Evaluate the sum $$S=\sum_{n=1}^{\infty} \frac{x \sin x}{\sqrt{1+n^2} (1+n x^2)}$$
Since $$\frac{1}{n(n+1) (1+x^2)} \le \frac{1}{\sqrt{1+n^2} (1+n x^2)} \le \frac{1}{n^2 x^2}$$ I found out that for $ x \sin x \neq 0$ $$\frac{1}{1+x^2}\le \frac{S}{x \sin x} \le \frac{\pi^2}{6 x^2} $$ and sum is convergent.
I also tried to use converting to integral using Riemann sum, Fourier series and complex numbers, But I was not able to find the sum. I would like to ask for your help on this problem.
You could also use for large values of $n$ $$\frac{1}{\sqrt{n^2+1} \left(n x^2+1\right)}=\frac{1}{n^2 x^2}-\frac{1}{n^3 x^4}+\frac{2-x^4}{2 n^4 x^6}+O\left(\frac{1}{n^5}\right)$$ which could provide tighter bounds (check it)