<s>[INST] Fourier series of $\cos\mu x$ Find the Fourier series for the $2\pi$-periodic function $F(x)=\cos\mu x$, $x\in[-\pi,\pi]$ and where  $\mu \notin \mathbb Z$. Simplify the result for $\mu=\frac{1}{4}$. Hence show that 
$$\pi\sqrt{2}=A-\sum^\infty_{n=1}\frac{(-1)^n}{Bn^2-\frac{1}{8}},$$
where $A$ and $B$are integers to be determined.

$a_n,a_0=0$
\begin{align}
b_n & =\frac{2}{\pi}\int^\pi_0 \cos\mu x\sin nx\, dx \\[10pt]
& =\frac{1}{\pi} \left[-\frac{\cos{(n-\mu)x}}{n-\mu}-\frac{\cos{(n+\mu)x}}{n+\mu}\right] \\[10pt]
& =\frac{1}{\pi} \left[\frac{(-1)^n(-2n)}{n^2-\mu^2}+\frac{2n}{n^2-\mu^2}\right] \\[10pt]
& =\frac{2n}{\pi(n^2-\mu^2)}[-(-1)^n+1]
\end{align}
$$-(-1)^n+1=2 \text{ if } n=2k+1$$
$$b_n=\frac{4(2k+1)}{\pi[(2k+1)^2-\mu^2]}$$
Putting $\mu=\frac{1}{4}$,
$$\cos{\frac{1}{4}x}=\frac{4}{\pi}\sum^{\infty}_{k=1} \frac{2k+1}{(2k+1)^2-\frac{1}{16}} \cdot \sin{nx}$$
This does not look like what I have to prove.
 [/INST] Hint: Once you get the correct Fourier cosine series
$$\cos(\mu x)=\frac{\sin(\mu \pi)}{\mu \pi}+\frac{2\mu\sin(\mu\pi)}{\pi}\sum_{n\geq 1}\frac{(-1)^n}{\mu^2-n^2}\cos(nx)\tag{1}$$
by setting $\mu=\frac{1}{4}$ you get
$$\cos\frac{x}{4}=\frac{2\sqrt{2}}{\pi}+\frac{\sqrt{2}}{4\pi}\sum_{n\geq 1}\frac{(-1)^n}{\frac{1}{16}-n^2}\cos(nx)\tag{2}$$
and by evaluating $(2)$ at $x=0$
$$ 1 = \frac{2\sqrt{2}}{\pi}+\frac{\sqrt{2}}{4\pi}\sum_{n\geq 1}\frac{(-1)^n}{\frac{1}{16}-n^2}\tag{3} $$
so:
$$\boxed{ \pi\sqrt{2}=\color{red}{4}-\sum_{n\geq 1}\frac{(-1)^n}{\color{red}{2}n^2-\frac{1}{8}}}\tag{4}$$
</s>