5.3 Consider the following geometric series: \( 4\left(\frac{1-k}{5}\right)+8\left(\frac{1-k}{5}\right)^{2}+16\left(\frac{1-k}{5}\right)^{3} \ldots \) Determine the values of \( k \) for which the series converges

To determine the values of \( k \) for which the given geometric series converges, we first need to identify the common ratio of the series. The series can be expressed as: \[ S = 4\left(\frac{1-k}{5}\right) + 8\left(\frac{1-k}{5}\right)^{2} + 16\left(\frac{1-k}{5}\right)^{3} + \ldots \] We can see that the first term \( a \) is \( 4\left(\frac{1-k}{5}\right) \) and the common ratio \( r \) can be determined from the second term divided by the first term: \[ r = \frac{8\left(\frac{1-k}{5}\right)^{2}}{4\left(\frac{1-k}{5}\right)} = 2\left(\frac{1-k}{5}\right) \] For a geometric series to converge, the absolute value of the common ratio must be less than 1: \[ |r| < 1 \] Substituting the expression for \( r \): \[ \left| 2\left(\frac{1-k}{5}\right) \right| < 1 \] Now, we can simplify this inequality: \[ \frac{2|1-k|}{5} < 1 \] Multiplying both sides by 5: \[ 2|1-k| < 5 \] Dividing both sides by 2: \[ |1-k| < \frac{5}{2} \] This absolute value inequality can be split into two inequalities: \[ -\frac{5}{2} < 1-k < \frac{5}{2} \] Now, we can solve these inequalities for \( k \): 1. From \( 1-k < \frac{5}{2} \): \[ -k < \frac{5}{2} - 1 \] \[ -k < \frac{3}{2} \] \[ k > -\frac{3}{2} \] 2. From \( -\frac{5}{2} < 1-k \): \[ -\frac{5}{2} < 1-k \] \[ k < 1 + \frac{5}{2} \] \[ k < \frac{7}{2} \] Combining these results, we find: \[ -\frac{3}{2} < k < \frac{7}{2} \] Thus, the values of \( k \) for which the series converges are: \[ k \in \left(-\frac{3}{2}, \frac{7}{2}\right) \]