Absolute Convergence (original) (raw)

Last Updated : 23 Jul, 2025

In mathematics, the concept of absolute convergence is crucial when dealing with infinite series. A series is said to converge absolutely if the series formed by taking the absolute values of its terms also converges. This is a stronger condition than simple convergence because it implies that the series will converge regardless of the order of its terms, which is not necessarily true for conditionally convergent series.

Absolute convergence is particularly useful because it allows us to apply various convergence tests, such as the ratio test and the root test, which typically require all terms to be positive.

Table of Content

• What is Absolute Convergence?
  • Definition of Absolute Convergence
• Examples of Absolutely Convergent Series
  • Example 1: Geometric Series
  • Example 2: p-Series with p > 1
  • Example 3: Alternating Harmonic Series
• Convergence Tests for Absolute Convergence
  • Comparison Test
  • Ratio Test
  • Root Test
  • Integral Test
• Absolute Vs Conditional Convergence
• Solved Examples on Absolute Convergence
• FAQs on Absolute Convergence

What is Absolute Convergence?

Absolute convergence is a concept in mathematics that pertains to the convergence of an infinite series. Specifically, a series \sum_{n=1}^{\infty} a_n​ is said to converge absolutely if the series of the absolute values of its terms, \sum_{n=1}^{\infty} a_n, also converges.

Absolute convergence is significant because if a series converges absolutely, it also converges in the usual sense (though the converse is not necessarily true).

Definition of Absolute Convergence

For a given series \sum_{n=1}^{\infty} a_n​, if the series of absolute values \sum_{n=1}^{\infty} |a_n| converges, then \sum_{n=1}^{\infty} a_n ​an​ is absolutely convergent.

Let's consider example of series with absolute and conditional convergence:

• **Absolutely Convergent Series: Consider the series \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}​. The absolute value series \sum_{n=1}^{\infty} \left|\frac{(-1)^n}{n^2}\right| = \sum_{n=1}^{\infty} \frac{1}{n^2} converges (by the p-series test with p = 2). Hence, the original series converges absolutely.
• **Conditionally Convergent Series: The alternating harmonic series \sum_{n=1}^{\infty} \frac{(-1)^n}{n} converges conditionally, meaning it converges but does not converge absolutely since the harmonic series \sum_{n=1}^{\infty} \frac{1}{n} diverges.

Examples of Absolutely Convergent Series

Some other general examples of absolutely convergent series:

• Geometric Series
• p-Series
• Alternating Harmonic Series

Example 1: Geometric Series

The geometric series with \sum_{n=0}^{\infty} ar^n is absolutely convergent.

For instance: \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots

Here, a = 1 and r = 1/2​. The series converges to: a/(1 − r) =1/(1 − 1/2) = 2

Since ∣(1/2)n∣ = (1/2)n, the series \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n is also absolutely convergent.

Example 2: p-Series with p > 1

The p-series: \sum_{n=1}^{\infty} \frac{1}{n^p} is absolutely convergent for p > 1.

For instance: \sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots

This series converges because it is a p-series with p = 2. The convergence of the series \sum_{n=1}^{\infty} \frac{1}{n^2}​ implies that it is absolutely convergent.

Example 3: Alternating Harmonic Series

The alternating harmonic series: \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots is conditionally convergent.

However, if we consider the absolute values: \sum_{n=1}^{\infty} \left|(-1)^{n+1} \frac{1}{n}\right| = \sum_{n=1}^{\infty} \frac{1}{n} the series \sum_{n=1}^{\infty} \frac{1}{n}​ is the harmonic series, which diverges.

Therefore, the alternating harmonic series is not absolutely convergent.

Convergence Tests for Absolute Convergence

• Comparison Test
• Ratio Test
• Root Test
• Integral Test

Comparison Test

The Comparison Test involves comparing the series in question to another series that is known to converge or diverge. There are two types of this comparison test:

• Direct Comparison Test
• Limit Comparison Test

**Direct Comparison Test

If 0 ≤ ∣an∣ ≤ bn​ for all n and ∑bn converges, then ∑∣an∣ also converges. Conversely, if ∑bn diverges and an ≥ bn​, then ∑∣an∣ diverges.

**Example: Check Convergence for series: \sum \frac{\sin(n)}{n^2}**​.

**Solution:

To test \sum \frac{\sin(n)}{n^2}​ for absolute convergence, compare it to ∑1/n2​, which converges by the p-series test.

Since \left|\frac{\sin(n)}{n^2}\right| \leq \frac{1}{n^2}, \sum \frac{\sin(n)}{n^2} converges absolutely.

**Limit Comparison Test

The Limit Comparison Test compares the given series to a known series using the limit of their terms.

If an ≥ 0, bn > 0 and lim⁡n→∞anbn = c where c is a positive finite number, then either both ∑an​ and ∑bn​ converge or both diverge.

**Example: Check convergence for series: ∑3 n 4 n

**Solution:

To test ∑3n4n for absolute convergence, compare it to ∑(3 · 4)n, a geometric series with ratio r = 3/4 < 1, which converges.

Since \lim_{n \to \infty} \frac{\frac{3^n}{4^n}}{\left(\frac{3}{4}\right)^n} = 1, \sum \frac{3^n}{4^n} converges absolutely.

Ratio Test

The Ratio Test uses the limit of the ratio of successive terms.

If \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L:

• If L < 1, the series ∑an​ converges absolutely.
• If L > 1 or L = ∞, the series diverges.
• If L = 1, the test is inconclusive.

Example: Check convergence for series \sum \frac{n!}{3^n}.

**Solution:

To test \sum \frac{n!}{3^n}​ for absolute convergence, compute \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{(n+1)! / 3^{n+1}}{n! / 3^n} = \lim_{n \to \infty} \frac{n+1}{3} = \infty.

Since L = ∞, the series diverges.

Root Test

The Root Test uses the limit of the nth root of the absolute value of the terms.

If \lim_{n \to \infty} \sqrt[n]{|a_n|} = L:

• If L < 1, the series ∑an​ converges absolutely.
• If L > 1 or L = ∞, the series diverges.
• If L = 1, the test is inconclusive.

**Example: Check the convergence for series: \sum \left(\frac{1}{2}\right)^n ****.**

**Solution:

To test \sum \left(\frac{1}{2}\right)^n for absolute convergence, compute \lim_{n \to \infty} \sqrt[n]{\left|\left(\frac{1}{2}\right)^n\right|} = \lim_{n \to \infty} \frac{1}{2} = \frac{1}{2}.

Since L < 1, the series converges absolutely.

Integral Test

The Integral Test relates the convergence of a series to the convergence of an improper integral.

If f(n) = an​ is a positive, decreasing, continuous function for n ≥ N and \int_{N}^{\infty} f(x) \, dx converges, then ∑an converges. If \int_{N}^{\infty} f(x) \, dx diverges, then ∑an​ diverges.

**Example: Check the convergence for series: \sum \frac{1}{n^2}

**Solution:

To test \sum \frac{1}{n^2} for absolute convergence, consider the improper integral \int_{1}^{\infty} \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_{1}^{\infty} = 1, which converges.

Hence, \sum \frac{1}{n^2} converges absolutely.

Absolute Vs Conditional Convergence

The key differences between absolute and conditional convergence are listed in the following table:

**Read More,

• **Sequence and Series
• **Sequence and Series Formulas
• **Conditional Convergence
• **Absolute and Conditional Convergence
• **Radius of Convergence

Solved Examples on Absolute Convergence

Example 1: Consider the geometric series \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n.

**Solution:

• The absolute value of the terms is \left|\left(\frac{1}{2}\right)^n\right| = \left(\frac{1}{2}\right)^n.
• This is a geometric series with common ratio r = 1/2​.

A geometric series ∑arn converges if ∣r∣ < 1. Here, ∣r∣ = 1/2, so the series converges absolutely.

Example 2: Consider the alternating series \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}.

**Solution:

• The absolute value of the terms is \left|\frac{(-1)^n}{n^2}\right| = \frac{1}{n^2}​​.
• The series \sum_{n=1}^{\infty} \frac{1}{n^2}​ is a p-series with p = 2 > 1.

A p-series ∑1/np converges if p > 1. Therefore, \sum_{n=1}^{\infty} \frac{1}{n^2}​ converges, and hence \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}​ converges absolutely.

Example 3: Consider the series \sum_{n=1}^{\infty} \frac{n!}{3^n}​.

**Solution:

Apply the Ratio Test: \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(n+1)! / 3^{n+1}}{n! / 3^n} \right| = \lim_{n \to \infty} \left| \frac{(n+1)}{3} \right| = \lim_{n \to \infty} \frac{n+1}{3} = \infty.

Since the limit is greater than 1, the series diverges by the Ratio Test.

Example 4: Consider the series \sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n.

**Solution:

Apply the Root Test: \lim_{n \to \infty} \sqrt[n]{\left|\left(\frac{1}{3}\right)^n\right|} = \lim_{n \to \infty} \left(\frac{1}{3}\right) = \frac{1}{3} < 1

Since the limit is less than 1, the series converges absolutely by the Root Test.

Example 5: Consider the series \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}.

**Solution:

Compare with \sum_{n=1}^{\infty} \frac{1}{n^3}:

• The absolute value of the terms is \left|\frac{(-1)^n}{n^3}\right| = \frac{1}{n^3}.
• The series \sum_{n=1}^{\infty} \frac{1}{n^3} is a p-series with p = 3 > 1.

Since \sum_{n=1}^{\infty} \frac{1}{n^3} converges, \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}​ converges absolutely by the Comparison Test.

Example 6: Consider the series \sum_{n=1}^{\infty} \frac{1}{n^2}.

**Solution:

Use the Integral Test: \int_{1}^{\infty} \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_{1}^{\infty} = \left(0 - (-1)\right) = 1.

Since the improper integral converges, the series \sum_{n=1}^{\infty} \frac{1}{n^2} converges absolutely by the Integral Test.