Series and Limits | 级数与极限

📚 Series and Limits | 级数与极限

In AQA A-Level Mathematics, series and limits connect many pure topics: arithmetic and geometric progressions, sigma notation, binomial expansions, and the behaviour of sequences and functions as n or x tends to infinity. This article explains the key definitions, formulas, convergence conditions and common exam techniques.

在 AQA A-Level 数学中，级数与极限连接了许多纯数学主题：等差与等比数列、求和符号、二项式展开，以及当 n 或 x 趋向无穷时数列与函数的行为。本文将解释关键定义、公式、收敛条件以及常见考试技巧。

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1. Sequences, Series and Key Terminology | 数列、级数与关键术语

ges if its terms get arbitrarily close to a fixed limit L as n → ∞; otherwise it diverges.

数列是按顺序排列的一列数，写作 u₁, u₂, u₃, ..., uₙ，其中 uₙ 是第 n 项。级数由数列的各项相加而成，因此第 n 个部分和为 Sₙ = u₁ + u₂ + ... + uₙ。如果数列的项在 n → ∞ 时无限接近一个固定的极限 L，则称该数列收敛；否则称其发散。

In AQA questions you may be asked to identify whether a sequence is arithmetic, geometric or neither, and to write down a formula for the nth term. You should also be comfortable moving between term notation and sum notation.

在 AQA 试题中，你可能需要判断一个数列是等差、等比还是两者都不是，并写出第 n 项公式。你还应能熟练地在项记号与求和记号之间进行转换。

Sₙ = u₁ + u₂ + u₃ + ... + uₙ

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2. Arithmetic Sequences: Finding the nth Term | 等差数列：求第 n 项

+ (n − 1)d. To find d, subtract any term from the next term; to find a, substitute a known term into the formula.

等差数列的相邻两项之间具有恒定的差 d，因此 uₙ₊₁ − uₙ = d。如果首项为 a，则第 n 项由 uₙ = a + (n − 1)d 给出。求 d 时，后一项减前一项；求 a 时，将已知项代入公式。

s a = 5 and d = 4, so uₙ = 5 + 4(n − 1) = 4n + 1. The 50th term is u₅₀ = 4 × 50 + 1 = 201.

例如，数列 5, 9, 13, 17, ... 的首项 a = 5，公差 d = 4，因此 uₙ = 5 + 4(n − 1) = 4n + 1。第 50 项为 u₅₀ = 4 × 50 + 1 = 201。

uₙ = a + (n − 1)d

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3. Arithmetic Series: Sum Formula | 等差数列求和公式

y Sₙ = n/2 (a + l), where l is the last term. Both forms are useful in AQA exam questions.

等差数列前 n 项和可以通过将首尾项配对来求解。公式为 Sₙ = n/2 [2a + (n − 1)d]，或等价地 Sₙ = n/2 (a + l)，其中 l 是末项。这两种形式在 AQA 考试中都很常用。

and d = 4, the sum of the first 30 terms is S₃₀ = 30/2 [2 × 5 + 29 × 4] = 15 × 126 = 1890.

对于数列 5, 9, 13, ...，a = 5，d = 4，前 30 项和为 S₃₀ = 30/2 [2 × 5 + 29 × 4] = 15 × 126 = 1890。

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

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4. Geometric Sequences: nth Term and Common Ratio | 等比数列：第 n 项与公比

ratio is found by dividing any term by the previous term, provided that term is not zero.

等比数列的相邻两项之间具有恒定的比值 r，因此 uₙ₊₁ = r uₙ。如果首项为 a，则第 n 项为 uₙ = arⁿ⁻¹。公比通过任意一项除以其前一项求得，前提是该项不为零。

r is negative, the terms alternate in sign; if 0 < r < 1, the terms decrease towards zero.

例如，3, 6, 12, 24, ... 的首项 a = 3，公比 r = 2，因此 uₙ = 3 × 2ⁿ⁻¹。如果 r 为负，各项符号交替变化；如果 0 < r < 1，则各项逐渐减小并趋向于零。

uₙ = arⁿ⁻¹, r = uₙ₊₁ / uₙ

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5. Geometric Series: Finite Sum | 等比数列有限和

subtracting rSₙ from Sₙ, which cancels most terms. If r = 1, the series is simply Sₙ = na.

等比数列前 n 项和为 Sₙ = a(1 − rⁿ)/(1 − r)，其中 r ≠ 1。这个公式通过计算 Sₙ − rSₙ 并消去大部分项得到。如果 r = 1，则该级数简化为 Sₙ = na。

up to 10 terms, a = 2 and r = 3, so S₁₀ = 2(1 − 3¹⁰)/(1 − 3) = 2(1 − 59049)/(−2) = 59048.

对于级数 2 + 6 + 18 + ... 的前 10 项，a = 2，r = 3，因此 S₁₀ = 2(1 − 3¹⁰)/(1 − 3) = 2(1 − 59049)/(−2) = 59048。

Sₙ = a(1 − rⁿ)/(1 − r), r ≠ 1

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6. Sum to Infinity of a Geometric Series | 等比级数无穷和

If the common ratio satisfies |r| y limiting process: the partial sums approach a fixed value, but never exceed it for a positive series with 0 < r < 1.

如果公比满足 |r| < 1，则当 n → ∞ 时 rⁿ → 0，因此无穷等比级数收敛。其无穷和为 S∞ = a/(1 − r)。这是一个关键的极限过程：部分和逐渐接近一个固定值，但对于 0 < r < 1 的正项级数而言不会超过该值。

If |r| zero, so the infinite series diverges and no finite sum exists. Always state the condition |r| < 1 before using S∞ = a/(1 − r) in an AQA answer.

如果 |r| ≥ 1 且 r ≠ 1，项不趋于零，因此无穷级数发散，不存在有限和。在 AQA 答案中，使用 S∞ = a/(1 − r) 之前一定要先写出条件 |r| < 1。

= 1/2, so S∞ = 8/(1 − 1/2) = 16. This means the partial sums get closer and closer to 16.

例如，级数 8 + 4 + 2 + 1 + ... 的 a = 8，r = 1/2，因此 S∞ = 8/(1 − 1/2) = 16。这意味着部分和会越来越接近 16。

S∞ = a/(1 − r), |r| < 1

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7. Sigma Notation and Properties | 求和符号及其运算性质

bᵣ) = ∑ aᵣ + ∑ bᵣ and ∑ c aᵣ = c ∑ aᵣ. These properties help simplify complicated series.

求和符号可以紧凑地表示一个和：∑ᵣ₌₁ⁿ aᵣ 表示 a₁ + a₂ + ... + aₙ。指标 r 可以改变，求和符号可以线性拆分或提取常数：∑ (aᵣ + bᵣ) = ∑ aᵣ + ∑ bᵣ，且 ∑ c aᵣ = c ∑ aᵣ。这些性质有助于简化复杂的级数。

and ∑ᵣ₌₁ⁿ r³ = n²(n + 1)²/4. These can be combined to evaluate sums such as ∑ (3r² − 2r).

对于多项式求和，AQA 考生应记住标准结果：∑ᵣ₌₁ⁿ 1 = n，∑ᵣ₌₁ⁿ r = n(n + 1)/2，∑ᵣ₌₁ⁿ r² = n(n + 1)(2n + 1)/6，以及 ∑ᵣ₌₁ⁿ r³ = n²(n + 1)²/4。这些结果可以组合起来计算如 ∑ (3r² − 2r) 的和。

∑ᵣ₌₁ⁿ r² = n(n + 1)(2n + 1)/6

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8. Binomial Expansion and Series Form | 二项式展开与级数形式

1 + x)ⁿ. For a positive integer n, the expansion is finite: (1 + x)ⁿ = 1 + nx + n(n − 1)/2! xⁿ. For rational or negative n, the expansion continues infinitely and is valid only when |x| < 1.

二项式展开推广了 (1 + x)ⁿ。当 n 为正整数时，展开式是有限的：(1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + ... + xⁿ。当 n 为有理数或负数时，展开式无限延续，并且仅在 |x| < 1 时有效。

The general term in the infinite expansion of (1 + x)ⁿ is n(n − 1)(n − 2)...(n − r + 1)/r! xʳ. In AQA exams you may need to find the first few terms or a specific coefficient, and state the validity condition.

(1 + x)ⁿ 无穷展开式中的一般项为 n(n − 1)(n − 2)...(n − r + 1)/r! xʳ。在 AQA 考试中，你可能需要求前几项或某个特定系数，并写出有效性条件。

(1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + n(n − 1)(n − 2)/3! x³ + ..., |x| < 1

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9. Introduction to Limits | 极限入门

if the terms aₙ can be made as close to L as desired by taking n sufficiently large. The sequence does not necessarily ever equal L.

极限描述的是当输入增大或接近某一特定点时，数列或函数所趋近的值。如果通过取足够大的 n 可以使项 aₙ 任意接近 L，则记作 limₙ→∞ aₙ = L。该数列不一定真的会等于 L。

1 as n → ∞, so limₙ→∞ aₙ = 1. This is an example of a convergent sequence; the difference |aₙ − 1| = 1/n tends to zero.

例如，数列 aₙ = (n + 1)/n = 1 + 1/n 在 n → ∞ 时趋近于 1，因此 limₙ→∞ aₙ = 1。这是一个收敛数列的例子；差值 |aₙ − 1| = 1/n 趋于零。

limₙ→∞ aₙ = L

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10. Limits of Rational Sequences | 有理数列的极限

To find the limit of a rational sequence such as (3n² + n − 1)/(2n² + 5), divide every term in the numerator and denominator by the highest power of n present. Terms of the form c/n or c/n² tend to zero, leaving the ratio of the leading coefficients.

求有理数列的极限，例如 (3n² + n − 1)/(2n² + 5)，可以将分子和分母中的每一项同时除以出现的 n 的最高次

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