Find the sum of the following infinite series. Enter DNE if the series diverges. \[ \sum_{n=0}^{\infty} \frac{1}{(\sqrt{3})^{n}} \] Sum:

To find the sum of the series \[ \sum_{n=0}^{\infty} \frac{1}{(\sqrt{3})^{n}}, \] we recognize this as a geometric series where the first term \( a = 1 \) (when \( n=0 \)) and the common ratio \( r = \frac{1}{\sqrt{3}} \). For a geometric series to converge, the absolute value of the common ratio must be less than 1, i.e., \( |r| < 1 \). Here, since \( \frac{1}{\sqrt{3}} < 1 \), the series converges. The sum of an infinite geometric series can be calculated using the formula: \[ S = \frac{a}{1 - r}, \] where \( a \) is the first term and \( r \) is the common ratio. Substituting \( a = 1 \) and \( r = \frac{1}{\sqrt{3}} \), we get: \[ S = \frac{1}{1 - \frac{1}{\sqrt{3}}} = \frac{1}{\frac{\sqrt{3}-1}{\sqrt{3}}} = \frac{\sqrt{3}}{\sqrt{3}-1}. \] Multiplying numerator and denominator by \( \sqrt{3}+1 \) to rationalize: \[ S = \frac{\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{\sqrt{3}(\sqrt{3}+1)}{3-1} = \frac{\sqrt{3}(\sqrt{3}+1)}{2} = \frac{3 + \sqrt{3}}{2}. \] Thus, the sum of the series is \[ \frac{3 + \sqrt{3}}{2}. \]