Let us consider the quadrature $Q_n(f)$ obtained by Lagrange interpolation to aproximate the integral $\int_{-1}^{1}f(x)dx$, using as nodes the $n+1$ roots of the Chebyshev's polynomial $T_{n+1}(x)$. I have to show that given $f(x)=\dfrac{1}{x+2}$ the we have that $\lim_{n\to\infty}\left(\int_{-1}^{1}f(x)dx - Q_n(f)\right) = 0$.
The first step I took was solving for the roots of $T_{n+1}(x) = \cos((n+1)\arccos (x))$, which are $x_i=cos\left(\dfrac{2i+1}{2n+2}\pi\right)$, for $i=0,\dots,n$.
After, if we work with the proposed limit, as $\int_{-1}^{1}f(x)dx$ does not depend on $n$, it turns out that we want to show: $$\lim_{n\to\infty}Q_n(f) = \int_{-1}^{1}f(x)dx = \int_{-1}^{1}\dfrac{1}{x+2}dx = \ln(3) $$
Now if we consider the Mclaurin expansion of the function $\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{x^n}{n}$, then for $x=2$ we have that: $$\ln(3)=\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{2^n}{n}$$
Also, the quadrature by Lagrange interpolation has the form: $$Q_n(f)=\sum_{i=0}^{n}\alpha_i f(x_i) =\sum_{i=0}^{n}\alpha_i \dfrac{1}{x_i+2}=\sum_{i=0}^{n}\alpha_i \dfrac{1}{\cos\left(\dfrac{2i+1}{2n+2}\pi\right)+2}$$
Therefore, I thought it was natural trying to show that the the former series was actually equivalent to the $\ln(3)$ expansion I mentioned above. Hence, I considered calculating the weights $\alpha_i$ by solving the linear system:
$$ \left( \begin{array}{} 1&1&\cdots&1 \\ \cos\left(\frac{1}{2n+2}\pi\right)&\cos\left(\frac{3}{2n+2}\pi\right)&\cdots&\cos\left(\frac{2n+1}{2n+2}\pi\right) \\ \vdots&\vdots&\ddots&\vdots \\ \cos^n\left(\frac{1}{2n+2}\pi\right)&\cos^n\left(\frac{3}{2n+2}\pi\right)&\cdots&\cos^n\left(\frac{2n+1}{2n+2}\pi\right) \end{array} \right) \left( \begin{array}{} \alpha_0\\ \alpha_1\\ \vdots\\ \alpha_n \end{array} \right) = \left( \begin{array}{} 2 \\ 0 \\ \vdots \\ \frac{1-(-1)^{n+1}}{n+1} \end{array} \right) $$
However, I am a little stuck here. I know the matrix of coefficients is Vandermonde-like, so by Cramer's rule: $$\alpha_i = \dfrac{\det(A_i)}{\det(A)} = \dfrac{\det(A_i)}{\prod_{0\leq i<j\leq n} (x_j-x_i)}=\dfrac{\det(A_i)}{\prod_{0\leq i<j\leq n} \left[\cos\left(\dfrac{2j+1}{2n+2}\pi\right)-\cos\left(\dfrac{2i+1}{2n+2}\pi\right)\right]} $$
But I don't know how to proceed with the $\det(A_i)$ terms.