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Concept 06 / 07Sequences and Series

Arithmetico-Geometric Series

Series whose terms are products of an AP and a GP, the multiply-by-r-and-subtract method for finite sums, the infinite sum formula when |r| < 1, and series such as sum of n x^n obtained by differentiating a geometric series.

An arithmetico-geometric progression (AGP) has terms that are the product of an AP and a GP: a,(a+d)r,(a+2d)r2,…a, (a + d)r, (a + 2d)r^{2}, \ldots, with general term Tn=[a+(n−1)d]rn−1T_{n} = [a + (n - 1)d]r^{n-1}. Examples are 1+2x+3x2+4x3+⋯1 + 2x + 3x^{2} + 4x^{3} + \cdots and 12+34+58+⋯\frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \cdots. Neither the AP formula nor the GP formula applies directly, but the GP trick almost works. The method: multiply by rr and subtract. Let S=a+(a+d)r+(a+2d)r2+⋯+[a+(n−1)d]rn−1S = a + (a + d)r + (a + 2d)r^{2} + \cdots + [a + (n - 1)d]r^{n-1}. Then rS=ar+(a+d)r2+⋯+[a+(n−2)d]rn−1+[a+(n−1)d]rnrS = ar + (a + d)r^{2} + \cdots + [a + (n - 2)d]r^{n-1} + [a + (n - 1)d]r^{n}. Subtracting and aligning equal powers of rr: (1−r)S=a+d(r+r2+⋯+rn−1)−[a+(n−1)d]rn(1 - r)S = a + d(r + r^{2} + \cdots + r^{n-1}) - [a + (n - 1)d]r^{n}. The AP part has turned into a constant dd, leaving a pure GP in the middle, which we can sum: Sn=a1−r+dr(1−rn−1)(1−r)2−[a+(n−1)d]rn1−rS_{n} = \frac{a}{1 - r} + \frac{dr(1 - r^{n-1})}{(1 - r)^{2}} - \frac{[a + (n - 1)d]r^{n}}{1 - r}, for r≠1r \neq 1. Rather than memorising this, reproduce the subtraction each time; it takes three lines and avoids sign errors. If the AP were of second degree (for example n2rnn^{2}r^{n}), you would multiply by rr and subtract twice. Infinite AGP. If ∣r∣<1|r| < 1, then rn→0r^{n} \to 0 and nrn→0nr^{n} \to 0 (exponential decay beats linear growth). So S∞=a1−r+dr(1−r)2S_{\infty} = \frac{a}{1 - r} + \frac{dr}{(1 - r)^{2}}. For ∣r∣≥1|r| \ge 1 the series diverges. The calculus viewpoint. Differentiating the geometric series ∑n≥0xn=11−x\sum_{n \ge 0}x^{n} = \frac{1}{1 - x} term by term (valid for ∣x∣<1|x| < 1) gives ∑n≥1nxn−1=1(1−x)2\sum_{n \ge 1}nx^{n-1} = \frac{1}{(1 - x)^{2}}, which is the simplest AGP. Differentiating again gives ∑n(n−1)xn−2=2(1−x)3\sum n(n - 1)x^{n-2} = \frac{2}{(1 - x)^{3}}. These are exactly the binomial expansions of (1−x)−2(1 - x)^{-2} and (1−x)−3(1 - x)^{-3}, so AGPs, differentiation of series, and the binomial theorem for negative index are three views of the same objects. From them, ∑nxn=x(1−x)2\sum nx^{n} = \frac{x}{(1 - x)^{2}} and ∑n2xn=x(1+x)(1−x)3\sum n^{2}x^{n} = \frac{x(1 + x)}{(1 - x)^{3}}. Recognising an AGP. Look for a coefficient that grows linearly multiplied by a power that grows geometrically. Sometimes the AP is disguised: 13+49+727+⋯\frac{1}{3} + \frac{4}{9} + \frac{7}{27} + \cdots has numerators 1,4,71, 4, 7 (AP with d=3d = 3) over powers of 33 (GP with r=13r = \frac{1}{3}). Key Takeaways: - AGP: Tn=[a+(n−1)d]rn−1T_{n} = [a + (n - 1)d]r^{n-1}. - Method: S−rSS - rS leaves a+d(r+⋯+rn−1)−Tnra + d(r + \cdots + r^{n-1}) - T_{n}r. - S∞=a1−r+dr(1−r)2S_{\infty} = \frac{a}{1 - r} + \frac{dr}{(1 - r)^{2}} for ∣r∣<1|r| < 1. - ∑n≥1nxn−1=1(1−x)2\sum_{n \ge 1}nx^{n-1} = \frac{1}{(1 - x)^{2}}; ∑nxn=x(1−x)2\sum nx^{n} = \frac{x}{(1 - x)^{2}}; ∑n2xn=x(1+x)(1−x)3\sum n^{2}x^{n} = \frac{x(1 + x)}{(1 - x)^{3}}. - For quadratic coefficients, subtract twice. - Pitfall: misaligning powers of rr when subtracting, especially losing the last term TnrT_{n}r. Brain-Teaser 1: Find the sum to infinity of 1+45+725+10125+⋯1 + \frac{4}{5} + \frac{7}{25} + \frac{10}{125} + \cdots. Solution: Here a=1a = 1, d=3d = 3 and r=15r = \frac{1}{5}. S∞=11−1/5+3×15(1−1/5)2=54+3/516/25=54+1516=3516S_{\infty} = \frac{1}{1 - 1/5} + \frac{3 \times \frac{1}{5}}{(1 - 1/5)^{2}} = \frac{5}{4} + \frac{3/5}{16/25} = \frac{5}{4} + \frac{15}{16} = \frac{35}{16}. Check by the method: S−S5=1+3(15+125+⋯ )=1+3×14=74S - \frac{S}{5} = 1 + 3(\frac{1}{5} + \frac{1}{25} + \cdots) = 1 + 3 \times \frac{1}{4} = \frac{7}{4}, so S=74×54=3516S = \frac{7}{4} \times \frac{5}{4} = \frac{35}{16}. Brain-Teaser 2: Find Sn=1+2⋅2+3⋅22+⋯+n⋅2n−1S_{n} = 1 + 2 \cdot 2 + 3 \cdot 2^{2} + \cdots + n \cdot 2^{n-1}. Solution: 2Sn=2+2⋅22+⋯+(n−1)2n−1+n2n2S_{n} = 2 + 2 \cdot 2^{2} + \cdots + (n - 1)2^{n-1} + n2^{n}. Subtract: Sn−2Sn=1+2+22+⋯+2n−1−n2n=(2n−1)−n2nS_{n} - 2S_{n} = 1 + 2 + 2^{2} + \cdots + 2^{n-1} - n2^{n} = (2^{n} - 1) - n2^{n}. So Sn=(n−1)2n+1S_{n} = (n - 1)2^{n} + 1. Check at n=3n = 3: 1+4+12=17=2×8+11 + 4 + 12 = 17 = 2 \times 8 + 1. Here r=2>1r = 2 > 1, so there is no infinite sum, but the finite method works for any r≠1r \neq 1. Brain-Teaser 3: Evaluate ∑n=1∞n22n\sum_{n=1}^{\infty}\frac{n^{2}}{2^{n}}. Solution: Use ∑n2xn=x(1+x)(1−x)3\sum n^{2}x^{n} = \frac{x(1 + x)}{(1 - x)^{3}} with x=12x = \frac{1}{2}: 12⋅3218=34×8=6\frac{\frac{1}{2} \cdot \frac{3}{2}}{\frac{1}{8}} = \frac{3}{4} \times 8 = 6. To derive the formula without calculus, let S=∑n2xnS = \sum n^{2}x^{n}, compute S−xS=∑(2n−1)xnS - xS = \sum(2n - 1)x^{n} (an AGP), sum that by the method, and divide by 1−x1 - x. The answer 66 is a classic, and the check 12+1+98+1+2532+⋯\frac{1}{2} + 1 + \frac{9}{8} + 1 + \frac{25}{32} + \cdots approaches 66 quickly.

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Arithmetico-Geometric Series — DPP

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