Compute each sum below. Give exact values, not decimal approximations. If the sum does not exist, click on "No sum". \[ \sum_{i=1}^{\infty}\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)^{2}+\left(-\frac{1}{2}\right)^{3}+\ldots=\square \]

To compute the sum \( \sum_{i=1}^{\infty}\left(-\frac{1}{2}\right)^i \), we recognize this as a geometric series. The general form of a geometric series is given by: \[ S = a + ar + ar^2 + ar^3 + \ldots \] where \( a \) is the first term and \( r \) is the common ratio. In this case, the first term \( a \) is \( -\frac{1}{2} \) and the common ratio \( r \) is also \( -\frac{1}{2} \). For a geometric series to converge, the absolute value of the common ratio must be less than 1, that is \( |r| < 1 \). Since \( |-\frac{1}{2}| = \frac{1}{2} < 1 \), the series converges. The formula for the sum of an infinite geometric series is: \[ S = \frac{a}{1 - r} \] Here, we have: \[ S = \frac{-\frac{1}{2}}{1 - \left(-\frac{1}{2}\right)} = \frac{-\frac{1}{2}}{1 + \frac{1}{2}} = \frac{-\frac{1}{2}}{\frac{3}{2}} = -\frac{1}{2} \cdot \frac{2}{3} = -\frac{1}{3} \] Thus, the sum is: \[ \boxed{-\frac{1}{3}} \]