Infinite Geometric Series | 无限几何级数

📚 Infinite Geometric Series | 无限几何级数

In the study of sequences and series, the concept of infinity often challenges our intuition. Sooner or later, every IB Mathematics student must face a deceptively simple question: what happens when we add up infinitely many terms of a geometric progression? If the terms are getting smaller and smaller, can we ever reach a finite total? The answer, given by infinite geometric series, is a resounding yes under certain conditions. This topic lies at the heart of the IB Analysis and Approaches syllabus, bridging algebraic manipulation, limits, and real-world applications such as repeating decimals, fractal geometry, and physical motion.

在序列与级数的学习中，无限的概念常常挑战我们的直觉。每一位IB数学学生迟早都要面对一个看似简单的问题：如果把一个等比数列的无穷多项加起来，会发生什么？如果这些项越来越小，我们能否得到一个有限的总和？无限几何级数给出的答案是肯定的，但需要满足特定的条件。这一主题位于IB分析与方法课程的核心位置，连接了代数运算、极限以及循环小数、分形几何和物理运动等实际应用。

1. Understanding Infinite Geometric Series | 理解无限几何级数

An infinite geometric series is formed by adding all terms of a geometric sequence without stopping. We write it as a + ar + ar² + ar³ + …, where a is the first term and r is the common ratio. Unlike a finite geometric series that has a fixed number of terms n, this sum goes on forever. Our primary concern is whether the sum approaches a unique finite value or grows without bound. This behaviour depends entirely on r.

无限几何级数是由一个等比数列的所有项无限相加而成。我们把它写作 a + ar + ar² + ar³ + …，其中 a 是首项，r 是公比。与项数有限的有限等比级数不同，这个和会一直延续下去。我们最关心的是这个和是趋近于一个唯一的有限值，还是会无限增长。这一行为完全取决于 r。

---

2. Convergence and the Common Ratio | 收敛与公比

al sums approach a fixed number. The necessary and sufficient condition for convergence is |r| < 1, i.e. –1 < r < 1. If |r| o infinity or oscillates without settling). The table below illustrates this with a = 100:

对于无限几何级数，收敛意味着当我们加入越来越多的项时，部分和趋近于一个固定的数。收敛的充分必要条件是 |r| < 1，即 –1 < r < 1。如果 |r| ≥ 1，级数发散（和趋向于无穷大，或者跳跃不停息）。下表用 a = 100 说明：

r	|r|	Behaviour	Sum to Infinity
0.5	0.5 < 1	Converges	200
–0.8	0.8 < 1	Converges	100/(1–(–0.8)) = 100/1.8 ≈ 55.56
1	1	Diverges	100 + 100 + … → ∞
–2	2 > 1	Diverges, oscillates	100, –200, 400, … no limit

The absolute value of r dictates everything. When |r| < 1, the terms shrink in magnitude, and their contribution becomes negligible after many terms. This fundamental idea is what makes summing to infinity possible.

r 的绝对值决定了一切。当 |r| < 1 时，各项的绝对值逐渐缩小，经过许多项之后它们的影响变得微乎其微。这个基本概念正是无限求和得以实现的关键。

---

3. Sum to Infinity Formula | 无限项求和公式

If a geometric series with first term a and common ratio r satisfies |r| < 1, the sum to infinity S_∞ is given by:

如果一个首项为 a、公比为 r 的几何级数满足 |r| < 1，其无限项之和 S_∞ 由下式给出：

S_∞ = a / (1 – r)

This remarkably compact formula allows us to compute the total sum without having to add infinitely many terms. It is vital to remember that the formula only holds when the convergence condition is met; applying it to a divergent series will yield a meaningless number.

这个极其简洁的公式使我们无需真的加上无穷多项，就能计算出总和。必须牢记，该公式仅在收敛条件满足时有效；对着一个发散级数套用，只会得到一个没有意义的数字。

---

4. Derivation of the Formula | 公式推导

The derivation rests on the formula for the sum of a finite geometric series and a limit argument. The sum of the first n terms is S_n = a(1 – rⁿ) / (1 – r), for r ≠ 1. To find S_∞, we let n approach infinity:

推导基于有限等比级数求和公式和一个极限论证。前 n 项之和为 S_n = a(1 – rⁿ) / (1 – r)，其中 r ≠ 1。为了求 S_∞，我们让 n 趋近于无穷大：

S_∞ = lim_n→∞ S_n = lim_n→∞ a(1 – rⁿ) / (1 – r).

If |r| < 1, then rⁿ → 0 as n → ∞. Therefore the expression simplifies to a(1 – 0) / (1 – r) = a / (1 – e gap between finite sums and an infinite total, provided the terms shrink fast enough. If |r| ≥ 1, rⁿ does not approach 0, so no finite sum exists.

如果 |r| < 1，那么当 n → ∞ 时，rⁿ → 0。因此表达式简化为 a(1 – 0) / (1 – r) = a / (1 – r)。这个简洁的极限在各项缩小得足够快时，填补了有限和与无限总和之间的鸿沟。若 |r| ≥ 1，rⁿ 不会趋向 0，因此不存在有限和。

---

5. First Examples | 基础例题

Let a = 8 and r = 0.25. Since |0.25| < 1, S_∞ = 8 / (1 – 0.25) = 8 / 0.75 = 32/3 ≈ 10.667. The infinite sum converges to just over ten and a half.

设 a = 8，r = 0.25。因为 |0.25| < 1，S_∞ = 8 / (1 – 0.25) = 8 / 0.75 = 32/3 ≈ 10.667。无限项之和收敛到刚好比十又一半多一点。

Now consider a = 5, r = –½. The negative ratio makes terms alternate in sign: 5, –2.5, 1.25, –0.625, … . Because |–½| = ½ < 1, the series converges: S_∞ = 5 / (1 – (– e around the final limit, illustrating that alternating series can also have a finite sum.

再考虑 a = 5，r = –½。负的公比使各项符号交替：5，–2.5，1.25，–0.625，…。由于 |–½| = ½ < 1，级数收敛：S_∞ = 5 / (1 – (–½)) = 5 / 1.5 = 10/3 ≈ 3.333。部分和在最终极限附近振荡，展示了交错级数同样可以拥有有限和。

An important check: if a question gives a divergent series like 3 + 6 + 12 + 24 + … formula. The sum to infinity does not exist, and you must state that the series diverges.

一个重要的检查：如果题目给出一个发散级数，比如 3 + 6 + 12 + 24 + … 且 r = 2，不要套用公式。无限项之和不存在，你必须指出级数发散。

---

6. Recurring Decimals as Infinite Series | 循环小数与无限级数

A classic application is converting recurring decimals to fractions without memorising rules. For instance, 0.666... = 0.6̅ can be expressed as an infinite geometric series with a = 0.6 and r = 0.1:

一个经典的应用是把循环小数化为分数，而不必死记硬背规则。例如，0.666… = 0.6̅ 可以表示为一个无限几何级数，其中 a = 0.6，r = 0.1：

0.6̅ = 0.6 + 0.06 + 0.006 + … = 0.6 / (1 – 0.1) = 0.6 / 0.9 = 6/9 = ⅔.

For a more complex decimal like 0.12̅, the repeating part starts after the hundredths place: 0.12̅ = 0.12 + 0.0012 + 0.000012 + … . Here a = 0.12, r = 0.01, so sum = 0.12 / (1 – 0.01) = 0.12 / 0.99 = 12/99 = 4/33.

对于更复杂的小数如 0.12̅，循环从百分位之后开始：0.12̅ = 0.12 + 0.0012 + 0.000012 + …。这里 a = 0.12，r = 0.01，因此总和 = 0.12 / (1 – 0.01) = 0.12 / 0.99 = 12/99 = 4/33。

If the decimal has a non-repeating part, such as 0.416… = 0.41 + 0.006 + 0.0006 + … you sum the terminating part and the infinite geometric series separately, then simplify.

如果小数有非循环部分，例如 0.416… = 0.41 + 0.006 + 0.0006 + …，你可以将有限部分与无限几何级数分别求和，然后化简。

---

广告 / Sponsored

7. Geometric Applications | 几何应用

Infinite geometric series appear naturally in problems involving repeated scaling. A well-known example is a bouncing ball. Suppose a ball is dropped from a height of 10 metres and rebounds to 60% of its previous height after each bounce. The total vertical distance travelled is:

无限几何级数自然地出现在涉及重复缩放的问题中。一个著名的例子是反弹的皮球。假设一个球从10米高处落下，每次反弹后高度恢复为前次的60%。则总的竖直运动距离为：

Downward: 10 + 10(0.6) + 10(0.6)² + … ; Upward: 10(0.6) + 10(0.6)² + … .

The total distance = 10 + 2 × (10 × 0.6 / (1 – . The key is to separate the first drop from the subsequent symmetric up-and-down pattern.

向下：10 + 10(0.6) + 10(0.6)² + … ；向上：10(0.6) + 10(0.6)² + …。总路程 = 10 + 2 × (10 × 0.6 / (1 – 0.6)) = 10 + 2 × (6 / 0.4) = 10 + 30 = 40 米。关键是将第一次下落与随后的对称上下模式分开处理。

1 has its middle quarter removed repeatedly, the total area removed is ¼ + ¼(¾) + ¼(¾)² + … = ¼ / (1 – ¾) = 1, confirming that the final fractal has zero area.

分形，例如谢尔宾斯基三角形，也提供了一个优雅的情境。如果一个面积为1的等边三角形被反复挖去中间四分之一，则被挖去的总面积是 ¼ + ¼(¾) + ¼(¾)² + … = ¼ / (1 – ¾) = 1，这证实了最终分形的面积为零。

---

8. Series with Non-Standard Starting Terms | 非标准起始项的级数

ot begin with the first term of the sequence. For example, you might see ar³ + ar⁴ + ar⁵ + … . You can factor out ar³ to get ar³(1 + r + r² + …) = ar³ / (1 – r) provided |r| < 1. Always identify the true first term of the infinite series that actually appears.

有时候一个无限几何级数并不是从序列的第一项开始的。比如，你可能看到 ar³ + ar⁴ + ar⁵ + …。你可以提取公因子 ar³，得到 ar³(1 + r + r² + …) = ar³ / (1 – r)，前提是 |r| < 1。一定要确认实际出现的无限级数的真正首项。

Another variation is a series defined by a sigma notation such as Σ (2 × 0.7^k k=5 to ∞. The first term (when k=5) is 2 × 0.7⁵, and r = 0.7, so the sum = (2 × 0.7⁵) / (1 – 0.7). Do not blindly use a = 2.

另一种变化形式是用西格玛符号定义的级数，比如 Σ (2 × 0.7^k)，k 从 5 到 ∞。首项（当 k=5 时）为 2 × 0.7⁵，而 r = 0.7，因此总和 = (2 × 0.7⁵) / (1 – 0.7)。不要盲目地用 a = 2。

---

9. Common Pitfalls | 常见错误

One of the most frequent mistakes is forgetting to check the convergence condition. Students often write S_∞ = a/(1 – 2 and obtain a negative sum, which is impossible for an all-positive series. Always verify |r| < 1 first.

最常见的错误之一是忘了检查收敛条件。学生经常在 r = 2 时写下 S_∞ = a/(1 – r)，得到一个负和，这对一个全正项的级数来说是不可能的。务必首先验证 |r| < 1。

Another pitfall is confusing the sequence with the series. A question might give you the n-th term formula u_n = 12 × 0.5ⁿ⁻¹ 1 to infinity. The correct a is 12 (not 0.5). Also, be careful when r is negative: the alternating nature may tempt you to incorrectly claim divergence, yet the condition is purely on absolute value.

另一个陷阱是把序列和级数弄混。题目可能给出第 n 项公式 u_n = 12 × 0.5ⁿ⁻¹，并要求从 n=1 到无穷大的所有项之和。正确的 a 是 12（而不是 0.5）。还要注意，当 r 为负数时，交错的性质可能会诱使你错误地声称发散，但条件只与绝对值有关。

or r. Write the decimal as a sum of place-value fractions explicitly to avoid errors: 0.2̅ = 0.2 + 0.02 + 0.002 + … clearly gives a = 0.2, r = 0.1.

最后，在处理循环小数时，学生有时会错误地识别 a 或 r。明确地将小数写为各位值分数之和可以避免错误：0.2̅ = 0.2 + 0.02 + 0.002 + … 显然给出 a = 0.2，r = 0.1。

---

10. Exam-Style Questions and Tips | 考试风格题目与技巧

Question 1: An infinite geometric series has first term 24 and sum to infinity 30. Find the common ratio.
Solution: 30 = 24 / (1 – r) → 1 – r = 24/30 = 0.8 → r = 0.2. Since |0.2| < 1, the series is convergent.

题目1：一个无限几何级数的首项为 24，无限项之和为30。求公比。
解答：30 = 24 / (1 – r) → 1 – r = 24/30 = 0.8 → r = 0.2。因为 |0.2| < 1，级数收敛。

Question 2: Express 1.27̅ as a fraction in its simplest form.
Approach: Separate as 1 + 0.27̅. Write 0.27̅ = 0.27 + 0.0027 + … with a = 0.27, r = 0.01. Sum = 0.27/0.99 = 27/99 = 3/11. So 1.27̅ = 1 + 3/11 = 14/11.

题目2：将 1.27̅ 化为最简分数。
方法：拆分为 1 + 0.27̅。写 0.27̅ = 0.27 + 0.0027 + …，其中 a = 0.27，r = 0.01。总和 = 0.27/0.99 = 27/99 = 3/11。因此 1.27̅ = 1 + 3/11 = 14/11。

Question 3: A ball is dropped from 8 m and bounces to ¾ of its previous height. Calculate the total time until it stops, given time of fall from height h is √2h/g. (Assume g = 10 m/s²)
This combines a geometric series for distance with kinematics. The total time involves summing times from each fall and rise, leading to a geometric series in √h. Such multi-step problems test deep understanding.

题目3：一个球从8 m落下，每次反弹到上次高度的¾。计算直到停止的总时间，假设从高度 h 下落的时间为 √2h/g。（取 g = 10 m/s²）
这结合了距离的几何级数与运动学。总时间涉及对每次下落和上升的时间求和，产生一个关于 √h 的几何级数。这类多步骤问题考查深刻理解。

IB exam tip: When a question states "the sum to infinity exists", it automatically tells you |r| < 1. Use that to set up inequalities for unknown constants.

IB考试技巧：当题目说明“无限项之和存在”时，它自动告诉你 |r| < 1。利用这一点为未知常数建立不等式。

---

11. Summary and Key Takeaways | 总结与关键要点

An infinite geometric series converges to a finite sum if and only if the absolute value of the common ratio is less than 1. The sum is elegantly given by S_∞ = a / (1 – metry, and modelling of repeated motion. Always begin by checking the convergence condition, correctly identify the first term and common ratio, and be particularly cautious with negative ratios or non-standard starting indices.

当且仅当公比的绝对值小于1时，无限几何级数收敛到一个有限和。这个和由 S_∞ = a / (1 – r) 优雅地给出，它源自有限求和公式并取 n → ∞ 的极限。这一工具可以解开循环小数的转换、分形几何的分析，以及重复运动的建模。务必从检查收敛条件开始，正确识别首项和公比，并特别留意负公比或非标准起始下标的情况。

Mastery of infinite geometric series is not only an essential skill for the IB exam but also a gateway to understanding convergence in calculus and beyond. With plenty of practice, the seemingly impossible task of adding infinitely many numbers becomes a straightforward algebraic procedure.

掌握无限几何级数不仅是IB考试的关键技能，也是通往理解微积分以及更广泛收敛概念的大门。只要勤加练习，将无穷多个数相加这一看似不可能的任务，便会成为一个直截了当的代数过程。

---

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）