📚 A-Level Mathematics: Sum of Infinite Geometric Series and Convergence Conditions | A-Level 数学：无穷等比级数求和公式与收敛条件
A geometric series is a sequence of terms where each term is obtained by multiplying the previous term by a fixed constant called the common ratio. An infinite geometric series is the sum of all terms of such a sequence. This topic appears frequently in A-Level mathematics, linking algebra, limits, and problem solving.
等比级数是指相邻两项之比为固定常数（称为公比）的数列所对应的和。无穷等比级数就是将这种数列的所有项相加。这一主题在 A-Level 数学中频繁出现，将代数、极限与解题技巧联系在一起。
1. Geometric Sequences and Series | 等比数列与等比级数
A geometric sequence has first term a and common ratio r. Its terms are:
等比数列的首项为 a，公比为 r，其项依次为：
a, ar, ar², ar³, …, arⁿ⁻¹, …
The n-th term is given by uₙ = a rⁿ⁻¹.
第 n 项为 uₙ = a rⁿ⁻¹。
e obtain a geometric series. In sigma notation, the infinite series can be written as ∑ₖ₌₀∞ a rᵏ or equivalently ∑ₖ₌₁∞ a rᵏ⁻¹.
若将这些项相加，则得到等比级数。用求和记号可写成 ∑ₖ₌₀∞ a rᵏ 或等价地 ∑ₖ₌₁∞ a rᵏ⁻¹。
2. The Finite Sum Formula | 有限项求和公式
For a finite geometric series with n terms, the sum Sₙ is:
对于项数为 n 的有限等比级数，其和 Sₙ 为：
Sₙ = a(1 − rⁿ) / (1 − r) (r ≠ 1)
This formula is derived by writing Sₙ and rSₙ, then subtracting to remove all middle terms:
该公式的推导方法是写出 Sₙ 与 rSₙ，然后相减以消去所有中间项：
Sₙ = a + ar + ar² + … + arⁿ⁻¹
rSₙ = ar + ar² + … + arⁿ
Sₙ − rSₙ = a − arⁿ
) = a(1 − rⁿ), which leads to the formula. If r = 1, then every term equals a, so Sₙ = na.
提取公因式得 Sₙ(1 − r) = a(1 − rⁿ)，从而得到公式。当 r = 1 时，每一项都等于 a，因此 Sₙ = na。
3. What Happens as n → ∞? | 当 n → ∞ 时会发生什么？
Consider a geometric series in which the terms become smaller and smaller. The sum may approach a finite number. For example:
观察一个项越来越小的等比级数，总和可能会逼近一个有限数。例如：
1 + 1/2 + 1/4 + 1/8 + … = ?
Here r = 1/2, so each new term halves the remaining distance to the limiting value 2.
这里 r = 1/2，每一项都会使与极限值 2 的距离减半。
verges. If r = −1, the partial sums alternate between a and 0, and no finite limit exists.
然而，若 r = 2，项无限增大，级数发散。若 r = −1，部分和在 a 与 0 之间交替，因而不存在有限极限。
4. The Convergence Condition | 收敛条件
An infinite geometric series converges to a finite value if and only if the common ratio satisfies:
无穷等比级数收敛到有限值的充要条件是公比满足：
|r| < 1
nitude of r is strictly less than 1. In this case rⁿ tends to 0 as n tends to infinity. If |r| ≥ 1, the series does not converge to a finite sum.
该条件意味着 r 的绝对值严格小于 1。此时当 n 趋于无穷大时，rⁿ 趋于 0。若 |r| ≥ 1，则该级数不收敛于有限和。
If r is negative but |r| < 1, the terms alternate in sign, yet the series still converges. For example, 1 − 1/2 + 1/4 − 1/8 + … converges.
若 r 为负但 |r| < 1，项会在正负之间交替，但级数仍然收敛。例如，1 − 1/2 + 1/4 − 1/8 + … 收敛。
5. The Sum to Infinity Formula | 无穷等比级数求和公式
When |r| < 1, the limit of the finite sum formula gives:
当 |r| < 1 时，对有限和公式取极限可得：
S∞ = a / (1 − r)
e rⁿ → 0, the term a rⁿ in the numerator disappears in the limit. More formally, as n → ∞:
因为 rⁿ → 0，分子中的 a rⁿ 项在极限中消失。更正式地，当 n → ∞ 时：
S∞ = limₙ→∞ Sₙ = limₙ→∞ a(1 − rⁿ) / (1 − r) = a / (1 − r)
This formula is quick to apply, but always verify that |r| < 1 before using it.
这个公式应用很快，但一定要先验证 |r| < 1 再使用。
6. Worked Example 1 — Basic Application | 例题 1：基本应用
Find the sum of the infinite geometric series 12 + 6 + 3 + 1.5 + …
求无穷等比级数 12 + 6 + 3 + 1.5 + … 的和。
Here a = 12 and r = 6 ÷ 12 = 1/2. Since |r| < 1, the sum to infinity is:
这里 a = 12，r = 6 ÷ 12 = 1/2。因为 |r| < 1，所以无穷和为：
S∞ = 12 / (1 − 1/2) = 24
Check: 12 + 6 + 3 + 1.5 + 0.75 + … approaches 24, and the partial sums are always below 24 for this positive series.
验证：12 + 6 + 3 + 1.5 + 0.75 + … 趋近于 24，并且对于这个正项级数，部分和始终小于 24。
Now consider the alternating series 4 − 2 + 1 − 1/2 + … Here a = 4 and r = −1/2, so:
再看交错级数 4 − 2 + 1 − 1/2 + …。这里 a = 4，r = −1/2，因此：
S∞ = 4 / (1 − (−1/2)) = 4 / (3/2) = 8/3
The result is finite because |r| = 1/2 < 1.
结果为有限值，因为 |r| = 1/2 < 1。
7. Worked Example 2 — Recurring Decimals | 例题 2：循环小数化分数
Write 0.272727… as a fraction.
将 0.272727… 化为分数。
Write it as a geometric series:
将其写成等比级数：
0.272727… = 27/100 + 27/10000 + 27/1000000 + …
The first term is a = 27/100 and the common ratio is r = 1/100. Therefore:
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