Find the sum of the infinite series,
`sinalpha+1/2 sin2alpha+1/2^2sin3alpha+1/2^3sin4alpha+.......oo.`

Answer

To find the sum of the infinite series

\[ S = \sin \alpha + \frac{1}{2} \sin 2\alpha + \frac{1}{2^2} \sin 3\alpha + \frac{1}{2^3} \sin 4\alpha + \ldots \]

we can use the properties of complex numbers and geometric series.

### Step 1: Rewrite the series using complex exponentials

We can express the sine function in terms of complex exponentials:

\[ \sin n\alpha = \frac{e^{i n \alpha} - e^{-i n \alpha}}{2i} \]

Thus, we can write the series as:

\[ S = \sum_{n=1}^{\infty} \frac{1}{2^{n-1}} \sin n\alpha = \sum_{n=1}^{\infty} \frac{1}{2^{n-1}} \cdot \frac{e^{i n \alpha} - e^{-i n \alpha}}{2i} \]

### Step 2: Split the series

This can be split into two separate series:

\[ S = \frac{1}{2i} \left( \sum_{n=1}^{\infty} \frac{e^{i n \alpha}}{2^{n-1}} - \sum_{n=1}^{\infty} \frac{e^{-i n \alpha}}{2^{n-1}} \right) \]

### Step 3: Recognize the geometric series

The series

\[ \sum_{n=1}^{\infty} x^n = \frac{x}{1-x} \quad \text{for } |x| < 1 \]

can be applied here. For the first series, let \( x = \frac{e^{i \alpha}}{2} \):

\[ \sum_{n=1}^{\infty} \left(\frac{e^{i \alpha}}{2}\right)^n = \frac{\frac{e^{i \alpha}}{2}}{1 - \frac{e^{i \alpha}}{2}} = \frac{e^{i \alpha}}{2 - e^{i \alpha}} \]

For the second series, let \( x = \frac{e^{-i \alpha}}{2} \):

\[ \sum_{n=1}^{\infty} \left(\frac{e^{-i \alpha}}{2}\right)^n = \frac{\frac{e^{-i \alpha}}{2}}{1 - \frac{e^{-i \alpha}}{2}} = \frac{e^{-i \alpha}}{2 - e^{-i \alpha}} \]

### Step 4: Substitute back into the expression for \( S \)

Now substituting these results back into the expression for \( S \):

\[ S = \frac{1}{2i} \left( \frac{e^{i \alpha}}{2 - e^{i \alpha}} - \frac{e^{-i \alpha}}{2 - e^{-i \alpha}} \right) \]

### Step 5: Simplify the expression

To simplify this, we can find a common denominator:

\[ S = \frac{1}{2i} \cdot \frac{e^{i \alpha}(2 - e^{-i \alpha}) - e^{-i \alpha}(2 - e^{i \alpha})}{(2 - e^{i \alpha})(2 - e^{-i \alpha})} \]

This simplifies to:

\[ S = \frac{1}{2i} \cdot \frac{2(e^{i \alpha} - e^{-i \alpha})}{(2 - e^{i \alpha})(2 - e^{-i \alpha})} \]

### Step 6: Recognize \( e^{i \alpha} - e^{-i \alpha} \)

Using \( e^{i \alpha} - e^{-i \alpha} = 2i \sin \alpha \):

\[ S = \frac{1}{2i} \cdot \frac{2 \cdot 2i \sin \alpha}{(2 - e^{i \alpha})(2 - e^{-i \alpha})} \]

This simplifies to:

\[ S = \frac{2 \sin \alpha}{(2 - e^{i \alpha})(2 - e^{-i \alpha})} \]

### Step 7: Final simplification

The denominator can be simplified further. We have:

\[ (2 - e^{i \alpha})(2 - e^{-i \alpha}) = 4 - 2(e^{i \alpha} + e^{-i \alpha}) + 1 = 5 - 4 \cos \alpha \]

Thus, the final result for the sum of the infinite series is:

\[ S = \frac{2 \sin \alpha}{5 - 4 \cos \alpha} \]