Let $f(x)= \sum_{n=-\infty}^\infty \frac {e^{inx}}{1+n^2}$ on $[-\pi,\pi]$. Prove $f(x)>0$ for $x\in[-\pi,\pi]$.
This is an review question for my Fourier course. I am not sure how to approach this problem. The series is uniformly convergent, so $f(x)$ is continuous. Also $f(0)>0$. Do I need to draw a contradiction for $f(x)\le 0$ at some $x$?
$$\overline{f(x)}=\sum_{n=-\infty}^\infty\frac{e^{-inx}}{1+n^2}\stackrel{m:=-n}=\sum_{m=\infty}^{-\infty}\frac{e^{imx}}{1+m^2}=f(x)$$
Now, writing $\;e^{inx}=\cos nx+i\sin nx\;$ , we get that (by absolute convergence)
$$\sum_{n=\infty}^\infty\frac{\sin nx}{1+n^2}=\sum_{n=-\infty}^{-1}\frac{\sin nx}{1+n^2}+\sum_{n=1}^\infty\frac{\sin nx}{1+n^2}=$$
$$=-\sum_{n=-\infty}^{-1}\frac{\sin(-nx)}{1+n^2}+\sum_{n=1}^\infty\frac{\sin nx}{1+n^2}=-\sum_{n=1}^\infty\frac{\sin nx}{1+n^2}+\sum_{n=1}^\infty\frac{\sin nx}{1+n^2}=0$$
and, of course, we got another proof of the fact that the function is real.
The problem now is just to evaluate
$$\sum_{n=-\infty}^\infty\frac{\cos nx}{1+n^2}=1+2\sum_{n=1}^\infty\frac{\cos nx}{1+n^2}$$
Positiviness::
$$1+2\sum_{n=1}^\infty\frac{\cos nx}{1+n^2}\ge1+2\sum_{n=1}^\infty\frac{\cos n\pi}{1+n^2}=1+2\sum_{n=1}^\infty\frac{(-1)^n}{1+n^2}>1+2\left(-\frac12\right)=0$$
Using the estimation of the sum of an alternating series