To find the product \( x_1 x_2 x_3 \ldots \), where \( x_n = \cos \frac{\pi}{3^n} + i \sin \frac{\pi}{3^n} \), we recognize that each \( x_n \) is a complex number in exponential form: \( x_n = e^{i\frac{\pi}{3^n}} \).
The product of these numbers can be expressed using properties of exponents:
\(x_1 x_2 x_3 \ldots = \prod_{n=1}^{\infty} e^{i\frac{\pi}{3^n}} = e^{i \sum_{n=1}^{\infty}\frac{\pi}{3^n}}\)
To evaluate this, we note that the series inside the exponent is a geometric series. The sum of an infinite geometric series is given by:
\(\sum_{n=1}^{\infty} r^n = \frac{r}{1-r}\)
Here, \( r = \frac{1}{3} \), so the sum becomes:
\(\sum_{n=1}^{\infty} \frac{\pi}{3^n} = \frac{\pi}{3} \cdot \frac{1}{1 - \frac{1}{3}} = \frac{\pi}{3} \cdot \frac{3}{2} = \frac{\pi}{2}\)
Thus, the product becomes:
\(x_1 x_2 x_3 \ldots = e^{i\frac{\pi}{2}}\)
From Euler's formula, we know:
\(e^{i\frac{\pi}{2}} = \cos \frac{\pi}{2} + i\sin \frac{\pi}{2} = 0 + i \cdot 1 = i\)
Therefore, the product \( x_1 x_2 x_3 \ldots \) is equal to \( i \).
The correct answer is \( i \), which corresponds to option 3.