# 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?

$$S = \frac{1}{2} \cdot \sin\alpha + \frac{1\cdot 3}{2 \cdot 4} \sin{2\alpha} + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \sin{3\alpha} + \cdots \ \text{ad inf}$$

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What did you try? You might want to check this. – Did Apr 23 '12 at 9:24
@Didier that link send me to edit your post – leo Apr 23 '12 at 22:49
@leo Sorry. Link. – Did Apr 23 '12 at 23:01

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.

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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, $$\sum_{n = 0}^{\infty} \frac{1}{4^n} {2n \choose n} \sin(n\alpha) = Im\left( \frac{1}{\sqrt{1 - e^{i\alpha}}} \right)$$

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My first post, so it took me some time to figure out the specifics of posting here. I realize that my answer is a duplicate of the earlier post. So, moderator can feel free to delete this post, if needed. – TenaliRaman Apr 23 '12 at 10:12
I found your answer much clearer. Thanks for a detailed answer. – nbubis Apr 23 '12 at 12:25