Compute $\sum^{2024}_{k=1} \frac{2023-2022 \cos \left(\frac{\pi(2k-1)}{2024} \right)}{2021-2020 \cos \left(\frac{\pi(2k-1)}{2024} \right)}$

Question

Question: Compute $$\sum^{2024}_{k=1} \frac{2023-2022 \cos \left(\frac{\pi(2k-1)}{2024} \right)}{2021-2020 \cos \left(\frac{\pi(2k-1)}{2024} \right)}$$

I began by rearranging the sum as follows:

$$\sum^{2024}_{k=1} \frac{2023-2022 \cos \left(\frac{\pi(2k-1)}{2024} \right)}{2021-2020 \cos \left(\frac{\pi(2k-1)}{2024} \right)}$$

$$=\sum^{2024}_{k=1} \left(1+\frac{2-2\cos \left(\frac{\pi(2k-1)}{2024} \right)}{2021-2020 \cos \left(\frac{\pi(2k-1)}{2024} \right)} \right)$$

$$=2024+\frac{1}{1010} \sum^{2024}_{k=1} \frac{2020-2020\cos \left(\frac{\pi(2k-1)}{2024} \right)}{2021-2020 \cos \left(\frac{\pi(2k-1)}{2024} \right)}$$

$$=2024+\frac{1}{1010} \sum^{2024}_{k=1} \left(1-\frac{1}{2021-2020 \cos \left(\frac{\pi(2k-1)}{2024} \right)} \right)$$

$$=\frac{1011 \cdot 1012}{505}-\frac{1}{1010} \sum^{2024}_{k=1} \frac{1}{2021-2020 \cos \left(\frac{\pi(2k-1)}{2024} \right)}$$

$$=\frac{1011 \cdot 1012}{505}-\frac{1}{2 \cdot 1010^2} \sum^{2024}_{k=1} \frac{1}{\frac{2021}{2020}-\cos \left(\frac{\pi(2k-1)}{2024} \right)}$$

So, I then turned my attention to trying to find $\sum^{2024}_{k=1} \frac{1}{\frac{2021}{2020}-\cos \left(\frac{\pi(2k-1)}{2024} \right)}$.

1. Complex numbers

I considered $z_k=e^{\frac{i \pi (2k-1)}{2024}}$ then observed $(z_k)^{2024}+1=0$. Then, I (mistakenly) believed that it was sufficient to find the real part of

$$\sum^{2024}_{k=1} \frac{1}{\frac{2021}{2020}-z_k}$$

For chuckles, I'll give a quick sketch of how I calculated this. Define $b_k$ to be the summand, and solve for $z_k$ in terms of $b_k$. Substitute this into the equation $(z_k)^{2024}+1=0$ and use Vieta's to find $\sum^{2024}_{k=1} b_k$.

2. Considering the derivative

Define $$f(x)=\prod^{2024}_{k=1} \left(x-\cos \left(\frac{(2k-1)\pi}{2024} \right) \right)$$

Then, we know that

$$\sum^{2024}_{k=1} \frac{1}{x-\cos \left(\frac{(2k-1)\pi}{2024} \right)}=\frac{f'\left(\frac{2021}{2020} \right)}{f \left(\frac{2021}{2020} \right)}$$

For the denominator, we can represent it in terms of Chebyshev polynomials as follows: $$f\left(\frac{2021}{2020} \right)=\frac{1}{2^{2022}} \left(T_{1012} \left(\frac{2021}{2020} \right) \right)^2$$

where $T_n$ is the $n$th Chebyshev polynomial of the first kind. It's annoying to calculate, but I think I could use the binomial theorem for this.

However, I have no idea how to compute $f'\left(\frac{2021}{2020} \right)$, so I'm at a dead end here as well.

Answer 1

The computations could be continued with a closed form of Chebyshev polynomials.

An (essentially equivalent) "algebraic" alternative is as follows. $$\bbox[5pt,border:2pt solid black]{\begin{aligned} \sum_{k=1}^{2024}\frac{2023-2022\cos\frac{(2k-1)\pi}{2024}}{2021-2020\cos\frac{(2k-1)\pi}{2024}}&=2024\left(1+\frac1{1010}\right)-\frac1{2\cdot 1010^2}f_{2024}\left(\frac{2021}{2020}\right),\\ f_n(z)&:=\sum_{k=1}^n\frac1{z-\cos[(2k-1)\pi/n]}=g_n\left(z\pm\sqrt{z^2-1}\right),\\ g_n(z)&:=\sum_{k=1}^n\frac{2z}{z^2-2z\cos[(2k-1)\pi/n]+1}=\frac{2nz}{z^2-1}\frac{z^n-1}{z^n+1}, \end{aligned}}$$ and the last equality is obtained from the factorization $$\color{gray}{\left[\prod_{k=1}^n(z-e^{\pm 2k\pi i/n})=z^n-1\implies\right]}\prod_{k=1}^n(z-e^{\pm(2k-1)\pi i/n})=z^n+1.$$

After $(z-e^{it})(z-e^{-it})=z^2-2z\cos t+1$, the product is $$\prod_{k=1}^n\left(z^2-2z\cos[(2k-1)\pi/n]+1\right)=(z^n+1)^2.$$ Taking the logarithmic derivative (w.r.t. $z$), we obtain $$\sum_{k=1}^n\frac{z-\cos[(2k-1)\pi/n]}{z^2-2z\cos[(2k-1)\pi/n]+1}=\frac{nz^{n-1}}{z^n+1}.$$

To get the claimed equality, multiply by $2z$, subtract $n$, and multiply by $\frac{2z}{z^2-1}$.