📚 Sum to Infinity of a Geometric Series | 几何级数的无穷项求和
In A-Level mathematics, particularly on the Edexcel specification, the idea of a geometric series extending to an infinite number of terms is a core topic. When a geometric series has a common ratio whose absolute value is less than 1, the series converges, and we can find an exact finite value for the sum of all infinitely many terms. This concept links algebra, limits, and real-world applications such as recurring decimals and financial models.
在A-level数学（尤其是Edexcel考纲）中，无穷项几何级数是一个核心主题。当几何级数的公比的绝对值小于1时，级数收敛，我们可以求出所有无穷项之和的精确有限值。这一概念将代数、极限与循环小数、金融模型等实际应用联系起来。
1. Geometric Series Recap | 几何级数回顾
A geometric series is the sum of the terms of a geometric sequence, where each term after the first is obtained by multiplying the previous term by a fixed non-zero number called the common ratio r. The general form is a + ar + ar2 + ar3 + ... , where a is the first term (a ≠ 0). The sum of the first n terms, denoted Sn, is given by Sn = a(1 - rn) / (1 - r) for r ≠ 1.
几何级数是几何数列各项的和，其中从第二项起每一项都是由前一项乘以一个固定的非零数（公比 r）得到。一般形式为 a + ar + ar2 + ar3 + ...，其中 a 是首项 (a ≠ 0)。前 n 项和记作 Sn，当 r ≠ 1 时由公式 Sn = a(1 - rn) / (1 - r) 给出。
2. The Concept of an Infinite Sum | 无穷项求和的概念
When we say “sum to infinity”, we mean the limit that the partial sums Sn approach as n becomes arbitrarily large. If the terms of the series get progressively smaller and the total sum approaches a finite limit, the series is convergent. In a geometric context this allows us to sum an endless number of terms and obtain a precise, finite result.
当我们说“无穷项求和”时，指的是部分和 Sn 在 n 无限增大时所趋近的极限。如果级数的项逐渐变小，且总和趋近于一个有限的极限，那么这个级数就是收敛的。在几何背景下，这让我们可以对无穷多项求和并得到一个精确的有限结果。
3. Formula for the Sum to Infinity | 无穷项求和公式
For a convergent geometric series with first term a and common ratio r, where |r| < 1, the sum to infinity S∞ is given by:
S∞ = a / (1 - r)
对于首项为 a，公比为 r 且满足 |r| < 1 的收敛几何级数，无穷项和 S∞ 由下式给出：
S∞ = a / (1 - r)
4. Derivation of the Formula | 公式推导
Starting from the finite sum formula Sn = a(1 - rn) / (1 - r), we examine what happens as n → ∞. If |r| < 1, the term rn approaches 0. Hence, the numerator a(1 - rn) tends towards a, and Sn tends to a / (1 - r). This limit is precisely the sum to infinity.
从有限和公式 Sn = a(1 - rn) / (1 - r) 出发，我们考察 n → ∞ 时的情况。如果 |r| < 1，那么 rn 趋近于 0。因此，分子 a(1 - rn) 趋近于 a，而 Sn 趋近于 a / (1 - r)。这个极限就是无穷项和。
It is crucial that we only apply this reasoning when the condition |r| < 1 holds. The derivation relies on rn → 0, which fails if |r| is highlights why the sum-to-infinity formula is exclusive to convergent geometric series.
关键是，我们仅在条件 |r| < 1 成立时应用这一推理。推导依赖于 rn → 0，若 |r| ≥ 1 则不成立。这就说明了为什么无穷项求和公式只适用于收敛的几何级数。
5. Convergence Condition: |r| < 1 | 收敛条件：|r| < 1
e usual sense. For example, if r = 1 the series becomes a + a + a + ..., which grows without bound. If r = -1, the partial sums oscillate between a and 0 and never settle to a limit.
公比的绝对值必须严格小于1，即 -1 < r < 1。如果 r 恰好为 1 或 -1，级数在通常意义下不收敛。例如，若 r = 1，级数变为 a + a + a +...，将无限增长。若 r = -1，部分和在 a 与 0 之间振荡，永不稳定于某个极限。
Some exam questions ask you to prove whether a given series has a sum to infinity. Always check |r| first. If the condition is met, you may then apply S∞ = a / (1 - r).
有些考题会要求你证明给定的级数是否有无穷项和。务必先检查 |r|。若条件满足，你才能应用 S∞ = a / (1 - r)。
6. Divergent Series: When |r| ≥ 1 | 发散级数：|r| ≥ 1 时
When the absolute value of the common ratio is greater than or equal to 1, the geometric series diverges. The terms either stay constant in magnitude or grow, and the partial sums do not approach a finite limit. In such cases we say the sum to infinity does not exist, or tends to plus or minus infinity if the terms are all positive or alternate in sign.
当公比的绝对值大于或等于1时，几何级数发散。各项的绝对值保持不变或不断增大，部分和不会趋近于有限的极限。此时我们说无穷项和不存在，或者根据各项恒正或正负交替而趋于正无穷或负无穷。
s not approach a single limiting value; its sum to infinity is undefined in the standard curriculum.
例如，2 + 4 + 8 + 16 +... 的 r = 2，因此级数无限增长。像 5 - 5 + 5 - 5 +...（r = -1）这样的级数不会趋近于一个极限值；在标准课程中，其无穷项和不存在。
7. Application: Recurring Decimals | 应用：循环小数化分数
he sum to infinity is converting a recurring decimal to a fraction. A decimal such as 0.2&# expressed as the geometric series 0.2 + 0.02 + 0.002 + ... with a = 0.2 and r = 0.1. Since |r| < 1, the sum is 0.2 / (1 - 0.1) = 0.2 / 0.9 = 2/9.
无穷项和最常见的应用之一是将循环小数转化为分数。像 0.2̇（0.222...）这样的循环小数可以表为几何级数 0.2 + 0.02 + 0.002 +...，其中 a = 0.2，r = 0.1。因为 |r| < 1，其和为 0.2 / (1 - 0.1) = 0.2 / 0.9 = 2/9。
For a more complex example, 0.12̇3&# r = 0.01. Its sum is 0.023 / 0.99 = 23/990, and adding the 0.1 (i.e. 99/990) gives 122/990, which simplifies to 61/495.
对于更复杂的例子，0.12̇3̇ 表示 0.1232323...。将其写作 0.1 + (0.023 + 0.00023 + 0.0000023 +...)。括号内的部分是一个无穷几何级数，a = 0.023，r = 0.01。其和为 0.023 / 0.99 = 23/990，再加上 0.1（即 99/990），得到 122/990，化简为 61/495。
8. Application: Financial Models and Depreciation | 应用：金融模型与折旧
centage each period forever, the total expected revenue is given by a sum to infinity. For example, a maintenance contract generates pound;1000 in year 1, and each subsequent year the amount is 80% of the previous year. Total revenue = 1000 + 1000(0.8) + 1000(0.8)2 + ... = 1000 / (1 - 0.8) = pound;5000.
某些金融情景可以用无穷几何级数建模。当一家公司永久性地在每个时期收到按固定百分比递减的定期付款时，总预期收入就用无穷项和表示。例如，一份维护合同第一年产生1000英镑，随后每年数额是前一年的80%。总收入 = 1000 + 1000(0.8) + 1000(0.8)2 + ... = 1000 / (1 - 0.8) = 5000英镑。
Similarly, for depreciation that continues indefinitely, the total loss in value can be summed. But be careful: many exam questions will require you to identify whether it is truly an infinite process or stops after a fixed number of years. The sum to infinity is only valid when the process goes on forever.
类似地，对于无限期折旧，价值的总损失也可以求和。但需注意：许多考题会要求你辨别这到底是一个无限的过程，还是经过固定年数就停止。无穷项和只有当过程永久持续时才有效。
9. Common Pitfalls and Mistakes | 常见易错点
Students frequently apply the sum-to-infinity formula without first verifying that |r| < 1. Using S∞ ent series leads to nonsense answers. Always check convergence before writing the formula.
学生们经常在没有首先验证 |r| < 1 的情况下就套用无穷项和公式。对发散级数使用 S∞ = a/(1 - r) 会导致荒谬的答案。务必在写出公式前先检查收敛性。
ther common error is misidentifying the first term a when the series does not start at the ‘ dealing with recurring decimals, forgetting to account for the non-repeating part or placing the decimal incorrectly can ruin an otherwise correct approach.
另一个常见错误是当级数并非以“自然”首项开始时，错认首项 a。像 12 + 6 + 3 +...，a = 12，而不是 6。此外，处理循环小数时，忘记考虑不循环的部分或点错小数点，也会使本该正确的方法功亏一篑。
Additionally, some learners try to find the sum to infinity when r = 1, writing S∞ = a/0, which is undefined. Recognise that division by zero signals a divergent series.
此外，有些学习者试图在 r = 1 时求无穷项和，写下 S∞ = a/0，但这是未定义的。要认识到除以零就意味着级数发散。
10. Worked Examples | 例题解析
he sum to infinity of the series 27 + 9 + 3 + 1 + ... . Here a = 27, r = 9/27 = 1/3. Since |1/3| < 1, S∞ = 27 / (1 - 1/3) = 27 / (2/3) = 40.5.
例1：求级数 27 + 9 + 3 + 1 +... 的无穷项和。这里 a = 27，r = 9/27 = 1/3。由于 |1/3| < 1，可得 S∞ = 27 / (1 - 1/3) = 27 / (2/3) = 40.5。
the common ratio. Using 20 = 8 / (1 - r) gives 1 - r = 8/20 = 0.4, so r = 0.6. Check that |0.6| < 1, confirming it is valid.
例2：某几何级数的无穷项和为20，首项为8。求公比。利用 20 = 8 / (1 - r)，得 1 - r = 8/20 = 0.4，因此 r = 0.6。验证 |0.6| < 1，确认其有效。
1 - r) = 27. Solving simultaneously yields r = 2/3 or r = 1/3. Since the series converges, |r| < 1, so both are possible; additional context would specify which is correct.
例3：一个收敛几何级数的第二项为6，无穷项和为27。求 a 和 r。已知 ar = 6，且 a/(1 - r) = 27。联立求解可得 r = 2/3 或 r = 1/3。因为级数收敛，|r| < 1，两者皆有可能；额外的条件会确定哪一个是正确的。
11. Connections to Other Topics | 与其他知识点的联系
The sum-to-infinity idea appears in several areas of A-Level Mathematics. It underpins the expansion of (1 - x)-1 as an infinite series 1 + x + x2 + ... for |x| < 1, which is a special case of binomial expansion. It also supports understanding the exponential function's Maclaurin series and iterative methods that converge to a limit.
无穷项和的概念出现在A-level数学多个领域。它是 (1 - x)-1 展开为无穷级数 1 + x + x2 + ... (|x| < 1) 的基础，这是二项式展开的特例。它还有助于理解指数函数的麦克劳林级数以及收敛至某个极限的迭代法。
Moreover, the idea of a limit that a sum approaches reinforces the formal definitions of convergence studied in later pure mathematics units. A strong grasp of S∞ for geometric series provides intuition for more advanced convergence tests.
此外，部分和趋近的极限这一概念，强化了后续纯数学单元中收敛的形式化定义。深刻理解几何级数的 S∞，将为更高级的收敛判别法提供直观基础。
12. Summary and Exam Tips | 总结与考试技巧
The sum to infinity of a geometric series is an elegant and powerful tool. Remember the crucial condition: |r| < 1. Use the formula S∞ gence. Be ready to apply it to recurring decimals, financial models, and problems where you need to find a or r from given information.
几何级数的无穷项和是一个简洁而强大的工具。牢记关键条件：|r| < 1。只有在检查收敛性之后才能使用公式 S∞ = a / (1 - r)。准备好将其应用于循环小数、金融模型以及需要根据给定信息求 a 或 r 的问题。
In the exam, show your steps: state |r|, confirm it is less than 1, then write the sum-to-infinity formula. Clear working not only earns method marks but also helps you avoid careless mistakes. Practice a variety of questions, including those that mix finite and infinite sums, to build complete confidence.
考试中要写出步骤：写出 |r|，确认其小于1，然后写下无穷项和公式。清晰的步骤不仅能获得方法分，还能帮助你避免粗心错误。练习各类题目，包括有限和与无限和混合的题，从而建立完全的自信。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图