Proving the given identity For every continuous periodic function $F(\theta)$ with period $2 \pi$ and for $0 < \rho < 1,$ prove that
$$\lim_{\rho \rightarrow 1}\frac{1}{4\pi \rho} \int_{-\pi}^{\pi} F(\theta ) \left( \frac{1-\rho^2}{1-2 \rho \cos n\theta + \rho^2} - \frac{1-\rho^2}{1+2 \rho \cos n\theta + \rho^2} \right)~d\theta=\frac{1}{2n}\sum_{k=1}^{2n}(-1)^k~F \left(\frac{k \pi }{n} \right).$$
Any suggestions of proving this is much appreciated.
 A: Using the generating function of the Chebyshev polynomials with $-1<\rho<1$,
$$ \frac{1-\rho^2}{1-2\rho\cos n\theta+\rho^2}=1+2\sum_{s=0}^\infty \rho^s\cos s\theta$$ the integral can be written as $$ I(\rho)=\frac{1}{\pi}\int_{-\pi}^\pi F(\theta)\sum_{p=0}^\infty\rho^{2p}\cos(n(2p+1)\theta)\,d\theta$$ Inverting the summations,$$I(\rho)=\frac{1}{\pi}\sum_{p=0}^\infty\rho^{2p}\int_{-\pi}^\pi F(\theta)\cos(n(2p+1)\theta)\,d\theta$$
 With the Fourier-cosine coefficients of $F$:$$I(\rho)=2\sum_{p=0}^\infty\rho^{2p}F_{n(2p+1)}$$
As $F$ is continuous and periodic, it can be represented by its Fourier series:$$F(\theta)=\sum_{-\infty}^\infty F_re^{ir\theta}$$ The limit of $I(\rho)$ for $\rho\rightarrow 1$ exists:$$I(1)=2\sum_{p=0}^\infty F_{n(2p+1)}$$ Now, using the Fourier representation, one can form the summation of the rhs of the question:$$\frac{1}{2n}\sum_{k=1}^{2n}(-1)^k\sum_{r=-\infty}^\infty F_re^{ir\theta}=\frac{1}{2n}\sum_{r=-\infty}^\infty F_r\sum_{k=1}^{2n}(-z)^k$$where $z=\exp(\frac{ir\pi}{n})$. Inner summation is zero except when $r=n(2p+1)$. In this case its value is $2n$. As the coefficient are even with respect to $r$, $$rhs=2\sum_{p=0}^\infty F_{n(2p+1)}$$
A: We can prove the identity by elementary methods.   
Since $F(\theta )$ and $\cos n\theta $ are periodic 
\begin{align}
&\int_{-\pi}^{\pi} F(\theta ) \left( \frac{1-\rho^2}{1-2 \rho \cos n\theta + \rho^2} - \frac{1-\rho^2}{1+2 \rho \cos n\theta + \rho^2} \right)~d\theta\\
=&\int_{0}^{2\pi} F(\theta ) \left( \frac{1-\rho^2}{1-2 \rho \cos n\theta + \rho^2} - \frac{1-\rho^2}{1+2 \rho \cos n\theta + \rho^2} \right)~d\theta
\end{align}
holds.
Making the change of variable $\varphi =n\theta $, we have
\begin{align}
I(\rho )&:=\frac{1}{2\pi }\int_{0}^{2\pi} F(\theta )\frac{1-\rho^2}{1-2 \rho \cos n\theta + \rho^2}d\theta \\
&= \frac{1}{2n\pi }\int_{0}^{2n\pi} F\left(\frac{\varphi }{n}\right)\frac{1-\rho^2}{1-2 \rho \cos \varphi  + \rho^2}d\varphi  \\
&=\frac{1}{2n\pi}\sum_{k=1}^n \int_{2(k-1)\pi}^{2k\pi} F\left(\frac{\varphi }{n}\right)\frac{1-\rho^2}{1-2 \rho \cos \varphi  + \rho^2}d\varphi  \\
&=\frac{1}{2n\pi}\sum_{k=1}^n \int_{0}^{2\pi} F\left(\frac{\varphi }{n}+\frac{2(k-1)\pi}{n}\right)\frac{1-\rho^2}{1-2 \rho \cos \varphi  + \rho^2}d\varphi  \\
&=\frac{1}{2\pi}\int_{0}^{2\pi} G(\varphi )\frac{1-\rho^2}{1-2 \rho \cos \varphi  + \rho^2}d\varphi , 
\end{align}
where $G(\varphi )=\frac{1}{n}\sum_{k=1}^n  F\left(\frac{\varphi }{n}+\frac{2(k-1)\pi}{n}\right).$
Since $G(\varphi )$ is continuous and periodic with period $2\pi$, letting $\rho $ to $1$ we have$$
\lim_{\rho \to 1} I(\rho )=G(0)=\frac{1}{n}\sum_{k=1}^n  F\left(\frac{2(k-1)\pi}{n}\right).
$$
Similarly we can prove \begin{align}
\frac{1}{2\pi }\int_{0}^{2\pi} F(\theta )\frac{1-\rho^2}{1+2 \rho \cos n\theta + \rho^2}d\theta
&=\frac{1}{2\pi}\int_{0}^{2\pi} G(\varphi )\frac{1-\rho^2}{1+2 \rho \cos \varphi  + \rho^2}d\varphi\\
&=\frac{1}{2\pi}\int_{0}^{2\pi} G(\varphi )\frac{1-\rho^2}{1-2 \rho \cos (\varphi-\pi)  + \rho^2}d\varphi
\\
&\to G(\pi)=\frac{1}{n}\sum_{k=1}^n  F\left(\frac{(2k-1)\pi}{n}\right).
\end{align}
Thus we have \begin{align}
&\lim_{\rho \to 1}\frac{1}{4\pi \rho} \int_{-\pi}^{\pi} F(\theta ) \left( \frac{1-\rho^2}{1-2 \rho \cos n\theta + \rho^2} - \frac{1-\rho^2}{1+2 \rho \cos n\theta + \rho^2} \right)~d\theta\\
&=\frac{1}{2n}\left(\sum_{k=1}^{n}  F\left(\frac{2(k-1)\pi}{n}\right)-\sum_{k=1}^{n}  F\left(\frac{(2k-1)\pi}{n}\right)\right)\\
&=\frac{1}{2n}\sum_{k=1}^{2n}(-1)^k~F \left(\frac{k \pi }{n} \right).
\end{align}
