Summing the trigonometric series $\sin(a)+\frac{1}{2}\sin(2a)+\frac{1}{2^2}\sin(3a)+\frac{1}{2^3}\sin(4a)....$ This is a problem of Loney's "Plane trigonometry, part 2"
$$\sin(a)+\frac{1}{2}\sin(2a)+\frac{1}{2^2}\sin(3a)+\frac{1}{2^3}\sin(4a)...=\sum_{n=0}^{+\infty}\frac{1}{2^n}\sin[(n+1)a]$$
Here is what I've got so far:
Let $S=\sin(a)+\frac{1}{2}\sin(2a)+\frac{1}{2^2}\sin(3a)+\frac{1}{2^3}\sin(4a)_....\tag{1}$
$C=1+\cos(a)+\frac{1}{2}\cos(2a)+\frac{1}{2^2}\cos(3a)+\frac{1}{2^3}\cos(4a)...\tag{2}$
$C+iS=1+[\cos(a)+i\sin(a)]+\frac{1}{2}[\cos(2a)+i\sin(2a)]+\frac{1}{2^2}[\cos(3a)+i\sin(3a)]+...\tag{3}$
$=1+e^{ia}+\frac{1}{2}(e^{ia})^2+\frac{1}{2^2}(e^{ia})^3+...\tag{4}$
I am stuck here. I am trying to find the general formula for $(4)$, $(4)$ resembles a geometric series, but not quite. I have looked into $-\log(1-x)$ with $x=e^{ia}$, but this doesn't work.
If I let $e^{ia}=x$, then I have a power series
$$1+x+\frac{1}{2}x^2+\frac{1}{2^2}x^3+...$$
But I can't come up with the general formula.
The answer at the back of the book says the sum of this series is $\frac{4\sin(a)}{5-4\cos(a)}$.
My guess with this problem is that you will arrive at a formula that is the sum of $C$ and $S$, then by equating the real and imaginary part, you will be able to arrive at the sum of this series.
 A: You're almost there, but you shouldn't add $1$ in the cosine summation. You can rather normalize the power of $1/2$:
\begin{align}
S&=\dfrac{1}{2}\sin(a)+\dfrac{1}{2^2}\sin(2a)+\dots+\dfrac{1}{2^n}\sin(na)+\dotsb
\\
C&=\dfrac{1}{2}\cos(a)+\dfrac{1}{2^2}\cos(2a)+\dots+\dfrac{1}{2^n}\cos(na)+\dotsb
\end{align}
Then your series is $2S$. You have
$$
C+iS=\sum_{n\ge1}\Bigl(\frac{e^{ia}}{2}\Bigr)^{\!n}=\frac{e^{ia}}{2}\frac{1}{1-e^{ia}/2}=\frac{e^{ia}}{2-e^{ia}}
$$
You can now rationalize, so the sum is
$$
\frac{e^{ia}(2-e^{-ia})}{(2-e^{ia})(2-e^{-ia})}
$$
The denominator is $5-4\cos a$, while the numerator is
$$
2e^{ia}-1=2\cos a+2i\sin a-1
$$
so twice the imaginary part is
$$
\frac{4\sin a}{5-4\cos a}
$$
A: $$\sum_{k=0}^\infty z^k = \frac{1}{1-z}, \quad |z| < 1$$ implies with the choice $z = e^{ia}/2$ which does have magnitude less than $1$, that $$\sum_{k=0}^\infty \left(\frac{e^{ia}}{2}\right)^k = \frac{1}{1-e^{ia}/2}.$$  Now split this into real and imaginary components; i.e. $$\frac{2}{2 - e^{ia}} = \frac{2}{(2 - \cos a) - i \sin a} = \frac{2((2 - \cos a) + i \sin a)}{(2 - \cos a)^2 + \sin^2 a},$$ and the rest is simple.  The missing factor of $2$ returns when you adjust the summation accordingly.
