Bound for the sum of a finite sequence

Question

Consider ${\bf c} = (a,b) \in \mathbb{R}^2$ with $0< \|{\bf c}\| < 1.$ Let $n \in \mathbb{N} $ and define \begin{align*} F_{n}(k) & := \frac{ [a + x_{n}(k)]^2}{ [a + x_{n}(k)]^2 + [b + y_{n}(k)]^2} \end{align*} where $\omega := 2\pi/n,$ $x_{n}(k) := \cos(2 k \pi/n)$ and $y_{n}(k) := \sin(2 k \pi/n).$ The question is to find an upper bound $h = h(n,{\bf c})$ on the sum $S_n$ such that $$ S_n \leq h(n,{\bf c}),$$ where $$S_n =\left( \frac{2 \pi}{n}\right) \sum_{k=0}^{n-1} F_{n}(k).$$ Is there any result that can help to find such a bound ?

Answer 1

One idea is the following:

Let $z_k=x_n(k)+iy_n(k)=e^{ik\omega}$ and $c=a+ib=re^{i\theta}$ with $0<r<1$. Then, $F_n(k)=\cos^2\theta_k,\ 0\le k\le n-1$ where $\theta_k=\angle (z_k+c)$ so that $F_n(k)=\pi+\frac{\pi}{n}\sum_{k=0}^{n-1}\cos 2\theta_k$.

I think this can be helpful in finding some bounds.

Edit: One can note that $\tan\theta_k=\frac{r\sin \theta+\sin k\omega}{r\cos \theta+\cos k\omega}$. Then, $$\cos2\theta_k=\frac{1-\tan^2{\theta_k}}{1+\tan^2\theta_k}=\frac{r^2\cos 2\theta+2r\cos(\theta+k\omega)+\cos2k\omega}{r^2+1+2r\cos(\theta-k\omega)}$$ So, $$S_n=\pi+\frac{\pi}{n}\sum_{k=0}^{n-1}\frac{r^2\cos2\theta+2r\cos(\theta+k\omega)+\cos2k\omega}{r^2+1+2r\cos(\theta-k\omega)}=\pi+\frac{\pi}{n}(A\cdot r^2\cos2\theta+B+C)$$

Now, we can write $$A=\frac{1}{r^2+1+2r\cos(\theta-k\omega)}=\frac{1}{r^2+1}\left(1-a\cos(\theta-k\omega)+a^2\cos^2(\theta-k\omega)+\cdots\right),\ a=\frac{2r}{r^2+1}\\=\frac{1}{r^2+1}\sum_{l=0}^{\infty}(-1)^la^l\cos^l(\theta-k\omega)\\=\frac{1}{r^2+1}\sum_{l=0}^{\infty}\sum_{m=0}^{\lfloor l/2\rfloor}\frac{a^l}{2^l}\binom{l}{2m}\cos((l-2m)(\theta-k\omega))$$ So that $$A=\frac{1}{r^2+1}\sum_{l\ge 0}\left(\frac{a^2}{4}\right)^{l}\binom{2l}{l}\\=\frac{1}{r^2+1}\frac{d}{dx}\left(xC(x)\right)\mid_{x=a^2/4}$$ where $C(x)$ is the generating function of Catalan numbers. Hence $\displaystyle xC(x)=(1-\sqrt{1-4x})/2\implies A=\frac{1}{(r^2+1)\sqrt{1-a^2}}=\frac{1}{1-r^2}$ With similar manipulation one can attempt to find exact expressions for $B,C$ and can find a suitable upper bound of $S_n$, though I guess those exact expressions will not be easy to handle.