Consider the following sum for large $N$
$$
f\left(N\right)=\frac1{2N}\sum_{n=1}^{N-1}\cos\left(\frac{\pi n}N\right)\cot\left(\frac{\pi n}N\right)
$$
Since it is like a Riemann sum, I converted to into an integral
$$
\frac12\int_0^\pi\cos\left(x\right)\cot\left(x\right)\,\mathrm dx=\infty
$$
But $\frac12\int_\epsilon^{\pi-\epsilon}\cos\left(x\right)\cot\left(x\right)\,\mathrm dx=-\log\left(\tan\left(\frac\epsilon2\right)\right)-\cos\left(\epsilon\right)=\log\left(\frac1\epsilon\right)+\log\left(2\right)-1+a\left(\epsilon\right)$ with $\lim_{\epsilon\to0}a\left(\epsilon\right)=0$. Since $\epsilon\sim\frac\pi N$ I guessed that $f\left(N\right)$ also diverges logarithmically.
I need to extract the large $N$ behaviour of $f\left(N\right)$, that is find a $g\left(N\right)$ such that $\lim_{N\to\infty}f\left(N\right)-g\left(N\right)=0$. My guess is that $g\left(N\right)=C\log\left(N\right)+ \text{others}$. An exact expression for $g$ with the non-logarithmic pieces would be great (that is, with all the relevant constant and subleading corrections). But any help would be greatly appreciated.
Best Answer
Let's use the integral approximation more carefully, by identifying the $N-1$ terms in $f(N)$ with the areas of $N-1$ rectangular approximations to strips of a finite integral, viz.$$f(N)\approx I_N:=\frac{1}{2\pi}\int_{\pi/(2N)}^{\pi-\pi/(2N)}\cos x\cot x dx.$$In particular, each area approximation uses the height at the strip's midpoint. Since$$\int\cos x\cot xdx=\int(\csc x-\sin x)dx=-\ln|\csc x+\cot x|+\cos x+C,$$we have$$\begin{align}I_N&=\frac{1}{2\pi}\left[-\ln|\csc x+\cot x|+\cos x\right]_{\pi/(2N)}^{\pi-\pi/(2N)}\\&=\frac{1}{\pi}\left(\ln\left(\csc\frac{\pi}{2N}+\cot\frac{\pi}{2N}\right)-\cos\frac{\pi}{2N}\right)\\&\approx\frac{1}{\pi}\left(\ln\frac{4N}{\pi}-1\right).\end{align}$$Since a quadratic approximation of a sufficiently nice function $f$ gives$$\int_a^bf(x)dx-(b-a)f\left(\frac{a+b}{2}\right)\approx\frac{(b-a)^3}{12}f^{\prime\prime}\left(\frac{a+b}{2}\right),$$and since the second derivative of $\csc x-\sin x$ is $\sin x-\csc x+2\csc^3x$,$$I_N-f(N)\approx\sum_{n=1}^{N-1}\frac{\pi^3}{12N^3}\left(\sin\frac{n\pi}{N}-\csc\frac{n\pi}{N}+2\csc^3\frac{n\pi}{N}\right).$$We can easily verify $\pm O(N^{-2}\ln N)$ bounds on everything apart from the cubed-cosecant contributions, which are more problematic; the $n=1$ term alone adds approximately $\frac16$, as does the $n=N-1$ term. The same technique as before gives$$\begin{align}\sum_n\frac{\pi^3}{6N^3}\csc^3\frac{n\pi}{N}&\approx\frac{\pi^2}{6N^2}\int_{\pi/(2N)}^{\pi-\pi/(2N)}\csc^3 xdx\\&=\frac{\pi^2}{48N^2}\left[\sec^2\frac{x}{2}-\csc^2\frac{x}{2}+4\ln\tan\frac{x}{2}\right]_{\pi/(2N)}^{\pi-\pi/(2N)}\\&=\frac{\pi^2}{24N^2}\left(\csc^{2}\frac{\pi}{4N}-\sec^{2}\frac{\pi}{4N}-4\ln\tan\frac{\pi}{4N}\right).\end{align}$$Asymptotically, this is $\frac23$. But since it's $O(1)$, getting the exact $O(1)$ error no doubt requires a finer approximation than the quadratic one attempted here.