Correct computation of Fourier Coefficient in WolframAlpha It's a very straightforward problem. I have to compute the Fourier series of 
$$\frac{3}{5-4\cos(x)}$$
I have done it using residues and I have got the following coefficients: 
$$c_n = 2 ^{-|n|}$$
I think they are correct, but when I checked with Wolfram using the FourierSeries function, the result seems to be wrong. But if I compute the integral directly in Wolfram, the output then is ok. Could it be that the function FourierSeries and FourierCoefficient have a bug? Anyone has used them?
Thanks a lot, because I have consumed a lot of time trying to figure out my mistake... 
For example, using this url.
And this one.
You can see a different result, and I think it should give the same output. 
EDIT: The people from Technical Support in Wolfram have confirmed by mail there was a bug and are trying to fix it. 
 A: There does seem to be an error with WA's Fourier Series calculator for the function $\frac{3}{5-4\cos(x)}$.  See this link. 
We can proceed to develop the Fourier coefficients by noting that
$$\begin{align}
\frac1{2\pi}\int_0^{2\pi}\frac{3}{5-4\cos(x)}e^{-inx}\,dx&=\frac1{2\pi i}\left(\oint_{|z|=1}\frac{z^{-n}}{z-1/2}\,dz-\oint_{|z|=1}\frac{z^{-n}}{z-2}\,dz \right)\tag1
\end{align}$$
For $n\le 0$, the first and second integrals on the right-hand side of $(1)$ have first order poles at $z=1/2$ and $z=2$, respectively.  Hence, we have from the residue theorem
$$\frac1{2\pi}\int_0^{2\pi}\frac{3}{5-4\cos(x)}e^{-inx}\,dx= \left(\frac12\right)^n$$ 
For $n\ge 1$, we can make use of symmetry (evenness and $2\pi$-periodicity) and write
$$\begin{align}
\frac1{2\pi}\int_0^{2\pi}\frac{3}{5-4\cos(x)}e^{-inx}\,dx&=\frac1{2\pi}\int_0^{2\pi}\frac{3}{5-4\cos(x)}e^{-i|n|x}\,dx\\\\
\end{align}$$
from which we immediately find that for $n\ge 1$
$$\frac1{2\pi}\int_0^{2\pi}\frac{3}{5-4\cos(x)}e^{-inx}\,dx= \left(\frac12\right)^{-n}$$
Therefore, we have 
$$\begin{align}
\frac{3}{5-4\cos(x)}&=\sum_{n=-\infty}^\infty \left(\frac12\right)^{|n|} e^{inx}\\\\
&=1 +2\sum_{n=1}^\infty \left(\frac12\right)^n\cos(nx)
\end{align}$$
A: A more direct computation without Fourier integrals uses partial fraction decomposition after inserting $\cos x = \frac12(e^{ix}+e^{-ix})$ and $e^{ix}=z$ and geometric series for the factors
\begin{align}
\frac{3}{5-2(z+z^{-1})}&=-\frac{(4z-2)-(z-2)}{(2z-1)(z-2)}=\frac{1}{2z-1}-\frac2{z-2}\\
&=\frac1{2z}\sum_{k=0}^\infty \left(\frac{1}{2z}\right)^k+\sum_{k=0}^\infty \left(\frac{z}{2}\right)^k\\
&=\frac12+\sum_{k=1}^\infty\frac{z^k+z^{-k}}{2^{k}}\\
&=\frac12+\sum_{k=1}^\infty\frac{\cos(kx)}{2^{k-1}}.
\end{align}
