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CASE 2: \( r<-1 \) \( \begin{array}{ll}\text { 1.3.1 } & \text { If } r=-1,001 \text {, determine the value of } r^{\prime \prime} \text { if } \\ \text { (a) } n=9999 \\ \text { (b) } & n=10000 \\ \text { 1.3.2 } & \text { Make a conjecture with regard to } r^{\prime \prime} \text { and } S_{n} \text { as } n \rightarrow \infty\end{array} \)

Ask by Hills Park. in South Africa
Mar 10,2025

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For \( r = -1.001 \): - When \( n = 9999 \), \( r^{\prime\prime} \approx -22,000 \). - When \( n = 10,000 \), \( r^{\prime\prime} \approx 22,000 \). As \( n \) becomes very large, \( r^{\prime\prime} \) grows without bound and alternates in sign, while the partial sum \( S_n \) does not approach a finite limit.

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Let’s break down the scenario when \( r = -1.001 \). For part (a), with \( n = 9999 \), we can derive the second term \( r^{\prime\prime} \). Since \( |r| > 1 \), the series diverges, which means that \( S_n \) grows without bound. Thus, \( r^{\prime\prime} \) tends to infinity as \( n \) approaches 9999. For part (b), when \( n = 10000 \), the same conclusion applies: \( r^{\prime\prime} \) remains undefined in the context of convergence and will also trend towards infinity. As for the conjecture regarding \( r^{\prime\prime} \) and \( S_n \) as \( n \rightarrow \infty \), it seems intuitive to predict that since the geometric series diverges, \( r^{\prime\prime} \) will continue to trend towards infinity while \( S_n \) expresses the same unbounded behavior. The relationship becomes clear as the negative growth of terms accentuates the divergence.

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