Sum the series: $ S = \frac{1}{2} \cdot \sin\alpha + \frac{1\cdot 3}{2 \cdot 4} \sin{2\alpha} + \cdots \ \text{ad inf}$ How do I sum the following series?
$$ S = \frac{1}{2} \cdot \sin\alpha + \frac{1\cdot 3}{2 \cdot 4} \sin{2\alpha} + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \sin{3\alpha} + \cdots \ \text{ad inf}$$
 A: The coefficients in your series can be written as,
$$\frac{1}{4^n} {2n \choose n}$$
This allows us to write your series a bit more compactly as,
$\sum_{n = 0}^{\infty} \frac{1}{4^n} {2n \choose n} \sin(n\alpha)$
Consider the following series instead,
$$\sum_{n = 0}^{\infty} \frac{1}{4^n} {2n \choose n} e^{in\alpha} = \sum_{n = 0}^{\infty} \frac{1}{4^n} {2n \choose n} \left( e^{\frac{i\alpha}{2}} \right)^{2n}$$
Now,
$$\frac{1}{\sqrt{1 - x^2}} = \sum_{n = 0}^{\infty} \frac{1}{4^n} {2n \choose n} x^{2n}.$$
So, our earlier series evaluates to,
$$\sum_{n = 0}^{\infty} \frac{1}{4^n} {2n \choose n} \left( e^{\frac{i\alpha}{2}} \right)^{2n} = \frac{1}{\sqrt{1 - e^{i\alpha}}}$$
Therefore,
$$\sum_{n = 0}^{\infty} \frac{1}{4^n} {2n \choose n} \sin(n\alpha) = Im\left( \frac{1}{\sqrt{1 - e^{i\alpha}}} \right)$$
A: Let $$ S = \frac{1}{2}\cdot \sin\alpha + \frac{1\cdot 3}{2 \cdot 4}\cdot \sin{2\alpha} + \frac{1\cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\cdot \sin{3\alpha} + \cdots \ \text{ad inf}$$ and $$ C = 1 + \frac{1}{2}\cdot \cos\alpha + \frac{1\cdot 3}{2 \cdot 4}\cdot \cos{2 \alpha} + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \cdot \cos{3\alpha} + \cdots \ \text{ad inf}$$
From this you have $C+iS = (1-e^{\alpha \cdot i})^{-1/2}$, if $\alpha \neq 2n\pi$. Again you have
\begin{align*}
C+iS &= \{1-\cos\alpha-i\: \sin\alpha\}^{-1/2} \\\ &= \biggl\{2 \sin\frac{\alpha}{2} \biggl(\sin\frac{\alpha}{2} - i\: \cos\frac{\alpha}{2}\biggr)\biggr\}^{-1/2} \\\ &= \biggl\{2 \sin\frac{\alpha}{2}\biggr\}^{-1/2} \ \biggl\{\cos\biggl(\frac{\alpha}{2}-\frac{\pi}{2}\biggr) + i \: \sin\biggl(\frac{\alpha}{2}-\frac{\pi}{2}\biggr)\biggr\}^{-1/2} \\\ &= \biggl\{2 \sin\frac{\alpha}{2}\biggr\}^{-1/2} \: \biggl\{\cos\biggl(\frac{\pi-\alpha}{4}\biggr) + i \: \sin\biggl(\frac{\pi-\alpha}{4}\biggr)\biggr\}
\end{align*}
Now equate real and imaginary parts to get the answer.
