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Let $a<0$ and $b>0$ be real numbers such that $a<-2b$. Let $n>1$ be a positive integer and consider the following partial sum: $$ f(n) = \frac{1}{(n+1)^2}\sum_{i=1}^{n}\sum_{j=1}^{n} (-1)^{i+j}\frac{(1-\cos(2x_i))(1-\cos(2x_j))}{-2a-2b(\cos(x_i)+\cos(x_j))}, \ \ \text{with }\ x_k:=\frac{k\pi}{n+1}. $$

Problem. Numerical simulations seem to suggest that $$\lim_{n\to\infty}\frac{f(n)}{c^{n}}= \infty$$ for small enough $c>0$. Is there a way to prove this claim?

I would like to stress that this is not an homework question, even if it may look so at a first glance. I found this fact in my research, and, after several unsuccessful proof attempts, I decided to post the problem here. If this is a trivial fact (it doesn't look so to me), please feel free to close this OP.

Let $a<0$ and $b>0$ be real numbers such that $a<-2b$. Let $n>1$ be a positive integer and consider the following partial sum: $$ f(n) = \frac{1}{(n+1)^2}\sum_{i=1}^{n}\sum_{j=1}^{n} (-1)^{i+j}\frac{(1-\cos(2x_i))(1-\cos(2x_j))}{-2a-2b(\cos(x_i)+\cos(x_j))}, \ \ \text{with }\ x_k:=\frac{k\pi}{n+1}. $$

Problem. Numerical simulations seem to suggest that $$\lim_{n\to\infty}\frac{f(n)}{c^{n}}= \infty$$ for small enough $c>0$. Is there a way to prove this claim? If so, is it possible to characterize the values of $c$ that satisfy the above condition?

I would like to stress that this is not an homework question, even if it may look so at a first glance. I encountered this fact in my research, and, after several unsuccessful proof attempts, I decided to post the problem here. If this is a trivial fact (it doesn't look so to me), please feel free to close this OP.

Let $a<0$ and $b>0$ be real numbers such that $a<-2b$. Let $n>1$ be a positive integer and consider the following partial sum: $$ f(n) = \frac{1}{(n+1)^2}\sum_{i=1}^{n}\sum_{j=1}^{n} (-1)^{i+j}\frac{(1-\cos(2x_i))(1-\cos(2x_j))}{-2a-2b(\cos(x_i)+\cos(x_j))} $$ where $x_k:=\frac{k\pi}{n+1}$.

Problem. Numerical simulations seem to suggest that $$\lim_{n\to\infty}\frac{f(n)}{c^{n}}= \infty$$ for small enough $c>0$. Is there a way to prove this claim?

I would like to stress that this is not an homework question, even if it may look so at a first glance. I found this fact in my research, and, after several unsuccessful proof attempts, I decided to post the problem here. If this is a trivial fact (it doesn't look so to me), please feel free to close this OP.

Let $a<0$ and $b>0$ be real numbers such that $a<-2b$. Let $n>1$ be a positive integer and consider the following partial sum: $$ f(n) = \frac{1}{(n+1)^2}\sum_{i=1}^{n}\sum_{j=1}^{n} (-1)^{i+j}\frac{(1-\cos(2x_i))(1-\cos(2x_j))}{-2a-2b(\cos(x_i)+\cos(x_j))}, \ \ \text{with }\ x_k:=\frac{k\pi}{n+1}. $$

Problem. Numerical simulations seem to suggest that $$\lim_{n\to\infty}\frac{f(n)}{c^{n}}= \infty$$ for small enough $c>0$. Is there a way to prove this claim?

I would like to stress that this is not an homework question, even if it may look so at a first glance. I found this fact in my research, and, after several unsuccessful proof attempts, I decided to post the problem here. If this is a trivial fact (it doesn't look so to me), please feel free to close this OP.

Let $a<0$ and $b>0$ be real numbers such that $a<-2b$. Let $n>1$ be a positive integer number and consider the following partial sum: $$ f(n) = \frac{1}{(n+1)^2}\sum_{i=1}^{n}\sum_{j=1}^{n} (-1)^{i+j}\frac{(1-\cos(2x_i))(1-\cos(2x_j))}{-2a-2b(\cos(x_i)+\cos(x_j))} $$ where $x_k:=\frac{k\pi}{n+1}$.

Problem. Numerical simulations seem to suggest that $$\lim_{n\to\infty}\frac{f(n)}{c^{n}}= \infty$$ for small enough $c>0$. Is there a way to prove this claim?

I would like to stress that this is not an homework question, even if it may look so at a first glance. I found this fact in my research, and, after several unsuccessful proof attempts, I decided to post the problem here. If this is a trivial fact (it doesn't look so to me), please feel free to close this OP.

Let $a<0$ and $b>0$ be real numbers such that $a<-2b$. Let $n>1$ be a positive integer and consider the following partial sum: $$ f(n) = \frac{1}{(n+1)^2}\sum_{i=1}^{n}\sum_{j=1}^{n} (-1)^{i+j}\frac{(1-\cos(2x_i))(1-\cos(2x_j))}{-2a-2b(\cos(x_i)+\cos(x_j))} $$ where $x_k:=\frac{k\pi}{n+1}$.

Problem. Numerical simulations seem to suggest that $$\lim_{n\to\infty}\frac{f(n)}{c^{n}}= \infty$$ for small enough $c>0$. Is there a way to prove this claim?

I would like to stress that this is not an homework question, even if it may look so at a first glance. I found this fact in my research, and, after several unsuccessful proof attempts, I decided to post the problem here. If this is a trivial fact (it doesn't look so to me), please feel free to close this OP.