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

Ask by Carrillo Washington. in South Africa
Mar 10,2025

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**1.3.1 Evaluating \( r^n \) for \( r = -1.001 \):** - **(a) For \( n = 9999 \):** The result is negative with magnitude \( (1.001)^{9999} \). - **(b) For \( n = 10000 \):** The result is positive with magnitude \( (1.001)^{10000} \). **1.3.2 Conjecture on \( r^n \) and \( S_n \) as \( n \rightarrow \infty \):** - **For \( r^n \):** Since \( r < -1 \), \( r^n \) oscillates between very large positive and negative values as \( n \) increases. - **For \( S_n \):** The series \( S_n \) does not converge as \( n \) approaches infinity because \( r^n \) does not approach zero. **Conclusion:** For \( r < -1 \), \( r^n \) oscillates with increasing magnitude, and the geometric series \( S_n \) diverges.

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When \( r = -1.001 \), calculating \( r^n \) for large \( n \) shows some interesting behavior. (a) For \( n = 9999 \), \( r^{9999} \) will be a very large negative number because its absolute value is greater than 1, meaning it will grow in magnitude as \( n \) increases. (b) For \( n = 10000 \), \( r^{10000} \) becomes a very large positive number since we’ve raised a negative base to an even power, flipping the sign. As \( n \) approaches infinity, our conjecture for \( r^n \) suggests that the magnitude of \( r^n \) will grow unbounded, alternating between negative and positive based on whether \( n \) is odd or even. Therefore, \( S_n \) (the sum of a geometric series) may not converge but rather oscillate or diverge, reflecting the growing values of \( r^n \).

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