Branden Valentine

2022-01-27

What is the value of: $\underset{n\to \mathrm{\infty }}{lim}\sum _{r=1}^{n-1}\frac{{\mathrm{cot}}^{2}\left(r\frac{\pi }{n}\right)}{{n}^{2}}$

ebbonxah

It would be naïve and incorrect to proceed as follows
$\sum _{k=1}^{n-1}\frac{{\mathrm{cot}}^{2}\left(\pi \frac{k}{n}\right)}{{n}^{2}}\underset{\text{WRONG!}}{\underset{⏟}{\approx }}\frac{1}{n}{\int }_{\frac{1}{n}}^{1-\frac{1}{n}}{\mathrm{cot}}^{2}\left(\pi x\right)dx$
$=\frac{1}{n}\left(-x-\frac{1}{\pi }\mathrm{cot}\left(\pi x\right)\right){\mid }_{\frac{1}{n}}^{1-\frac{1}{n}}$
$=\frac{2}{{n}^{2}}-\frac{1}{n}+\frac{2}{n\pi }\mathrm{cot}\left(\frac{\pi }{n}\right)$
$\to \frac{2}{{\pi }^{2}}$
Instead, we use ${\mathrm{cot}}^{2}\left(x\right)={\mathrm{csc}}^{2}\left(x\right)-1$ to write
$\sum _{k=1}^{n-1}\frac{{\mathrm{cot}}^{2}\left(\pi \frac{k}{n}\right)}{{n}^{2}}=\frac{1}{n}-\frac{1}{{n}^{2}}+\sum _{k=1}^{n-1}\frac{1}{{n}^{2},{\mathrm{sin}}^{2}\left(\pi \frac{k}{n}\right)}$
$=\frac{1}{n}-\frac{1}{{n}^{2}}+2\sum _{k=1}^{|n/2|-1}\frac{1}{{n}^{2}\phantom{\rule{0.167em}{0ex}}{\mathrm{sin}}^{2}\left(\pi k/n\right)}$
Next, we note that for Hence, we have

coolbananas03ok

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

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