3.1 Consider \( \sum_{k=0}^{n}\left(3^{s-k}\right)=\frac{1093}{3} \) 3.1.1 \( \quad \) Write down the first three terms of the series. 3.1.2 \( \begin{array}{ll}\text { 3.1.3 } & \text { Does the series converge? Motivate your answer. }\end{array} \).

The first three terms of the series \( \sum_{k=0}^{n}\left(3^{s-k}\right) \) can be written as follows: when \( k = 0 \), the term is \( 3^s \); when \( k = 1 \), it becomes \( 3^{s-1} \); and when \( k = 2 \), it is \( 3^{s-2} \). Therefore, the first three terms are \( 3^s, 3^{s-1}, \) and \( 3^{s-2} \). To determine if the series converges, we must analyze the behavior of the terms as \( n \) approaches infinity. If \( s > 0 \), the terms grow without bounds since \( 3^{s-k} \) will eventually result in significant positive values as \( k \) increases. Conversely, if \( s < 0 \), the terms shrink and approach zero, suggesting convergence. If \( s = 0 \), the series terms become constant (i.e., 1), leading to divergence. Thus, convergence depends on the value of \( s \)!