A contour integration approach to infinite series

We will evaluate the trigonometric series \(\sum_{n=0}^\infty \frac{\cot\left(\frac{2n+1}{2}\pi\sqrt{2} \right)}{(2n+1)^3}\) by integrating a meromorphic function around a sequence of expanding squares. The residues encode the desired sum, while equidistribution helps us choose a useful subsequence of contours. Define the meromorphic function

\[ f(z) = \frac{\pi \tan(\pi z)\tan(\pi z \theta)}{z^3} \]

where the parameter \(\theta\) is a positive irrational number. Let \(C_N\) denote the positively oriented square with vertices \((N+1)(1+i),\; (N+1)(-1+i),\; (N+1)(-1-i)\) and \((N+1)(1-i)\). The irrationality of \(\theta\) ensures that no pole on the real axis lies on the boundary of this square.

Estimating the contour integral

Put \(m=N+1\). We will estimate the horizontal and vertical sides separately. We use the elementary identities

\[ |\tan(u+iv)|^2 =\frac{\sin^2u+\sinh^2v}{\cos^2u+\sinh^2v} \]

and, whenever \(\tan u\) is finite,

\[ |\tan(u+iv)|^2 =\frac{\tan^2u+\tanh^2v} {1+\tan^2u\,\tanh^2v}. \]

The first identity, including its limiting cases when \(\cos u=0\), shows that for \(v\neq0\),

\[ \tanh|v|\leq|\tan(u+iv)|\leq\coth|v|, \]

and, if \(u\) is fixed while \(v\) varies,

\[ |\tan(u+iv)|\leq\max\{1,|\tan u|\}. \]

For the second bound, put \(a=\tan^2u\) and \(s=\tanh^2v\). Then the square of the modulus is \((a+s)/(1+as)\), which lies between \(a\) and \(1\) as \(s\) ranges from \(0\) to \(1\). This proves the claimed estimate.

On either horizontal side, write \(z=x\pm im\) with \(-m\leq x\leq m\). Since \(|z|\geq m\), we have

\[ |\tan(\pi z)|\leq\coth(\pi m), \qquad |\tan(\pi\theta z)|\leq\coth(\pi\theta m). \]

The combined length of the two horizontal sides is \(4m\). Therefore,

\[ \int_{\text{horizontal sides}}|f(z)|\,|dz| \leq \frac{4\pi}{m^2}\coth(\pi m)\coth(\pi\theta m). \]

On either vertical side, write \(z=\pm m+iy\) with \(-m\leq y\leq m\). Because \(m\) is an integer,

\[ |\tan(\pi z)|=|\tan(\pm m\pi+i\pi y)| =|\tanh(\pi y)|\leq1. \]

Set

\[ T_m:=|\tan(\pi m\theta)|. \]

The second consequence of the tangent identity gives

\[ |\tan(\pi\theta(\pm m+iy))| \leq\max\{1,T_m\}. \]

Again \(|z|\geq m\), and the two vertical sides have combined length \(4m\), so

\[ \int_{\text{vertical sides}}|f(z)|\,|dz| \leq\frac{4\pi}{m^2}\max\{1,T_m\}. \]

Combining the four sides yields the explicit estimate

\[ \begin{aligned} \left|\int_{C_N} f(z)\,dz\right| &\leq\int_{C_N}|f(z)|\,|dz|\\ &\leq\frac{4\pi}{m^2} \left( \max\{1,T_m\} +\coth(\pi m)\coth(\pi\theta m) \right). \end{aligned} \]

Choosing the contours by equidistribution

It remains to keep \(T_m\) away from the poles of the tangent. By Weyl's equidistribution theorem, the fractional parts \(\{m\theta\}\) are equidistributed in \([0,1)\). In particular,

\[ \lim_{M\to\infty} \frac{1}{M} \#\left\{1\leq m\leq M:\{m\theta\}\in \left[0,\frac14\right]\right\} =\frac14. \]

Thus there are infinitely many integers \(m\) for which \(\{m\theta\}\in[0,1/4]\). Enumerate them as an increasing sequence \(m_1<m_2<\cdots\). Since tangent has period \(\pi\),

\[ T_{m_k} =|\tan(\pi\{m_k\theta\})| \leq1. \]

Let \(N_k=m_k-1\). Substituting this bound into the contour estimate gives

\[ \left|\int_{C_{N_k}}f(z)\,dz\right| \leq\frac{4\pi}{m_k^2} \left(1+\coth(\pi m_k)\coth(\pi\theta m_k)\right). \]

Because \(m_k\to\infty\) and both hyperbolic cotangents tend to \(1\), the expression on the right is asymptotic to \(8\pi/m_k^2\). Hence

\[ \lim_{k\to \infty}\int_{C_{N_k}} f(z)\; dz = 0. \]

On the other hand, the residue theorem gives

\[ \frac{1}{2\pi i}\int_{C_{N_k}} f(z)\,dz = \substack{\displaystyle \operatorname{Res} \\ z=0}f(z) + \sum_{|i+\frac12|<m_k} \substack{\displaystyle \operatorname{Res} \\ z=\frac{2i+1}{2}}f(z) + \sum_{|j+\frac12|<\theta m_k} \substack{\displaystyle \operatorname{Res} \\ z=\frac{2j+1}{2\theta}}f(z). \]

A simple calculation gives the following residues:

\[ \begin{aligned} \substack{\displaystyle \text{Res} \\ z=0}f(z) &= \pi^3 \theta \\ \substack{\displaystyle \text{Res} \\ z=\frac{2n+1}{2}}f(z) &= -8\frac{\tan\left(\frac{\pi \theta}{2}(2n+1) \right)}{(2n+1)^3}, \quad n\in \{0,1,2,\cdots\}\\ \substack{\displaystyle \text{Res} \\ z=\frac{2n+1}{2\theta}}f(z) &= -8\theta^2 \frac{\tan\left(\frac{\pi}{2\theta}(2n+1) \right)}{(2n+1)^3} , \quad n\in \{0,1,2,\cdots\} \end{aligned} \]

The nonzero poles occur in pairs \(z\) and \(-z\), and the two residues in each pair are equal. Letting \(k\to\infty\) in the residue identity therefore gives the following relation, provided the two series converge:

\[ \sum_{n=0}^\infty \frac{\tan\left(\frac{\pi \theta}{2}(2n+1) \right)}{(2n+1)^3} + \theta^2 \sum_{n=0}^\infty \frac{\tan\left(\frac{\pi}{2\theta}(2n+1) \right)}{(2n+1)^3} = \frac{\pi^3 \theta}{16} \]

Equidistribution was used only to choose contours on which the boundary integral is small; by itself it does not guarantee convergence of the two series. For the value needed here, \(\theta=\sqrt2+1\), convergence presents no problem. A quadratic irrational is badly approximable, so there is a constant \(c>0\) such that

\[ |q\theta-p|\geq\frac{c}{q} \]

for all integers \(p\) and positive integers \(q\). The same is true of \(1/\theta=\sqrt2-1\). Consequently, the tangent factors in the two series are \(O(2n+1)\), even when their arguments approach a pole. After division by \((2n+1)^3\), the summands are \(O((2n+1)^{-2})\), and both series converge absolutely.

Finally, substitute \(\theta=\sqrt2+1\). Since \(2n+1\) is odd and \(1/\theta=\sqrt2-1\), periodicity gives

\[ \tan\left(\frac{\pi\theta}{2}(2n+1)\right) =\tan\left(\frac{\pi}{2\theta}(2n+1)\right) =-\cot\left(\frac{\pi\sqrt2}{2}(2n+1)\right). \]

Moreover,

\[ \frac{\theta}{1+\theta^2} =\frac{1}{\theta+\theta^{-1}} =\frac{1}{2\sqrt2}. \]

The relation above therefore reduces to

\[ \boxed{\sum_{n=0}^\infty \frac{\cot\left(\frac{2n+1}{2}\pi\sqrt{2} \right)}{(2n+1)^3} = -\frac{\pi^3}{32\sqrt{2}}} \]