# Evaluate $\lim\limits_{n\to +\infty}\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right)$

How to evaluate: $$\lim\limits_{n\rightarrow +\infty}\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right)$$

Personally, i always have trouble to evaluate sum of trignometric series, if you know a paper to recommend, or part of a book, about this, please let me know.

Anyway

$$\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right) = \frac{1}{n^2}\left(\sum_{k=1}^{n} \frac{1}{\sin^2\left(\frac{\pi k}{2n+1}\right)} - \sum_{k=1}^{n} 1\right)$$

What now?

• Use this identity (also here). $$\sum\limits_{k=1}^{m} \cot^2\dfrac{\pi k}{2m + 1} = \dfrac{m \left( 2m - 1 \right)}{3}$$ Commented Dec 24, 2020 at 0:47

Let $$P_n(X)=\frac{(X+i)^{2n+1}-(X-i)^{2n+1}}{2i}$$, then \begin{aligned} P_n(z)=0&\iff \left(\frac{z+i}{z-i}\right)^{2n+1}=1 \\ &\iff \exists k\in[\![1,2n+1]\!],\frac{z+i}{z-i}=e^{\frac{2ik\pi}{2n+1}} \\ &\iff\exists k\in[\![1,2n+1]\!],z=i\frac{e^{\frac{2ik\pi}{2n+1}}+1}{e^{\frac{2ik\pi}{2n+1}}-1} \\ &\iff\exists k\in[\![1,2n+1]\!],z=\cot\left(\frac{k\pi}{2n+1}\right) \end{aligned} Therefore the roots of $$P_n$$ are $$\alpha_k:=\cot\left(\frac{k\pi}{2n+1}\right)$$. Moreover, using the binomial theorem, we have $$P_n(X)=\frac{1}{2i}\sum_{k=0}^{2n+1}\binom{2n+1}{k}\left(i^k-(-i)^k\right)X^{2n+1-k}=\sum_{k=0}^n\binom{2n+1}{2k+1}(-1)^k X^{2n-2k}$$ because the even terms cancel each other. Let $$\displaystyle Q_n(X)=\sum_{k=0}^n\binom{2n+1}{2k+1}(-1)^k X^{n-k}$$, then $$Q_n(X^2)=P_n(X)$$ and the roots of $$Q_n$$ are the $$\alpha_k^2=\cot^2\left(\frac{k\pi}{2n+1}\right)$$. But since $$\alpha_{2n+1-k}=-\alpha_k$$ it remains only $$n=\deg Q_n$$ roots which are the $$\alpha_k^2$$ for $$k\in[\![1,n]\!]$$. Therefore, the lead coefficient of $$Q_n$$ being $$2n+1$$, we have $$Q_n(X)=(2n+1)\prod_{k=1}^n(X-\alpha_k^2)$$ Thus, the sum $$\sum_{k=1}^n\alpha_k^2$$ is the coefficient of degree $$n-1$$ in $$Q_n$$ divided by $$-\frac{1}{2n+1}$$, that is $$\sum_{k=1}^n\alpha_k^2=\frac{1}{2n+1}\binom{2n+1}{3}=\frac{n(2n-1)}{3}$$ Finally, $$\lim\limits_{n\rightarrow +\infty}\frac{1}{n^2}\sum_{k=1}^n\cot^2\left(\frac{k\pi}{2n+1}\right)=\frac{2}{3}$$
It is possible to prove, by studying the function $$f: x\mapsto 1-x^{2}\cot^{2}{x}-\frac{2 x^{2}}{3}$$ on $$\mathcal{D}=\left]-\frac{\pi}{2},\frac{\pi}{2}\right[\setminus\left\lbrace 0\right\rbrace$$, that for any $$x\in\mathcal{D}$$, we have : $$\left|1-x^{2}\cot^{2}{x}\right|\leq\frac{2x^{2}}{3}$$
Let $$n \in\mathbb{N}^{*}$$, and $$k\leq n$$. Setting $$x\leftarrow \frac{k\pi}{2n+1}$$, we get : $$\left|1-\frac{k^{2}\pi^{2}}{\left(2n+1\right)^{2}}\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}\right|\leq\frac{2k^{2}\pi^{2}}{3\left(2n+1\right)^{2}}$$
Using the previous inequality, we can write the following : \begin{aligned} \left|\frac{1}{\left(2n+1\right)^{2}}\sum_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}-\frac{1}{\pi^{2}}\sum_{k=1}^{n}{\frac{1}{k^{2}}}\right|&\leq\frac{1}{\pi^{2}}\sum_{k=1}^{n}{\frac{1}{k^{2}}\left|1-\frac{k^{2}\pi^{2}}{\left(2n+1\right)^{2}}\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}\right|}\\ &\leq \frac{2n}{3\left(2n+1\right)^{2}}\underset{n\to +\infty}{\longrightarrow}0\end{aligned}
Thus, the sequence $$\left(\frac{1}{\left(2n+1\right)^{2}}\sum\limits_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}\right)_{n\in\mathbb{N}^{*}}$$ does converge and : $$\lim_{n\to +\infty}{\frac{1}{\left(2n+1\right)^{2}}\sum_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}}=\lim_{n\to +\infty}{\frac{1}{\pi^{2}}\sum_{k=1}^{n}{\frac{1}{k^{2}}}}=\frac{1}{6}$$
Hence : \begin{aligned}\lim_{n\to +\infty}{\frac{1}{n^{2}}\sum_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}}&=\lim_{n\to +\infty}{\left(\frac{2n+1}{n}\right)^{2}\times\frac{1}{\left(2n+1\right)^{2}}\sum_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}}\\ &=4\times\frac{1}{6}\\ \lim_{n\to +\infty}{\frac{1}{n^{2}}\sum_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}}&=\frac{2}{3}\end{aligned}