无穷级数之和公式

无穷级数公式用于求无穷多个项之和，假设项呈无限几何级数，且公比的绝对值小于 1。这是因为，只有当公比小于1，和将收敛到一个确定的值，否则和的绝对值将趋于无穷大。

公式

For a geometric series, we can express the sum as,

a + ar + ar² + ar³ + … + (infinite terms) = a/(1 – r)

where, 

a = first term of the geometric series

r = common ratio, where -1 < r < 1

状况：

• 该系列应呈几何级数。
• 公比的绝对值应小于 1。

公式的推导

Let’s consider,

a = first term of the geometric series

r = common ratio, where -1 < r < 1

Let us consider the sum of the geometric progression be S.

Then we can write,

S = a + ar + ar²+ ar³ + …             —– (i)

Multiplying both sides of the equation by r, we get,

Sr = ar + ar² + ar³ + ar⁴ + …        —– (ii)

Subtracting Eq. (ii) from Eq. (i), we get

S – Sr = (a + ar + ar²+ ar³) + … – (ar + ar² + ar³ + ar⁴ + …)

S(1 – r) = a

S = a/(1 – r)

Hence, the sum of infinite series of a geometric progression is a/(1 – r)

笔记：

If the absolute value of the common ratio ‘r’ is greater than 1, then the sum will not converge.

Thus, the absolute value of the sum will tend to infinity. Thus, if r > 1,

| S | = | a + ar + ar² + ar³ + … | = ∞

示例问题

问题 1. 求第一项为 4 且公比为 1/2 的无穷级数之和。

解决方案：

Given, the first term a = 4

The common ratio r = 1/2

Thus, we can write the series as,

S = 4 + 4 × (1/2) + 4 × (1/2)² + …

So, the sum will stand as

S = 4/(1 – (1 / 2)) = 4/(1/2) = 4 × 2 = 8

S = 8

So, the sum of the series is equal to 8.

问题 2. 求无穷级数 1 + (1/2) + (1/2) ² + (1/2) ³ + ... 的和。

解决方案：

Given, the first term of the series a =  1.

The common ratio is  r = 1/2.

Since the absolute value of the common ratio is less than 1, we can apply the general formula.

So, the sum is,

S = 1/(1 – (1/2)) = 2

So, the sum of the given infinite series is 2.

问题 3. 计算总和 2 + 4 + 8 + 16 + ... 。

解决方案：

We can write the sum of the given series as,

S = 2 + 2² + 2³ + 2⁴ + …

We can observe that it is a geometric progression with infinite terms and first term equal to 2 and common ratio equals 2.

Thus, r = 2.

Since, the value of r > 1, the sum will not converge and tend to infinity. Thus,

S = + ∞

问题 4. 求级数 2 – 1/5 + 1 – 1/25 + 1/2 – 1/125 + ... 的和。

解决方案：

We can write the sum of the series as the difference of two infinite series as:

S = (2 + 1 + 1/2 + 1/2² + …) – (1/5 + 1/25 + 1/125 + … )

S = (2 + 1 + 1/2 + 1/2² + …) – (1/5 + 1/5² + 1/5³ + …)

S = S1 – S2

where,

S1 = 2 + 1 + 1/2 + 1/2² + …

S2 = 1/5 + 1/5² + 1/5³ + …

Here, we can see both S1 and S2 are infinite summation of geometric series, where,

a1 = 2, r1 = 1/2

a2 = 1/5, r2 = 1/5

Thus, we can write,

S1 = 2/(1 – (1/2)) = 2/(1/2) = 4

S2 = (1/5)/(1 – (1/5)) = (1/5) / (4/5) = 1/4

So, the summation S stands as,

S = S1 – S2 = 4 – 1/4 = (16 – 1)/4 = 15/4 = 3.75

S = 3.75

Thus, the sum of the given series is 3.75 .