \cot\left(\frac \pi2 - \frac{\pi a}2\cdot n\right) = \cot\left(\frac \pi2\cdot n - \frac{\pi a}2\cdot n\right) = \cot\left(\frac {\pi(1-a)}2\cdot n \right)\ , $$ so $G(a)$ can be expressed in terms of $F((a-1)/2)$ and $F(a)$, from $$ \small \begin{aligned} G(a) &= \sum_{\substack{n\ge 1\\n\text{ odd}}}\frac 1{n^3}\tan\left(\frac{\pi a}2\cdot n\right) \\ &= \sum_{\substack{n\ge 1\\n\text{ odd}}}\frac 1{n^3}\cot\left(\frac {\pi(1-a)}2\cdot n \right) = \left(\sum_{n\ge 1} - \sum_{\substack{n\ge 1\\n\text{ even}}}\right) \frac 1{n^3}\cot\left(\frac {\pi(1-a)}2\cdot n \right) \\ &=+F\left(\frac{1-a}2\right) -\frac 18F(1-a)\\ &=-F\left(\frac{a-1}2\right) +\frac 18F(a-1)\\ &=-F\underbrace{\left(\frac{a+1}2\right)}_{=:b} +\frac 18F(a)\ . \end{aligned} $$

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In this block we consider the special case $N=4m^2-1$ for an integer $m$, and $a=\sqrt N$. It is convenient to introduce $M=2m-1$.

Then $a$ and $b:=(a+1)/2$ have the continued fractions expressions: $$ \begin{aligned} a &=[M;1,2M,1,2M,1,2M,\dots] =\left[M;\overline{1,2M}\right]\ ,\\ b:=\frac{a+1}2 &=[m;2,M,2,M,2,M,\dots] =\left[m;\overline{2,M}\right]\ ,\\ \end{aligned} $$ From here we manufacture quickly relations that determine $F(a)$ and $F(b)$. We introduce $a_0=a+M=[2M;1,2M,1,2M,\dots]$ and $a_1=[1;2M,1,2M,1,\dots]=1/(a-M)$. Then $F(a)=F(a_0)$ by $1$-periodicity. By the same argument $F(1/a_0)=F(a_1)$ and $F(1/a_1)=F(a_0)$. So: $$ \left\{ \begin{aligned} F(a_0) + a_0^2 F(a_1) &= R(a_0)\ ,\\ a_1^2 F(a_0) + F(a_1) &= R(a_1)\ , \end{aligned} \right. $$ which gives the value for $F(a)=F(a_0)$ by Cramer: $$ F(a) = F(a_0)= \frac{R(a_0)-a_0^2R(a_1)}{1-a_0^2a_1^2} = -\frac{\pi^3}{180a}(4M^2+8M-3) = -\frac{\pi^3}{180a}(4a^2-7) \ . $$ The same game for $b$, we are using the numbers (fully periodic as continued fractions) $b_0=b+(m-1)=[M;2,M,2,M,\dots]$ and $b_1=1/(b-m)=[2;M,2,M,2,\dots]$, which generate a system $$ \left\{ \begin{aligned} F(b_0) + b_0^2F(b_1) &= R(b_0)\ ,\\ b_1^2F(b_0) + F(b_1) &= R(b_1)\ , \end{aligned} \right. $$ which gives the value for $F(b)=F(b_0)$ by Cramer: $$ F(b) = F(b_0)= \frac{R(b_0)-b_0^2R(b_1)}{1-b_0^2b_1^2} = -\frac{\pi^3}{360a}(M^2+2M-12)\ . $$ Putting all together: $$ \begin{aligned} G(a) &= -F(b) +\frac 18 F(a) =\frac{\pi^3}a\cdot\frac1{180\cdot 8} \Big[+4(M^2+2M-12)-(4M^2+8M-3)\Big] \\ &=\frac{\pi^3}a\cdot\frac{-48+3}{180\cdot 8} =\frac{\pi^3}a\cdot\bbox[yellow]{\left(-\frac1{32}\right)}\ . \end{aligned} $$ $\square$

We have the wanted result from Conjecture 1.

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The same applies for the case $a=\sqrt N$ with $N=4m^2+1$.

We also use $M=2m-1$.

Then $a$ and $b:=(a+1)/2$ have the continued fractions expressions: $$ \begin{aligned} a &=[2m;4m,4m,4m,4m,\dots] =\left[2m;\overline{4m}\right]\ ,\\ b:=\frac{a+1}2 &=[m;1,1,M,1,1,M,1,1,M,\dots] =\left[m;\overline{1,1,M}\right]\ ,\\ \end{aligned} $$ From here we manufacture quickly relations that determine $F(a)$ and $F(b)$. We introduce $a_0=a+2m=[4m;4m,4m,\dots]$. Then $1/a_0$ is $a_0-4m$. Then $F(a)=F(a_0)=F(1/a_0)$ by $1$-periodicity, and the relation $(*)$ computed in $a_0$ gives $$ F(a) + a_0^2 F(a) = R(a_0)\ . $$ We immediately isolate $F(a)$: $$ F(a) = \frac{R(a_0)}{1+a_0^2} = \frac{\pi^3}{180a}(16m^2-3) = \frac{\pi^3}{180a}(4a^2-7) \ . $$ The same game for $b$, we are using the numbers (fully periodic as continued fractions)

• $b_0=b+(m-1)=[M;1,1,M,1,1,M,\dots]$,
• $b_1=1/(b_0-M)=[1;1,M,1,1,M,1,\dots]$,
• $b_2=1/(b_1-1)=[1;M,1,1,M,1,1,\dots]$, which generate a system with three equations and three unknowns, $F(b_0)$, $F(b_1)$, $F(b_2)$, $$ \left\{ \begin{aligned} F(b_0) \qquad\qquad + b_0^2F(b_2) &= R(b_0)\ ,\\ b_1^2F(b_0) + F(b_1) \qquad\qquad &= R(b_1)\ ,\\ \qquad\qquad b_2^2F(b_1) + F(b_2) &= R(b_2)\ , \end{aligned} \right. $$ which gives the value for $F(b)=F(b_0)$ by Cramer: $$ F(b) = F(b_0)= \frac { \begin{vmatrix} R(b_0) & 0 & b_0^2 \\ R(b_1) & 1 & 0 \\ R(b_2) & b_2^2 & 1 \end{vmatrix}} { \begin{vmatrix} 1 & 0 & b_0^2 \\ b_1^2 & 1 & 0 \\ 0 & b_2^2 & 1 \end{vmatrix}} = \frac{\pi^3}{90a}(m^2-3) \ . $$ Putting all together: $$ \begin{aligned} G(a) &= -F(b) +\frac 18 F(a) =\frac{\pi^3}a\cdot\frac1{180\cdot 8} \Big[-16(m^2-3)+(16m^2-3)\Big] \\ &=\frac{\pi^3}a\cdot\frac{48-3}{180\cdot 8} =\frac{\pi^3}a\cdot\bbox[yellow]{\frac1{32}}\ . \end{aligned} $$ $\square$

We have also the wanted result from Conjecture 2.

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Note: The computations were supported by CAS (sage and pari/gp). If wanted, in between steps can be shown, quick numerical checks for the explicit formulas for $F(a), F(b)$ can be added.