Basic Concepts of Infinite Series

Concept of Series

Definition of Series

Given a sequence $\{a_n\}$ , the expression:

\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots + a_n + \cdots

is called an infinite series, or simply a series. Here $a_n$ is called the general term of the series.

符号说明

Symbol	Type	Pronunciation/Explanation	Meaning in This Article
$\{a_n\}$	Mathematical symbol	Sequence notation	Represents a sequence, $a_n$ is the $n$ -th term
$\sum$	Greek letter	Sigma	Summation symbol, representing series
$\infty$	Mathematical symbol	Infinity	Represents infinite series, infinite number of terms

Partial Sums

Definition of Partial Sums

The sum of the first $n$ terms of the series $\sum_{n=1}^{\infty} a_n$ :

S_n = a_1 + a_2 + \cdots + a_n = \sum_{k=1}^n a_k

is called the $n$ -th partial sum of the series.

Convergence

The diagram below shows the partial sums of a classic convergent geometric series (where each term is halved). The horizontal axis represents the number of terms $n$ , and the vertical axis represents the partial sum of the first $n$ terms. You can see that the curve approaches a finite value quickly, indicating that the series converges rapidly.

Definition of Series Convergence

If the sequence $\{S_n\}$ converges (convergence), meaning there exists a finite limit:

\lim_{n \to \infty} S_n = S

then the series $\sum_{n=1}^{\infty} a_n$ is said to converge (convergence), and $S$ is called the sum of the series, denoted as:

\sum_{n=1}^{\infty} a_n = S

Divergence

When the sequence of partial sums grows continuously without a finite limit, we say the corresponding series diverges. The chart below specifically shows the partial sums of the harmonic series ( $\sum \frac{1}{n}$ ), which continues to rise as the number of terms increases, clearly demonstrating the characteristic of divergence.

Definition of Series Divergence

If the sequence $\{S_n\}$ diverges (divergence), then the series $\sum_{n=1}^{\infty} a_n$ is said to diverge (divergence).

Properties of Series

Linearity Property

If both series $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ converge, then:

Linearity Property of Series

$\sum_{n=1}^{\infty} (a_n + b_n) = \sum_{n=1}^{\infty} a_n + \sum_{n=1}^{\infty} b_n$

$\sum_{n=1}^{\infty} (ca_n) = c \sum_{n=1}^{\infty} a_n$

where $c$ is a constant.

Necessary Condition for Convergence

If the series $\sum_{n=1}^{\infty} a_n$ converges (convergence), then:

Necessary Condition for Series Convergence

If the series $\sum_{n=1}^{\infty} a_n$ converges, then its general term must approach 0:

\lim_{n \to \infty} a_n = 0

证明

Proof Strategy: Utilize the definition of series convergence and properties of sequence limits.

Detailed Proof:

Assume the series converges: Let the series $\sum_{n=1}^{\infty} a_n$ converge to sum $S$ , that is: $\lim_{n \to \infty} S_n = S$ where $S_n = a_1 + a_2 + \cdots + a_n$
Express the general term: Note that $a_n = S_n - S_{n-1}$ (for $n \geq 2$ )
Compute the limit: $\lim_{n \to \infty} a_n = \lim_{n \to \infty} (S_n - S_{n-1}) = \lim_{n \to \infty} S_n - \lim_{n \to \infty} S_{n-1} = S - S = 0$
Conclusion: Therefore $\lim_{n \to \infty} a_n = 0$

Proof Complete: If the series converges, its general term must approach zero.

Note: This is a necessary condition, not a sufficient condition. That is, $\lim_{n \to \infty} a_n = 0$ does not guarantee that the series converges (convergence).

Exercises

Exercise 1

Determine the convergence of the series $\sum_{n=1}^{\infty} \frac{n}{2^n}$ .

Reference Answer (2 个标签)

series convergence ratio test

Problem-solving approach: Use the ratio test to compute the limit of the ratio between consecutive terms.

Detailed steps:

Let $a_n = \frac{n}{2^n}$
Calculate the ratio: $\frac{a_{n+1}}{a_n} = \frac{n+1}{2^{n+1}} \cdot \frac{2^n}{n} = \frac{n+1}{2n}$
Find the limit: $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{n+1}{2n} = \frac{1}{2} < 1$
Determine convergence: Ratio is less than 1, so the series converges

Answer: The series converges (convergence).

Exercise 2

Determine the convergence of the series $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$ .

Reference Answer (3 个标签)

series convergence absolute convergence p-series

Problem-solving approach: First check for absolute convergence; if absolutely convergent, the original series converges.

Detailed steps:

Consider the absolute value series: $\sum_{n=1}^{\infty} \frac{1}{n^2}$
This is a $p$ -series with $p = 2 > 1$ , so it converges absolutely
Since it converges absolutely, the original series converges

Answer: The series converges (convergence).

Exercise 3

Determine the convergence of the series $\sum_{n=1}^{\infty} \frac{1}{n\ln n}$ .

Reference Answer (2 个标签)

series convergence integral test

Problem-solving approach: Use the integral test to compare the series with an integral.

Detailed steps:

Let $f(x) = \frac{1}{x\ln x}$ , then $a_n = f(n)$
Compute the integral: $\int_2^{+\infty} \frac{1}{x\ln x} dx = \int_2^{+\infty} \frac{1}{\ln x} d(\ln x) = \ln(\ln x) \big|_2^{+\infty} = +\infty$
The integral diverges, so the series diverges

Answer: The series diverges (divergence).

Exercise 4

Determine the convergence of the series $\sum_{n=1}^{\infty} \frac{1}{n^2 + n}$ .

Reference Answer (3 个标签)

series convergence comparison test p-series

Problem-solving approach: Use the comparison test to compare with a known convergent series.

Detailed steps:

Since $\frac{1}{n^2 + n} < \frac{1}{n^2}$ , and $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is a convergent $p$ -series ( $p = 2 > 1$ )
By the comparison test, $\sum_{n=1}^{\infty} \frac{1}{n^2 + n}$ converges

Answer: The series converges (convergence).

Exercise 5

Determine the convergence of the series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{1/2}}$ .

Reference Answer (6 个标签)

series convergence absolute convergence conditional convergence alternating series Leibniz test p-series

Problem-solving approach: First check for absolute convergence; if the absolute series diverges, check for conditional convergence.

Detailed steps:

Consider the absolute value series: $\sum_{n=1}^{\infty} \frac{1}{n^{1/2}}$
This is a $p$ -series with $p = \frac{1}{2} \leq 1$ , so it diverges
Since the absolute series diverges, further testing is needed
The original series is an alternating series with $a_n = \frac{1}{n^{1/2}}$
Check Leibniz test conditions:
- $a_n > 0$ ✓
- $a_{n+1} < a_n$ ✓ (since $\frac{1}{(n+1)^{1/2}} < \frac{1}{n^{1/2}}$ )
- $\lim_{n \to \infty} a_n = 0$ ✓
Satisfies Leibniz test conditions, so the series converges

Answer: The series converges (conditionally) (conditional convergence).

Summary

Symbols Used in This Article

Symbol	Type	Pronunciation/Explanation	Meaning in This Article
$\{a_n\}$	Mathematical symbol	Sequence notation	Represents a sequence, $a_n$ is the $n$ -th term
$S_n$	Mathematical symbol	Partial sum	Sum of the first $n$ terms of the series
$\lim$	Mathematical symbol	Limit	Represents limit of sequence or function
$\rho$	Greek letter	Rho	Represents limit value in series convergence tests

Chinese-English Glossary

Chinese Term	English Term	IPA Pronunciation	Explanation
级数	series	/ˈsɪəriːz/	Sum of infinite terms, denoted as $\sum_{n=1}^{\infty} a_n$
无穷级数	infinite series	/ˈɪnfɪnɪt ˈsɪəriːz/	Series with infinite number of terms
通项	general term	/ˈdʒenərəl tɜːm/	The $n$ -th term $a_n$ in the series
部分和	partial sum	/ˈpɑːʃəl sʌm/	Sum of first $n$ terms $S_n$ of the series
收敛	convergence	/kənˈvɜːdʒəns/	Partial sums sequence has a finite limit
发散	divergence	/daɪˈvɜːdʒəns/	Partial sums sequence has no finite limit
和	sum	/sʌm/	Limit value of convergent series
线性性质	linearity property	/ˈlɪniəriti ˈprɒpəti/	Linear characteristics of series operations
必要条件	necessary condition	/nɪˈsesəri kənˈdɪʃən/	Condition that must be satisfied for series convergence
充分条件	sufficient condition	/səˈfɪʃənt kənˈdɪʃən/	Condition that guarantees series convergence
条件收敛	conditional convergence	/kənˈdɪʃənəl kənˈvɜːdʒəns/	Series converges but absolute series diverges

PreviousHistorical Background of Infinite Series NextConvergence TestsLearn various convergence tests for series, including tests for positive-term series, alternating series, and general series.

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