1.3.2 Make a conjecture with regard to \( r^{n} \) and \( S_{n} \) as \( n \rightarrow \infty \) CASE 3: \( r=1 \) 1.4.1 What is the sum of the geometric series \( S_{n}=a+a r+a r^{2}+\ldots a r^{n-1} \) if \( r=1 \) ? 1.4.2 Make a conjecture with regard to \( r^{n} \) and \( S_{n} \) as \( n \rightarrow \infty \)

We begin by recalling that for a geometric series with first term a and common ratio r, the partial sum is   Sₙ = a + a·r + a·r² + … + a·rⁿ⁻¹. A. For the general case (not necessarily r = 1), you may observe that:  • If |r| < 1, then as n → ∞ the term rⁿ → 0, and the partial sums Sₙ approach a finite value (namely a/(1 – r)).  • If |r| > 1, then rⁿ grows without bound in absolute value and so do the partial sums Sₙ (they diverge). B. CASE 3: r = 1 1.4.1 When r = 1, each term in the series is simply a. There are n terms, so the sum is   Sₙ = a + a + a + … + a  (n times) = n·a. 1.4.2 As n → ∞:  • Notice that rⁿ = 1ⁿ = 1 for all n. That is, the term rⁿ remains constant (equal to 1) and does not approach 0, unlike the case when |r| < 1.  • The sum Sₙ = n·a grows without bound for a ≠ 0; more specifically, it increases linearly as n increases. Conjecture Summary:  – In general, the behavior of rⁿ as n → ∞ depends on the magnitude of r:   • If |r| < 1, then rⁿ → 0.   • If |r| = 1 (here r = 1), then rⁿ remains 1 for all n.   • If |r| > 1, then rⁿ diverges (grows in magnitude without bound).  – Correspondingly, the partial sum Sₙ of the geometric series approaches a finite limit only when |r| < 1; when r = 1, Sₙ = n·a increases without bound (if a ≠ 0), and similarly for |r| > 1 the sum diverges. Thus, for r = 1 we conclude that while rⁿ is a constant 1 for all n, the sum Sₙ = n·a grows indefinitely as n → ∞.