Series and Limits | 级数与极限

📚 Series and Limits | 级数与极限

In AQA A-level Mathematics, Series and Limits covers arithmetic and geometric sequences, summation techniques, recurrence relations, binomial series and limiting behaviour. The topic tests both routine formula use and the ability to justify convergence in infinite cases.

在 AQA A-level 数学中，"级数与极限"涵盖等差与等比数列、求和技巧、递推关系、二项级数以及极限行为。本专题既考查公式的常规使用，也考查在无穷情形下证明收敛性的能力。

Questions on this topic appear in both Pure papers and often carry a 'show that' or 'prove that' instruction, so clear algebra and correct notation matter as much as the final value. Many marks are lost through missing conditions, such as failing to state that |r| < 1 before using a sum-to-infinity formula.

本专题的题目会出现在两份 Pure 试卷中，且常带 "show that" 或 "prove that" 指令，因此清晰的代数推导与正确的记号，与最终答案同样重要。很多失分源于遗漏条件，例如在使用无穷和公式前未说明 |r| < 1。

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1. Sequences and Notation | 数列与记号

A sequence is an ordered list of numbers u₁, u₂, u₃, ... produced by a rule. The nth term is written uₙ or aₙ, and the first term is often denoted u₁ or a. The order matters: 1, 3, 5 is a different sequence from 5, 3, 1.

数列是按规则产生的一列有序数字 u₁、u₂、u₃ ……。第 n 项写作 uₙ 或 aₙ，首项通常记为 u₁ 或 a。顺序很重要：1、3、5 与 5、3、1 是不同的数列。

is written Sₙ = u₁ + u₂ + ... + uₙ. The symbol ∑ is used as a compact summation notation.

级数是数列各项之和。前 n 项和写作 Sₙ = u₁ + u₂ + ... + uₙ。符号 ∑ 用于紧凑地表示求和。

term, n = 2 the second, and so on; the expression uₙ is called the position-to-term rule.

例如，若 uₙ = 2n + 1，则数列开头为 3、5、7、9 ……，且 S₄ = 3 + 5 + 7 + 9 = 24。代入 n = 1 得首项，n = 2 得第二项，依此类推；uₙ 这一表达式称为通项公式。

Sequences may be classified as increasing, decreasing, periodic or neither. A sequence is increasing if uₙ₊₁ > uₙ for every n, decreasing if uₙ₊₁ < uₙ, and periodic if it repeats a fixed block of values.

数列可分为递增、递减、周期或其他类型。若对每个 n 都有 uₙ₊₁ > uₙ，则数列递增；若 uₙ₊₁ < uₙ，则递减；若重复出现固定的一组值，则称周期数列。

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2. Arithmetic Series | 等差级数

+ (n − 1)d. The difference is found by subtracting any term from the term that follows it.

等差数列的相邻项之差 d 为常数。若首项为 a，则第 n 项为 uₙ = a + (n − 1)d。公差可由任意一项与其后一项相减得到。

The sum of the first n terms is given by two equivalent forms, one using the last term L:

前 n 项和有两种等价形式，其中一种使用末项 L：

Sₙ = (n/2)(a + L) = (n/2)[2a + (n − 1)d]

first form is useful when the last term is known, while the second is used when d is given.

其中 L = a + (n − 1)d 是末项。当已知末项时，第一种形式更实用；当已知公差 d 时，则用第二种。

ir has the same total a + L, and there are n/2 such equal pairs, giving Sₙ = (n/2)(a + L).

该公式来自将首尾项配对：(a + L)、(a + d + L − d) 等。每一对的和都等于 a + L，这样的等和配对共有 n/2 对，于是得到 Sₙ = (n/2)(a + L)。

Worked example: re a = 3, d = 4 and n = 50, so S₅₀ = (50/2)[2(3) + 49(4)] = 25(6 + 196) = 25 × 202 = 5050.

例题：求 3、7、11、…… 前 50 项之和。此处 a = 3，d = 4，n = 50，所以 S₅₀ = (50/2)[2(3) + 49(4)] = 25(6 + 196) = 25 × 202 = 5050。

If d is negative the series decreases, and the same formulas still apply. To find n when a, d and Sₙ are given, form a quadratic in n, solve it, and reject any non-positive or non-integer result.

若 d 为负，级数递减，公式依然适用。当已知 a、d 与 Sₙ 求 n 时，可整理出关于 n 的二次方程，解出后舍去非正数或非整数解。

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3. Geometric Series | 等比级数

rm a, the nth term is uₙ = a rⁿ⁻¹, and r is found by dividing any term by its predecessor.

等比数列的相邻项之比 r 为常数，即每一项都是前一项的 r 倍。若首项为 a，则第 n 项为 uₙ = a rⁿ⁻¹，r 可由任意一项除以其前一项得到。

The sum of the first n terms, for r ≠ 1, is:

当 r ≠ 1 时，前 n 项和公式为：

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

Both forms are correct; the second is often neater when r > 1 because numerator and denominator are then positive.

两种形式都正确；当 r > 1 时第二种更整洁，因为此时分子与分母都是正数。

Worked example: 2(243 − 1)/2 = 242. If instead the ratio were r = ½, the terms would shrink towards zero.

例题：对 2、6、18、…… 有 a = 2、r = 3，故 S₅ = 2(3⁵ − 1)/(3 − 1) = 2(243 − 1)/2 = 242。若比值为 r = ½，则各项会逐渐趋近于零。

, since n appears in the exponent. Take logs of both sides and rearrange, remembering that r must be positive for the log step.

等比数列问题常需用对数解形如 a rⁿ⁻¹ = 某值 的方程，因为 n 出现在指数上。两边取对数并整理，注意取对数要求 r 为正数。

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4. Sum to Infinity and Convergence | 无穷和与收敛性

An infinite geometric series converges to a finite sum only when |r| 1 < r < 1. In that case the term rⁿ tends to 0 as n → ∞, and the sum to infinity is:

无穷等比级数只有在 |r| < 1（即 −1 < r < 1）时才收敛到有限和。此时 rⁿ 随 n → ∞ 趋于 0，无穷和为：

S∞ = a/(1 − r), for |r| < 1

ws from letting n → ∞ in Sₙ = a(1 − rⁿ)/(1 − r), since rⁿ → 0 and the numerator becomes a.

推导方法是在 Sₙ = a(1 − rⁿ)/(1 − r) 中令 n → ∞，由于 rⁿ → 0，分子变为 a。

If |r| 1 the partial sums oscillate between two values; when r < −1 the terms grow in size and alternate in sign.

若 |r| ≥ 1，则级数发散。当 r > 1 时各项无限增大；当 r = −1 时部分和在两个值之间振荡；当 r < −1 时各项绝对值增大且符号交替。

= 0.7 + 0.07 + 0.007 + ... has a = 0.7 and r = 0.1, so S∞ = 0.7/(1 − 0.1) = 0.7/0.9 = 7/9.

循环小数可写成等比级数。例如 0.777... = 0.7 + 0.07 + 0.007 + ……，其中 a = 0.7、r = 0.1，故 S∞ = 0.7/(1 − 0.1) = 0.7/0.9 = 7/9。

e the condition for convergence before quoting S∞. Writing the value without the condition |r| < 1 usually costs a mark, even if the arithmetic is perfect.

考试题常要求在使用 S∞ 之前先写出收敛条件。若只写数值而不写 |r| < 1，即使计算完全正确，通常也会被扣分。

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5. Sigma Notation and Summation Techniques | 求和记号与求和技巧

₁ⁿ uᵣ means u₁ + u₂ + ... + uₙ, where r is the counter, 1 the lower limit and n the upper limit.

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求和记号 ∑ 可把整串求和压缩成一个表达式。一般形式 ∑ᵣ₌₁ⁿ uᵣ 表示 u₁ + u₂ + ... + uₙ，其中 r 为计数变量，1 为下限，n 为上限。

Three standard results should be memorised, together with the constant rule ∑ᵣ₌₁ⁿ k = kn:

应牢记三个标准结论，以及常数规则 ∑ᵣ₌₁ⁿ k = kn：

∑ r = n(n + 1)/2    ∑ r² = n(n + 1)(2n + 1)/6    ∑ r³ = [n(n + 1)/2]²

low sums such as ∑(3r + 2) to be split into 3∑r + ∑2 and evaluated directly in terms of n.

借助这些结论，形如 ∑(3r + 2) 的求和可拆成 3∑r + ∑2，并直接用 n 表示结果。

The key rules are linearity: a constant factor can be taken outside the sigma sign, and a sum of terms can be separated term by term. Limits may be adjusted by adding and subtracting the missing terms.

核心法则是线性性：常数因子可提到 ∑ 之外，各项之和可逐项拆开。当下限不为 1 时，可通过加上或减去缺少的项来调整上下限。

Example: ∑ᵣ₌₁ⁿ (3r + 2) = 3 × n(n + 1)/2 + 2n = (3n² + 3n + 4n)/2 = (3n² + 7n)/2 = n(3n + 7)/2.

例题：∑ᵣ₌₁ⁿ (3r + 2) = 3 × n(n + 1)/2 + 2n = (3n² + 3n + 4n)/2 = (3n² + 7n)/2 = n(3n + 7)/2。

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6. Recurrence Relations | 递推关系

e first value given is called the initial condition and is needed before any term can be generated.

递推关系用一个或多个前面的项来定义后一项，例如 uₙ₊₁ = 3uₙ − 1，u₁ = 2。给出的第一个值称为初始条件，没有它就无法生成任何后续项。

5, u₃ = 3(5) − 1 = 14, u₄ = 3(14) − 1 = 41, and the sequence grows rapidly without limit.

由该规则可得：u₂ = 3(2) − 1 = 5，u₃ = 3(5) − 1 = 14，u₄ = 3(14) − 1 = 41，数列迅速增长且无上界。

uces a geometric one. A recurrence such as uₙ₊₁ = 1/(2 − uₙ) can give a periodic sequence.

简单的递推描述熟悉的数列：uₙ₊₁ = uₙ + d 生成等差数列，uₙ₊₁ = r uₙ 生成等比数列。形如 uₙ₊₁ = 1/(2 − uₙ) 的递推则可能给出周期数列。

Exam questions may ask whether a sequence is increasing, decreasing or periodic, or for the value of a specific term. Working term by term, and showing each substitution, is safer than trying to spot a pattern.

考试可能要求判断数列递增、递减还是周期，或求某一项的值。逐项计算并写出每次代入过程，比试图猜测规律更稳妥。

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7. Binomial Expansion and Binomial Series | 二项展开与二项级数

For a positive integer n, the binomial theorem expands (a + b)ⁿ as a finite sum. The coefficients come from Pascal's triangle or from the formula for combinations:

当 n 为正整数时，二项式定理将 (a + b)ⁿ 展开成有限项之和。系数来自杨辉三角，或由组合数公式给出：

(a + b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ    where   ⁿCᵣ = n!/[r!(n − r)!]

The general term is ⁿCᵣ aⁿ⁻ʳ bʳ, so the term in x⁴, or the term containing a particular power, can be written down directly without expanding everything.

通项为 ⁿCᵣ aⁿ⁻ʳ bʳ，因此 x⁴ 项或任何含特定次幂的项都能直接写出，不必完整展开。

When n is not a positive integer, the expansion becomes an infinite binomial series:

当 n 不是正整数时，展开式成为无穷二项级数：

(1 + x)ⁿ = 1 + nx + [n(n − 1)/2!]x² + [n(n − 1)(n − 2)/3!]x³ + ...

This series converges only for |x| e stated whenever the expansion is used. For example (1 + x)⁻¹ = 1 − x + x² − x³ + ... for |x| < 1.

该级数仅在 |x| < 1 时收敛，使用展开式时必须写出这一条件。例如 (1 + x)⁻¹ = 1 − x + x² − x³ + ……，其中 |x| < 1。

a few terms. Questions often ask for the range of values for which the approximation is valid.

该级数用于近似计算，例如代入 x = 0.02 并在若干项后截断来估算 1.02⁻¹。题目常要求给出近似成立的有效取值范围。

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8. Limits of Sequences and Limit Laws | 数列极限与极限运算法则

arily large, we write uₙ → L as n → ∞, or lim uₙ = L, and say the sequence converges to L.

若当 n 无限增大时数列各项趋近于同一个值 L，就记作 n → ∞ 时 uₙ → L，或 lim uₙ = L，并称该数列收敛于 L。

Standard limits worth knowing are 1/n → 0, 1/n² → 0, rⁿ → 0 for |r| < 1, and (1 + 1/n)ⁿ → e ≈ 2.718. A constant sequence k, k, k, ... converges to k.

值得掌握的标准极限有：1/n → 0，1/n² → 0，当 |r| < 1 时 rⁿ → 0，以及 (1 + 1/n)ⁿ → e ≈ 2.718。常数列 k、k、k、…… 收敛于 k。

→ L and vₙ → M, then uₙ + vₙ → L + M, uₙvₙ → LM, kuₙ → kL, and uₙ/vₙ → L/M provided M ≠ 0.

极限运算法则指出：若 uₙ → L 且 vₙ → M，则 uₙ + vₙ → L + M，uₙvₙ → LM，kuₙ → kL，且当 M ≠ 0 时 uₙ/vₙ → L/M。

present, then apply the standard limits. For instance (3n² + 5)/(2n² − n) → 3/2 as n → ∞.

对于关于 n 的有理式，先用分母中 n 的最高次幂同除分子与分母，再套用标准极限。例如当 n → ∞ 时，(3n² + 5)/(2n² − n) → 3/2。

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9. Convergence Tests for Series | 级数收敛判别法

The simplest check is the nth term test: if the individual terms uₙ do not tend to 0, the series cannot converge. This test can only prove divergence, never convergence.

最简单的检验是通项判别法：若各项 uₙ 不趋于 0，则级数不可能收敛。该检验只能证明发散，不能证明收敛。

For geometric series the decisive condition is |r| < 1 for convergence, with sum a/(1 − r). This is the main convergence result required at A-level.

对等比级数而言，收敛的决定性条件是 |r| < 1，其和为 a/(1 − r)。这是 A-level 要求掌握的主要收敛结论。

Comparison arguments are also useful: if every term of a positive series is no larger than the corresponding term of a known convergent series, then it converges too. Comparing with the geometric series is the usual approach.

比较法也很有用：若正项级数的每一项都不大于某个已知收敛级数的对应项，则该级数也收敛。通常的做法是与等比级数作比较。

An alternating series whose terms decrease to 0 in size converges. The classic example is 1 − ½ + ⅓ − ¼ + ..., which converges even though the corresponding positive series diverges.

若交错级数的各项绝对值单调递减且趋于 0，则该级数收敛。经典例子是 1 − ½ + ⅓ − ¼ + ……，它收敛，而对应的正项级数却发散。

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10. Proof by Induction and Series Results | 数学归纳法与级数结论

n(n + 1)(2n + 1)/6 are proved by mathematical induction, because they claim something true for every positive integer n.

像 ∑r = n(n + 1)/2 与

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