Find the fourier series of the function Find the fourier series of the function
$g(x) = \sum\limits_{n=1}^\infty \frac{sin(nx)}{6^n sin(x)}$ for $x \not= k\pi$, and $g(k\pi) = \lim_{x\to k\pi} g(x)$, $(k \in \mathbb{Z})$
 A: $g(x)=\sum\limits_{n=1}^\infty \frac{\alpha^n}{sin(x)}(sin(n-2)x cos2x + cos(n-2)x sin2x)$
$ = \sum\limits_{n=1}^\infty \alpha^n\frac{sin(n-2)x}{sin(x)} + 2\sum\limits_{n=1}^\infty \alpha^n cos(n-1)x$
$= -\alpha + \alpha^2\sum\limits_{n=1}^\infty\alpha^n \frac{sin(nx)}{sinx} + 
2\sum\limits_{n=1}^\infty\alpha^n cos(n-1)x$
where we can find $g(x)$ to be:
$g(x) = \frac{\alpha}{\alpha^2-1} + 2\sum\limits_{n=1}^\infty \frac{\alpha^n}{1-\alpha^2}cos(n-1)x$
$ = \frac{\alpha}{1-\alpha^2} + 2\alpha\sum\limits_{n=1}^\infty\frac{\alpha^n}{1-\alpha^2} cos(nx)$
now in our case $\alpha = \frac{1}{6}$
A: The problem is stated to be
\begin{align}
g(x) = \sum\limits_{n=1}^\infty \frac{sin(nx)}{6^n sin(x)} \mbox{  for  } x \not= k\pi, \mbox{   and  }  g(k\pi) = \lim_{x\to k\pi} g(x), \hspace{5mm} k \in \mathbb{Z}
\end{align}
Consider the following.
\begin{align}
g(x,t) &= \sum_{n=1}^{\infty} \frac{\sin(nx) \, t^{n}}{\sin(x)} \\
&= \sum_{n=1}^{\infty} \frac{t^{n} \, ( \cos(2x) \, \sin((n-2)x) + \sin(2x) \, \cos((n-2)x)}{\sin(x)} \\
&= - \cos(2x) \, t + \cos(2x) \, \sum_{n=2}^{\infty} \frac{t^{n} \, \sin((n-2)x)}{\sin(x)} + 2 \, \cos(x) \, \sum_{n=1}^{\infty} t^{n} \, \cos((n-2)x) \\
&= - \cos(2x) \, t + \cos(2x) \, t^{2} \, g(x,t) + 2 \cos^{2}(x) t + 2 \cos(x) \, t \, \sum_{n=0}^{\infty} \cos(nx) \, t^{n} \\
&= t + \cos(2x) \, t^{2} \, g(x,t) + 2 \, t \, \cos^{2}(x) \, \sum_{n=0}^{\infty} \cos(nx) \, t^{n}
\end{align}
This leads to
\begin{align}
g(x,t) = \frac{t}{1 - \cos(2x) \, t^{2}} \left[ 1 + 2 \cos^{2}(x) \, \sum_{n=0}^{\infty} \cos(nx) \, t^{n} \right].
\end{align}
In general
\begin{align}
g(x,t) &= \frac{t}{1 - \cos(2x) \, t^{2}} \left[ 1 + \frac{2 \cos^{2}(x) \, (1 - \cos(x) \, t)}{1 - 2 \cos(x) \, t + t^{2}} \right] \\
&= \frac{t}{1 - \cos(2x) \, t^{2}} \cdot \frac{1 + 2 \cos^{2}(x) - 2 \cos x (1 + \cos^{2} x) \, t}{1 - 2 \cos x \, t + t^{2}}.
\end{align}
For the case $x \to k \pi$ then 
\begin{align}
g(k \pi, t) = \frac{t}{1 - t^{2}} \cdot \frac{3 - 4 (-1)^{k} \, t}{1 - 2 (-1)^{k} \, t + t^{2}}
\end{align}
When $t = 1/6$ this becomes
\begin{align}
g\left(k \pi, \frac{1}{6}\right) = \frac{72}{35} \, \frac{9 - 2(-1)^{k}}{37 - 12 (-1)^{k}}.
\end{align}
