What is the value of: $\lim_{n\rightarrow \infty}\sum_{r=1}^{n-1}\frac{\cot^2(r\pi/n)}{n^2}$

Question

$$\lim_{n\rightarrow \infty}\sum_{r=1}^{n-1}\frac{\cot^2(r\pi/n)}{n^2}$$

How can I calculate the value of this trigonometric function where limits tends to infinity? I have thought and tried various of methods like using:

1. Tan(x)/x property
2. Converting lim to Integration
3. Sandwich theorem
4. Trigonometric series

But none worked out for me.

So please tell me a good and easy approach for this question :)

Answer 1

It would be naïve and incorrect to proceed as follows

$$\begin{align} \sum_{k=1}^{n-1}\frac{\cot^2(\pi k/n)}{n^2}&\underbrace{\approx}_{\text{WRONG!}} \frac1n\int_{1/n}^{1-1/n}\cot^2(\pi x)\,dx\\\\ &=\frac1n\left.\left(-x-\frac1\pi \cot(\pi x)\right)\right|_{1/n}^{1-1/n}\\\\ &=\frac2{n^2}-\frac1n +\frac2{n\pi}\cot(\pi/n)\\\\ &\to \frac2{\pi^2} \end{align}$$

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Instead, we use $\cot^2(x)=\csc^2(x)-1$ to write

$$\begin{align} \sum_{k=1}^{n-1}\frac{\cot^2(\pi k/n)}{n^2}&=\frac1n-\frac1{n^2}+\sum_{k=1}^{n-1}\frac{1}{n^2\,\sin^2(\pi k/n)}\\\\ &=\frac1n-\frac1{n^2}+2\sum_{k=1}^{\lfloor n/2\rfloor-1}\frac{1}{n^2\,\sin^2(\pi k/n)} \end{align}$$

Next, we note that for $\pi/2>x>0$, $\left(x-\frac16 x^3\right)^2\le \sin^2(x)\le x^2$. Hence, we have

$$\begin{align} \frac2{\pi^2}\sum_{k=1}^{\lfloor n/2\rfloor-1}\frac1{k^2} \le \frac2{n^2}\sum_{k=1}^{\lfloor n/2\rfloor-1}\frac{1}{n^2\,\sin^2(\pi k/n)}&\le \frac2{\pi^2}\sum_{k=1}^{\lfloor n/2\rfloor-1}\frac1{k^2\left(1-\frac16\frac{\pi^2k^2}{n^2}\right)^2}\\&=\frac2{\pi^2}\sum_{k=1}^{\lfloor n/2\rfloor-1}\frac1{k^2}+O\left(\frac1n\right) \end{align}$$

whence letting $n\to \infty$ and applying the squeeze theorem yields the coveted limit

$$\lim_{n\to\infty}\sum_{k=1}^{n-1}\frac{\cot^2(k\pi/n)}{n^2}=\frac13$$

Answer 2

By applying Vieta's formulas to Chebyshev polynomials of the second kind we have $$ \sum_{r=1}^{n-1}\cot^2\left(\frac{\pi r}{n}\right)=\frac{(n-1)(n-2)}{3} $$ (compare Cauchy's proof of $\zeta(2)=\frac{\pi^2}{6}$ in his Cours d'Analyse) hence the wanted limit is clearly $\frac{1}{3}$.

Answer 3

Here I came up with a solution involving only elementary calculation, I hope it's appreciable...

(Sorry as it's in image form)