Geometric Series

Definition of a Geometric Series

Geometric series

The series $\sum_{n=0}^{\infty} ar^n$ is called a geometric series, where $a \neq 0$.

Convergence

Convergence of a geometric series

When $|r| < 1$, the series converges and the sum is

$\sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r}$

When $|r| \geq 1$, the series diverges.

Proof

Let $S_n = a + ar + ar^2 + \cdots + ar^{n-1}$

Then $rS_n = ar + ar^2 + ar^3 + \cdots + ar^n$

Subtracting gives $(1-r)S_n = a - ar^n = a(1-r^n)$

When $|r| < 1$, $\lim_{n \to \infty} r^n = 0$

So $\lim_{n \to \infty} S_n = \frac{a}{1-r}$

Examples

Example 1

Determine the convergence of $\sum_{n=0}^{\infty} \frac{1}{2^n}$ and find its sum.

Solution: This is a geometric series with $a = 1, r = \frac{1}{2}$.

Since $|r| = \frac{1}{2} < 1$, the series converges.

Sum: $S = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{2}} = 2$

Example 2

Determine the convergence of $\sum_{n=0}^{\infty} \frac{1}{3^n}$ and find its sum.

Solution: This is a geometric series with $a = 1, r = \frac{1}{3}$.

Since $|r| = \frac{1}{3} < 1$, the series converges.

Sum: $S = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2}$

Example 3

Determine the convergence of $\sum_{n=0}^{\infty} 2^n$.

Solution: This is a geometric series with $a = 1, r = 2$.

Since $|r| = 2 \geq 1$, the series diverges.

练习题

练习 1

Determine the convergence of $\sum_{n=0}^{\infty} \frac{1}{4^n}$ and find its sum.

参考答案

思路：Identify it as a geometric series and compare $|r|$ with 1.

步骤：

1. Series type: $\sum_{n=0}^{\infty} \frac{1}{4^n}$ is geometric
2. Parameters: $a = 1, r = \frac{1}{4}$
3. Convergence: $|r| = \frac{1}{4} < 1$ so it converges
4. Sum: $S = \frac{a}{1-r} = \frac{1}{1-\frac{1}{4}} = \frac{4}{3}$

答案：Convergent, sum $\frac{4}{3}$.

练习 2

Determine the convergence of $\sum_{n=0}^{\infty} \frac{2}{5^n}$ and find its sum.

参考答案

思路：Geometric series; check $|r|$.

步骤：

1. Series type: $\sum_{n=0}^{\infty} \frac{2}{5^n}$ is geometric
2. Parameters: $a = 2, r = \frac{1}{5}$
3. Convergence: $|r| = \frac{1}{5} < 1$ so it converges
4. Sum: $S = \frac{a}{1-r} = \frac{2}{1-\frac{1}{5}} = \frac{5}{2}$

答案：Convergent, sum $\frac{5}{2}$.

练习 3

Determine the convergence of $\sum_{n=0}^{\infty} (-1)^n$.

参考答案

思路：Geometric with ratio $r=-1$.

步骤：

1. Series type: $\sum_{n=0}^{\infty} (-1)^n$ is geometric
2. Parameters: $a = 1, r = -1$
3. Convergence: $|r| = 1 \geq 1$, so it diverges

答案：Divergent.

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总结

本文出现的符号

符号	类型	读音/说明	在本文中的含义
$\sum$	希腊字母	Sigma（西格玛）	求和符号，表示级数
$\infty$	数学符号	无穷大	表示无穷级数，项数无限
$r$	数学符号	公比	几何级数中相邻两项的比值
$a$	数学符号	首项	几何级数的首项
$S_n$	数学符号	部分和	级数的前 $n$ 项和
$\lim$	数学符号	极限	表示数列或函数的极限

中英对照

中文术语	英文术语	音标	说明
几何级数	geometric series	/dʒiːəˈmetrɪk ˈsɪəriːz/	形如 $\sum_{n=0}^{\infty} ar^n$ 的级数
公比	common ratio	/ˈkɒmən ˈreɪʃiəʊ/	几何级数中相邻两项的比值 $r$
首项	first term	/fɜːst tɜːm/	几何级数的第一项 $a$
收敛	convergence	/kənˈvɜːdʒəns/	级数部分和序列有有限极限
发散	divergence	/daɪˈvɜːdʒəns/	级数部分和序列无有限极限
和	sum	/sʌm/	收敛级数的极限值
部分和	partial sum	/ˈpɑːʃəl sʌm/	级数前 $n$ 项的和 $S_n$

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