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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?

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  • $\begingroup$ Use this identity (also here). $$\sum\limits_{k=1}^{m} \cot^2\dfrac{\pi k}{2m + 1} = \dfrac{m \left( 2m - 1 \right)}{3}$$ $\endgroup$
    – rtybase
    Dec 24 '20 at 0:47
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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, $$ \lim\limits_{n\rightarrow +\infty}\frac{1}{n^2}\sum_{k=1}^n\cot^2\left(\frac{k\pi}{2n+1}\right)=\frac{2}{3} $$

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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}

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