Lorelei Stanton

2022-03-22

Solve the equation:
$f\left(n\right)={\mathrm{cot}}^{2}\left(\frac{\pi }{n}\right)+{\mathrm{cot}}^{2}\left(\frac{2\pi }{n}\right)+\dots +{\mathrm{cot}}^{2}\left(\frac{\left(n-1\right)\pi }{n}\right)$
then how to find limit of $\frac{3f\left(n\right)}{\left(n+1\right)\left(n+2\right)}$ as $n\to \mathrm{\infty }$ ?

### Answer & Explanation

kattylouxlvc

Recall the expansion:
$\mathrm{tan}nx=\frac{\left(\begin{array}{c}n\\ 1\end{array}\right)\mathrm{tan}x-\left(\begin{array}{c}n\\ 3\end{array}\right){\mathrm{tan}}^{3}x+\dots }{1-\left(\begin{array}{c}n\\ 2\end{array}\right){\mathrm{tan}}^{2}x+\left(\begin{array}{c}n\\ 4\end{array}\right){\mathrm{tan}}^{4}x+\dots }$
$=\frac{\left(\begin{array}{c}n\\ 1\end{array}\right){\mathrm{cot}}^{n-1}x-\left(\begin{array}{c}n\\ 3\end{array}\right){\mathrm{cot}}^{n-3}x+\dots }{{\mathrm{cot}}^{n}-\left(\begin{array}{c}n\\ 2\end{array}\right){\mathrm{cot}}^{n-2}x+\left(\begin{array}{c}n\\ 4\end{array}\right){\mathrm{cot}}^{n-4}+\dots }$
Now ${\left[\frac{k\pi }{n}\right]}_{k=1}^{n-1}$ are the roots of the equation: $\mathrm{tan}nx=0$
Thus, $\left(\begin{array}{c}n\\ 1\end{array}\right){\mathrm{cot}}^{n-1}x-\left(\begin{array}{c}n\\ 3\end{array}\right){\mathrm{cot}}^{n-3}x+\dots =0$ whenever, $x=\frac{k\pi }{n}$ for $1\le k\le n-1$
Thus, using Vieta's formula we might write:
$\sum _{k=1}^{n-1}{\mathrm{cot}}^{2}\frac{k\pi }{n}={\left(\sum _{k=1}^{n-1}\mathrm{cot}\frac{k\pi }{n}\right)}^{2}-2\sum _{1\le {k}_{1}<{k}_{2}\le n-1}\mathrm{cot}\frac{{k}_{1}\pi }{n}\mathrm{cot}\frac{{k}_{2}\pi }{n}$
$=0+2\frac{\left(\begin{array}{c}n\\ 3\end{array}\right)}{\left(\begin{array}{c}n\\ 1\end{array}\right)}$
$=\frac{\left(n-1\right)\left(n-2\right)}{3}$

Do you have a similar question?

Recalculate according to your conditions!