\( \sum _ { k = 1 } ^ { \infty } \frac { 2 k - 1 } { 2 ^ { k } } , \quad S _ { n } = 3 - \frac { 2 n + 3 } { 2 ^ { n } } \)

To tackle the infinite series \( \sum_{k=1}^{\infty} \frac{2k - 1}{2^k} \), we can use the formula for the sum of a series of the form \( \sum_{k=1}^{\infty} kx^k \), which is \( \frac{x}{(1-x)^2} \). For \( x = \frac{1}{2} \), this gives us a way to find \( S = \sum_{k=1}^{\infty} k\left(\frac{1}{2}\right)^k = \frac{\frac{1}{2}}{(1 - \frac{1}{2})^2} = 2 \). Now, looking at \( \sum_{k=1}^{\infty} \frac{1}{2^k} \), we recognize it as a geometric series which sums to 1. Combining these results, we calculate: \[ \sum_{k=1}^{\infty} \frac{2k}{2^k} - \sum_{k=1}^{\infty} \frac{1}{2^k} = 2(2) - 1 = 4 - 1 = 3. \] Thus, the infinite series sums to 3. In reference to the equation \( S_n = 3 - \frac{2n + 3}{2^n} \), this establishes a finite sum for the series up to \( n \). As \( n \) approaches infinity, the term \( \frac{2n + 3}{2^n} \) converges to zero, affirming that \( S_n \) indeed approaches 3, corroborating our previous calculation of the infinite series.