Sum to Infinity | 无穷级数求和

📚 Sum to Infinity | 无穷级数求和

In Edexcel A-Level Mathematics, 'sum to infinity' tics, and understanding when and how to use the formula S∞ = a / (1 − r) is essential for exam success.

在 Edexcel A-Level 数学中，"无穷级数求和"通常指收敛等比级数部分和的极限值。该主题出现在纯数学的数列与级数章节中。理解何时以及如何使用公式 S∞ = a / (1 − r)，对考试成功至关重要。

This article restates the whole topic from first principles, then works through the standard exam question types: direct substitution, finding a and r from given terms, inequality conditions on an unknown, recurring decimals, and applied modelling questions.

本文从基本原理出发完整梳理该主题，然后逐一演练标准考题类型：直接代入、由已知项反求 a 与 r、含未知数的取值范围条件、循环小数化分数，以及实际建模应用题。

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1. What Is an Infinite Series? | 什么是无穷级数

constant ratio, the series is called geometric. For example, 1 + 1/2 + 1/4 + 1/8 + ... is an infinite geometric series with first term a = 1 and common ratio r = 1/2.

无穷级数是一个无穷数列所有项的和，记作 u₁ + u₂ + u₃ + ...。如果每一项都是前一项乘以一个常数比值得到，该级数就称为等比级数。例如，1 + 1/2 + 1/4 + 1/8 + ... 是一个首项 a = 1、公比 r = 1/2 的无穷等比级数。

. + uₙ, and the sum to infinity is the limit of Sₙ as n tends to infinity, provided that this limit exists and is finite.

严格来说，无穷级数是借助部分和来定义的。第 n 个部分和为 Sₙ = u₁ + u₂ + ... + uₙ，而无穷和就是当 n 趋于无穷时 Sₙ 的极限——前提是该极限存在且为有限值。

Not all infinite series have a finite sum. A series can only have a sum to infinity when its terms approach zero sufficiently fast. The series 1 + 2 + 4 + 8 + ... grows without bound, and even 1 + 1/2 + 1/3 + 1/4 + ... (the harmonic series) diverges, so the speed at which terms shrink matters, not merely the fact that they shrink.

并非所有无穷级数都有有限和。只有当级数各项以足够快的速度趋近于零时，它才可能有无穷和。级数 1 + 2 + 4 + 8 + ... 无限增大，而即使是 1 + 1/2 + 1/3 + 1/4 + ...（调和级数）也发散，因此关键在于各项缩小的速度，而不只是"它们在缩小"这一事实。

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2. Geometric Series Recap | 等比级数回顾

given by uₙ = arⁿ⁻¹, and each term is obtained from the previous one by multiplying by r.

等比级数具有 a + ar + ar² + ar³ + ... 的形式，其中 a 是首项，r 是公比。第 n 项为 uₙ = arⁿ⁻¹，每一项都由前一项乘以 r 得到。

The sum of the first n terms is

前 n 项的和为

Sₙ = a(1 − rⁿ) / (1 − r), r ≠ 1

1). Both forms are correct; the first is the one used in Edexcel mark schemes most often.

等价的写法是 Sₙ = a(rⁿ − 1) / (r − 1)，当 r 为负数时较为好用。两种形式都正确；第一种在 Edexcel 评分标准中出现得最多。

The key question for sum to infinity is what happens to rⁿ in that formula when n becomes very large. This single observation drives the whole topic.

求无穷和的关键问题在于：当 n 变得非常大时，公式中的 rⁿ 会怎样变化。正是这一观察支撑起整个主题。

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3. Convergence and Divergence | 收敛与发散

Whether a geometric series has a sum to infinity depends entirely on the value of r. If |r| nfinity, the partial sums settle down to a fixed value, and the series is said to converge. If |r| he partial sums do not settle down, and the series diverges, so it has no sum to infinity.

等比级数是否存在无穷和完全取决于 r 的取值。若 |r| < 1，则当 n 趋于无穷时 rⁿ 趋于 0，部分和稳定于某个固定值，此时称级数收敛。若 |r| ≥ 1，部分和不会稳定下来，级数发散，因此没有无穷和。

The convergence condition can be written in three equivalent ways: |r| < 1, or −1 < r < 1, or "r is strictly between −1 and 1".

收敛条件可以写成三种等价形式：|r| < 1，或 −1 < r < 1，或"r 严格介于 −1 与 1 之间"。

Value of r / r 的取值	Behaviour of rⁿ / rⁿ 的变化	Sum to infinity? / 是否有无穷和
|r| < 1	rⁿ → 0	Converges / 收敛
r = 1	rⁿ = 1 always / 恒为 1	Diverges (unless a = 0) / 发散
r = −1	Oscillates ±1 / 在 ±1 之间振荡	Diverges / 发散
|r| > 1	|rⁿ| → ∞	Diverges / 发散

ember that r may be negative. The series 8 − 4 + 2 − 1 + ... has r = −1/2, which satisfies |r| < 1, so it converges despite the terms alternating in sign.

请记住 r 可以是负数。级数 8 − 4 + 2 − 1 + ... 的公比 r = −1/2，满足 |r| < 1，因此尽管各项正负交替，它仍然收敛。

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4. The Sum to Infinity Formula | 无穷和公式

When |r| < 1, the geometric series converges and its sum to infinity is given by the standard formula.

当 |r| < 1 时，等比级数收敛，其无穷和由标准公式给出。

S∞ = a / (1 − r), valid only when |r| < 1

In this formula a is the first term of the series and r is the common ratio. The formula must never be used without first checking, or stating, that |r| < 1.

公式中 a 是级数首项，r 是公比。使用该公式前必须先检验或说明 |r| < 1，绝不可省略。

(1/2) = 2. Adding up the first ten terms gives 1.998047, which is already very close to 2 — a good numerical check on any answer.

将其用于 1 + 1/2 + 1/4 + 1/8 + ...，得 a = 1、r = 1/2，因此 S∞ = 1 / (1 − 1/2) = 1 / (1/2) = 2。把前十项相加得到 1.998047，已非常接近 2——这是检验答案的一个很好的数值方法。

8 / (1 + 1/2) = 8 / (3/2) = 16/3 ≈ 5.33. Note the sign change: 1 − (−1/2) becomes 1 + 1/2.

对于公比为负的级数，如 8 − 4 + 2 − 1 + ...，有 a = 8、r = −1/2，故 S∞ = 8 / (1 + 1/2) = 8 / (3/2) = 16/3 ≈ 5.33。注意符号变化：1 − (−1/2) 变为 1 + 1/2。

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5. Where the Formula Comes From | 公式的由来

The formula is not something to accept blindly; the derivation is short and is itself an examinable proof. Start from the finite sum formula and let n tend to infinity.

这个公式不应盲目接受；其推导很短，而且本身就是可考的证明。从有限和公式出发，令 n 趋于无穷即可。

Sₙ = a(1 − rⁿ) / (1 − r) = a / (1 − r) − arⁿ / (1 − r)

If |r| as n → ∞, so the second fraction vanishes and Sₙ tends to a / (1 − r). That is exactly S∞.

若 |r| < 1，则当 n → ∞ 时 rⁿ → 0，第二项趋于消失，Sₙ 趋于 a / (1 − r)，这正是 S∞。

If |r| e term arⁿ / (1 − r) does not vanish, so no finite limit exists. This is why the condition |r| ears inside the formula itself: it is not a technicality but the reason the formula works.

若 |r| > 1 或 r = −1，则 arⁿ / (1 − r) 不会消失，因此不存在有限极限。这正是条件 |r| < 1 要写在公式中的原因：它不是技术性附加条件，而是公式成立的根本原因。

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A rigorous answer to a "prove" question should mention: write Sₙ in the standard form, state that rⁿ → 0 because |r| < 1, and conclude that Sₙ → a / (1 − r).

回答"证明"类题目时，严谨的步骤是：把 Sₙ 写成标准形式，指出因 |r| < 1 故 rⁿ → 0，最后得出 Sₙ → a / (1 − r)。

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6. Worked Example 1: Direct Substitution | 例题一：直接代入

Question: find the sum to infinity of the geometric series 24 + 18 + 13.5 + ... .

题目：求等比级数 24 + 18 + 13.5 + ... 的无穷和。

irst identify a and r. The first term is a = 24, and the ratio is r = 18 / 24 = 3/4. Since |3/4| < 1, the series converges and the formula applies.

先确定 a 与 r。首项 a = 24，公比 r = 18 / 24 = 3/4。因为 |3/4| < 1，级数收敛，公式适用。

S∞ = 24 / (1 − 3/4) = 24 / (1/4) = 96

So the sum to infinity is 96. A useful sanity check: the first term alone is 24 and the terms are positive, so the answer must exceed 24; it does.

因此无穷和为 96。一个有用的合理性检验是：仅首项就是 24，且各项为正，所以答案必须大于 24；结果确实如此。

. Using uₙ = arⁿ⁻¹ gives 12 = a(1/2)² = a/4, so a = 48, and then S∞ = 48 / (1 − 1/2) = 96.

若题目给出的是第三项，例如 u₃ = 12 且 r = 1/2，则必须先求回 a。由 uₙ = arⁿ⁻¹ 得 12 = a(1/2)² = a/4，故 a = 48，进而 S∞ = 48 / (1 − 1/2) = 96。

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7. Worked Example 2: Finding a and r | 例题二：求首项与公比

Question: a geometric series has first term 18 and sum to infinity 27. Find the common ratio.

题目：一个等比级数的首项为 18，无穷和为 27，求公比。

the formula: 27 = 18 / (1 − r). Rearranging gives 1 − r = 18 / 27 = 2/3, so r = 1/3. Since |1/3| this value is consistent with the existence of a sum to infinity, so the answer is valid.

代入公式：27 = 18 / (1 − r)。整理得 1 − r = 18 / 27 = 2/3，故 r = 1/3。因为 |1/3| < 1，与无穷和存在的前提一致，所以答案有效。

A harder variation: a geometric series has second term 6 and sum to infinity 32. Find the two possible values of r and the corresponding first terms.

难度更高的变式：一个等比级数的第二项为 6，无穷和为 32。求 r 的两个可能取值及相应的首项。

es to 32r(1 − r) = 6, then 32r² − 32r + 6 = 0, and dividing by 2 gives 16r² − 16r + 3 = 0.

由 u₂ = ar = 6 得 a = 6 / r。代入求和公式：32 = (6 / r) / (1 − r)，整理得 32r(1 − r) = 6，即 32r² − 32r + 6 = 0，两边除以 2 得 16r² − 16r + 3 = 0。

r = (16 ± √256 − 192) / 32 = (16 ± 8) / 32

So r = 24/32 = 3/4 or r = 8/32 = 1/4. Both satisfy |r| , if the question demands it, verify with the original data: 8 × 3/4 = 6 and 24 × 1/4 = 6.

因此 r = 24/32 = 3/4 或 r = 8/32 = 1/4，两者都满足 |r| < 1。对应首项为 a = 6 / (3/4) = 8 与 a = 6 / (1/4) = 24。务必用收敛条件检验两个根，必要时再用原始数据验证：8 × 3/4 = 6，24 × 1/4 = 6。

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8. Worked Example 3: Inequalities and Unknowns | 例题三：不等式与未知数

Questions frequently hide an unknown inside the common ratio. The series 1 + (2x + 1) + (2x + 1)² + (2x + 1)³ + ... converges only when the ratio satisfies |2x + 1| < 1.

题目常把未知数藏在公比里。级数 1 + (2x + 1) + (2x + 1)² + (2x + 1)³ + ... 仅在公比满足 |2x + 1| < 1 时收敛。

Solving the modulus inequality: −1 < 2x + 1 < 1, so −2 < 2x < 0, giving −1 < x < 0. Any x in this open interval makes the series convergent.

解这个绝对值不等式：−1 < 2x + 1 < 1，于是 −2 < 2x < 0，得 −1 < x < 0。该开区间内的任何 x 都使级数收敛。

rval and gives S∞ = −1 / (−1/2) = 2, which matches the series 1 + 1/2 + 1/4 + ... exactly.

当它收敛时，其和为 S∞ = 1 / (1 − (2x + 1)) = 1 / (−2x) = −1 / (2x)。例如 x = −1/4 落在区间内，得 S∞ = −1 / (−1/2) = 2，恰好与级数 1 + 1/2 + 1/4 + ... 一致。

of values of k for which 2 + 6k + 18k² + ... converges. Here r = 3k, so the requirement is |3k| , that is −1/3 < k < 1/3. Note how the coefficient must be taken inside the modulus.

另一种考法是求使 2 + 6k + 18k² + ... 收敛的 k 的取值范围。此处 r = 3k，要求 |3k| < 1，即 −1/3 < k < 1/3。注意系数必须一并放进绝对值内。

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9. Recurring Decimals as Fractions | 循环小数化分数

fraction. For 0.444... , write it as 4/10 + 4/100 + 4/1000 + ... with a = 0.4 and r = 0.1.

每个循环小数本质上都是一个收敛的等比级数，这为化分数提供了完全严谨的方法。对于 0.444...，可写成 4/10 + 4/100 + 4/1000 + ...，其中 a = 0.4，r = 0.1。

S∞ = 0.4 / (1 − 0.1) = 0.4 / 0.9 = 4/9

/ (1 − 1/100) = (27/100) / (99/100) = 27/99 = 3/11, after cancelling the common factor 9.

对于两位循环节如 0.272727...，公比为 1/100，首项为 27/100。因此 S∞ = (27/100) / (1 − 1/100) = (27/100) / (99/100) = 27/99 = 3/11（约去公因数 9）。

The general pattern is that a recurring block of k digits corresponds to a geometric series with ratio 10⁻ᵏ. This explains the familiar "9s and 0s" shortcut used at GCSE, and shows why it works.

一般规律是：k 位循环节对应的等比级数公比为 10⁻ᵏ。这解释了 GCSE 阶段常用的"分母写 9 或 0"的技巧，并说明它为何成立。

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10. Applied Modelling Questions | 实际应用题

A ball is dropped from a height of 3 m and after each bounce it rises to 0.6 of its previous height. Find the total vertical distance travelled before it comes to rest.

一个球从 3 米高处落下，每次弹起后上升到前一次高度的 0.6 倍。求它静止前所经过的竖直总路程。

m. So the total is 3 + 2(1.8 + 1.08 + ...) = 3 + 2 × 1.8 / (1 − 0.6) = 3 + 2 × 4.5 = 12 m.

下落路程构成 3, 1.8, 1.08, ...，a = 3，r = 0.6。上升路程构成同样的级数但去掉首项。因此总路程为 3 + 2(1.8 + 1.08 + ...) = 3 + 2 × 1.8 / (1 − 0.6) = 3 + 2 × 4.5 = 12 米。

Other common contexts include depreciation of an asset by a fixed percentage each year, repeated dilution of a solution, and the total area of an infinite nested pattern of squares or triangles.

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