1.3.2 Make a conjecture with regard to \( r^{n} \) and \( S_{n} \) as \( n \rightarrow \infty \) (2) 1.4 CASE 3: \( r=1 \) 1.4.1 What is the sum of the geometric series \[ S_{n}=a+a r+a r^{2}+\ldots a r^{n-1} \text { if } r=1 \text { ? } \] 1.4.2 Make a conjecture with regard to \( r^{n} \) and \( S_{n} \) as \( n \rightarrow \infty \) (2) 1.5 CASE 4: \( r=-1 \) 1.5.1 What is the sum of the geometric series \[ S_{n}=a+a r+a r^{2}+\ldots a r^{n-1} \text { if } r=-1 ? \] 1.5.2 Do the sums above approach some finite particular number as \( n \rightarrow \infty \) i.e. is the sequence divergent or convergent? 1.6 CASE 5: \( -1

Ask by Chandler Page. in South Africa

Feb 27,2025

Below is one sample solution that addresses each part of the question. You may write your answers on an A4 paper using the “provided grid” and following the cutting/folding instructions. Notice that some parts ask for a written “conjecture” – your answer may be phrased in words with a brief explanation. ────────────────────────────── 1.3.2 Conjecture (Geometric Series for arbitrary r) • Conjecture for rⁿ as n → ∞:  – If |r| < 1, then rⁿ becomes smaller and approaches 0.  – If |r| > 1, then rⁿ grows without bound (tends to infinity or minus infinity depending on the sign).  – If r = 1, then rⁿ always equals 1.  – If r = –1, then rⁿ oscillates between 1 and –1. • Conjecture for Sₙ, the sum of the first n terms:  – When |r| < 1, the finite sum Sₙ = a(1 – rⁿ)/(1 – r) approaches a/(1 – r) as n → ∞ (because rⁿ → 0).  – When |r| > 1 or r = –1, the sum does not settle toward a fixed finite number, so we say the series is divergent (except in the trivial case a = 0).  – When r = 1 the sum Sₙ = a + a + … + a = n · a, so if a ≠ 0 it grows without bound. ────────────────────────────── 1.4 CASE 3: r = 1 1.4.1 Sum of the series when r = 1 The series is   Sₙ = a + a·1 + a·1² + ⋯ + a·1^(n–1). Since 1 raised to any power is 1, we have:   Sₙ = a + a + a + ⋯ + a   (n terms), so   Sₙ = n · a. 1.4.2 Conjecture for r = 1 as n → ∞ Since r = 1, rⁿ = 1 for every n. Also, Sₙ = n · a:  – If a ≠ 0, then as n increases, n · a becomes arbitrarily large (or arbitrarily negative if a < 0).  – Thus, the sum does not approach a finite number but diverges.  – In summary, when r = 1 with a ≠ 0, the series diverges; the n‑th term stays constant while the sum increases without bound. ────────────────────────────── 1.5 CASE 4: r = –1 1.5.1 Sum of the series when r = –1 The series is   Sₙ = a + a(–1) + a(–1)² + ⋯ + a(–1)^(n–1). Notice the pattern:  – For n = 1, S₁ = a.  – For n = 2, S₂ = a + (–a) = 0.  – For n = 3, S₃ = a – a + a = a.  – For n = 4, S₄ = a – a + a – a = 0. Thus, the sum alternates between a (when n is odd) and 0 (when n is even). 1.5.2 Convergence or divergence for r = –1? Because Sₙ oscillates between two values (a and 0) and does not approach one single number as n → ∞, the sequence of partial sums does not converge. Therefore, the geometric series (for nonzero a) is divergent in this case. ────────────────────────────── 1.6 CASE 5: –1 < r < 1 1.6.1 Write THREE possible values of r Examples of r in the interval (–1, 1) include:  • r = 0.5  • r = –0.75  • r = 0.2 1.6.2 Step 1 (Paper cut): • Begin with one A4 paper. • Cut the A4 paper along its longest side into two equal rectangles. • Each resulting rectangle is defined to have an area of 16 unit².   (Thus, the whole A4 paper has an area of 32 unit².) 1.6.3 Step 2 (Paper cut): • Place one of the two 16 unit² rectangles on the desktop. • Take the other 16 unit² rectangle and cut it along its longest side into two equal rectangles. • Each of these two new rectangles has an area of 16/2 = 8 unit². 1.6.4 Step 3 (Paper cut): • Now take one of the 8 unit² rectangles and place it on the desktop. • Cut the remaining 8 unit² rectangle along its longest side into two equal parts. • Each of these pieces has an area of 8/2 = 4 unit². 1.6.5 Step 4 (Continue the process): • Continue the procedure: at each step, set aside one piece and cut the last piece you hold (which always has been the “other half”) into two equal rectangles. • Repeat this until the paper becomes too small to continue the folding/cutting process. 1.6.6 Step 5 (List the areas): The procedure shows that:  – The first piece placed on the desktop has an area equal to ½ of the A4 paper, that is, 16 unit².  – The second piece placed (from the first cut of the other half) has an area of ¼ of the A4 paper, i.e. 8 unit².  – The next three pieces will have areas that are half of the previous piece each time. Thus:  • Third piece: 1/8 of the A4, which is 16/4 = 4 unit².  • Fourth piece: 1/16 of the A4, which is 16/8 = 2 unit².  • Fifth piece: 1/32 of the A4, which is 16/16 = 1 unit². (Alternatively, you can express these as fractions of the A4 area: 1/2, 1/4, 1/8, 1/16, 1/32.) 1.6.7 Explanation: Why these areas form a geometric sequence Each time you cut the piece of paper, the area of the piece you produce is exactly one-half the area of the piece that was just cut. In other words, if you denote the area of the first piece by A (with A = 1/2 of A4), each subsequent area is obtained by multiplying by the constant ratio r = 1/2. Therefore, the areas form a geometric sequence where:   A, A·(1/2), A·(1/2)², A·(1/2)³, etc. Because the ratio between successive terms is equal throughout (namely, 1/2), this is the definition of a geometric sequence. ────────────────────────────── Additional Notes • In summary, these examples illustrate the behavior of a geometric series for different choices of r:  – |r| < 1 leads to convergence of the series to a/(1–r), and the individual term rⁿ → 0.  – r = 1 gives a constantly repeated term (thus a linear sum in n).  – r = –1 produces oscillation.  – |r| > 1 (though not shown in the paper-cut activity) results in divergence. You may now record your concise answers on the provided A4 grid.