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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.
What is Absolute Convergence?Absolute convergence is a concept in mathematics that pertains to the convergence of an infinite series. Specifically, a series [Tex]\sum_{n=1}^{\infty} a_n[/Tex] is said to converge absolutely if the series of the absolute values of its terms, [Tex]\sum_{n=1}^{\infty} a_n[/Tex], 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 ConvergenceFor a given series [Tex]\sum_{n=1}^{\infty} a_n[/Tex], if the series of absolute values [Tex]\sum_{n=1}^{\infty} |a_n|[/Tex] converges, then [Tex]\sum_{n=1}^{\infty} a_n[/Tex] an is absolutely convergent.
Let’s consider example of series with absolute and conditional convergence:
- Absolutely Convergent Series: Consider the series [Tex]\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}[/Tex]. The absolute value series [Tex]\sum_{n=1}^{\infty} \left|\frac{(-1)^n}{n^2}\right| = \sum_{n=1}^{\infty} \frac{1}{n^2}[/Tex] converges (by the p-series test with p = 2). Hence, the original series converges absolutely.
- Conditionally Convergent Series: The alternating harmonic series [Tex]\sum_{n=1}^{\infty} \frac{(-1)^n}{n}[/Tex] converges conditionally, meaning it converges but does not converge absolutely since the harmonic series [Tex]\sum_{n=1}^{\infty} \frac{1}{n}[/Tex] diverges.
Examples of Absolutely Convergent SeriesSome other general examples of absolutely convergent series:
- Geometric Series
- p-Series
- Alternating Harmonic Series
Example 1: Geometric SeriesThe geometric series with [Tex]\sum_{n=0}^{\infty} ar^n[/Tex] is absolutely convergent.
For instance: [Tex]\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots[/Tex]
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 [Tex]\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n[/Tex] is also absolutely convergent.
Example 2: p-Series with p > 1The p-series: [Tex]\sum_{n=1}^{\infty} \frac{1}{n^p}[/Tex] is absolutely convergent for p > 1.
For instance: [Tex]\sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots[/Tex]
This series converges because it is a p-series with p = 2. The convergence of the series [Tex] \sum_{n=1}^{\infty} \frac{1}{n^2}[/Tex] implies that it is absolutely convergent.
Example 3: Alternating Harmonic SeriesThe alternating harmonic series: [Tex] \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n} = 1 – \frac{1}{2} + \frac{1}{3} – \frac{1}{4} + \cdots[/Tex] is conditionally convergent.
However, if we consider the absolute values: [Tex]\sum_{n=1}^{\infty} \left|(-1)^{n+1} \frac{1}{n}\right| = \sum_{n=1}^{\infty} \frac{1}{n}[/Tex] the series [Tex]\sum_{n=1}^{\infty} \frac{1}{n}[/Tex] 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 TestThe 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: [Tex]\sum \frac{\sin(n)}{n^2}[/Tex].
Solution:
To test [Tex]\sum \frac{\sin(n)}{n^2}[/Tex] for absolute convergence, compare it to ∑1/n2, which converges by the p-series test.
Since [Tex] \left|\frac{\sin(n)}{n^2}\right| \leq \frac{1}{n^2}, \sum \frac{\sin(n)}{n^2}[/Tex] 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 limn→∞anbn = c where c is a positive finite number, then either both ∑an and ∑bn converge or both diverge.
Example: Check convergence for series: ∑3n4n
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 [Tex]\lim_{n \to \infty} \frac{\frac{3^n}{4^n}}{\left(\frac{3}{4}\right)^n} = 1[/Tex], [Tex]\sum \frac{3^n}{4^n}[/Tex] converges absolutely.
Ratio TestThe Ratio Test uses the limit of the ratio of successive terms.
If [Tex]\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L[/Tex]:
- 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 [Tex]\sum \frac{n!}{3^n}[/Tex].
Solution:
To test [Tex]\sum \frac{n!}{3^n}[/Tex] for absolute convergence, compute [Tex]\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[/Tex].
Since L = ∞, the series diverges.
Root TestThe Root Test uses the limit of the nth root of the absolute value of the terms.
If [Tex]\lim_{n \to \infty} \sqrt[n]{|a_n|} = L[/Tex]:
- 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:[Tex] \sum \left(\frac{1}{2}\right)^n [/Tex].
Solution:
To test [Tex] \sum \left(\frac{1}{2}\right)^n [/Tex] for absolute convergence, compute [Tex]\lim_{n \to \infty} \sqrt[n]{\left|\left(\frac{1}{2}\right)^n\right|} = \lim_{n \to \infty} \frac{1}{2} = \frac{1}{2}[/Tex].
Since L < 1, the series converges absolutely.
Integral TestThe 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 [Tex]\int_{N}^{\infty} f(x) \, dx[/Tex] converges, then ∑an converges. If [Tex]\int_{N}^{\infty} f(x) \, dx[/Tex] diverges, then ∑an diverges.
Example: Check the convergence for series: [Tex]\sum \frac{1}{n^2}[/Tex]
Solution:
To test [Tex]\sum \frac{1}{n^2}[/Tex] for absolute convergence, consider the improper integral [Tex]\int_{1}^{\infty} \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_{1}^{\infty} = 1[/Tex], which converges.
Hence, [Tex]\sum \frac{1}{n^2}[/Tex] converges absolutely.
Absolute Vs Conditional ConvergenceThe key differences between absolute and conditional convergence are listed in the following table:
| Feature | Absolute Convergence | Conditional Convergence |
|---|
| Definition | A series ∑an converges absolutely if:
∑|an| < ∞ | A series ∑an converges conditionally if:
The series ∑an converges.
The series ∑|an| converges. | | Rearrangement Property | The series remains convergent and sums to the same value regardless of the order of terms. | The series can be rearranged to converge to different values or even diverge. | | Example | Geometric series ∑(1/2)n. | Alternating harmonic series ∑(−1)n+1(1/n). | | Common Tests/n | Ratio Test, Root Test, Comparison Test. | Alternating Series Test, Dirichlet’s Test, Abel’s Test. | | Implication of Convergence | Implies both absolute and conditional convergence. | Does not imply absolute convergence. | | Behavior of Terms | Terms decrease rapidly enough in magnitude. | Alternating terms with decreasing magnitude but not rapidly enough for absolute convergence. | | Impact of Positive and Negative Terms | Positive and negative terms do not affect convergence as much. | Positive and negative terms significantly impact convergence. |
Read More,
Solved Examples on Absolute ConvergenceExample 1: Consider the geometric series [Tex]\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n[/Tex].
Solution:
- The absolute value of the terms is [Tex]\left|\left(\frac{1}{2}\right)^n\right| = \left(\frac{1}{2}\right)^n[/Tex].
- 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 [Tex]\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}[/Tex].
Solution:
- The absolute value of the terms is [Tex] \left|\frac{(-1)^n}{n^2}\right| = \frac{1}{n^2}[/Tex].
- The series [Tex]\sum_{n=1}^{\infty} \frac{1}{n^2}[/Tex] is a p-series with p = 2 > 1.
A p-series ∑1/np converges if p > 1. Therefore, [Tex]\sum_{n=1}^{\infty} \frac{1}{n^2}[/Tex] converges, and hence [Tex] \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}[/Tex] converges absolutely.
Example 3: Consider the series [Tex]\sum_{n=1}^{\infty} \frac{n!}{3^n}[/Tex].
Solution:
Apply the Ratio Test: [Tex]\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[/Tex].
Since the limit is greater than 1, the series diverges by the Ratio Test.
Example 4: Consider the series [Tex]\sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n[/Tex].
Solution:
Apply the Root Test: [Tex]\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[/Tex]
Since the limit is less than 1, the series converges absolutely by the Root Test.
Example 5: Consider the series [Tex]\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}[/Tex].
Solution:
Compare with [Tex]\sum_{n=1}^{\infty} \frac{1}{n^3}[/Tex]:
- The absolute value of the terms is [Tex]\left|\frac{(-1)^n}{n^3}\right| = \frac{1}{n^3}[/Tex].
- The series [Tex]\sum_{n=1}^{\infty} \frac{1}{n^3}[/Tex] is a p-series with p = 3 > 1.
Since [Tex] \sum_{n=1}^{\infty} \frac{1}{n^3}[/Tex] converges, [Tex]\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}[/Tex] converges absolutely by the Comparison Test.
Example 6: Consider the series [Tex]\sum_{n=1}^{\infty} \frac{1}{n^2}[/Tex].
Solution:
Use the Integral Test: [Tex]\int_{1}^{\infty} \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_{1}^{\infty} = \left(0 – (-1)\right) = 1[/Tex].
Since the improper integral converges, the series [Tex]\sum_{n=1}^{\infty} \frac{1}{n^2}[/Tex] converges absolutely by the Integral Test.
FAQs on Absolute ConvergenceDefine absolute convergence.Absolute convergence of a series ∑an means that the series of absolute values ∑∣an∣ converges. If a series converges absolutely, then it converges regardless of the order in which the terms are summed.
How is absolute convergence different from conditional convergence?Absolute convergence occurs when the series of absolute values ∑∣an∣ converges. Conditional convergence occurs when ∑an converges, but ∑∣an∣ does not. In other words, a series is conditionally convergent if it converges only because of the specific arrangement of its terms.
Why is absolute convergence important?Absolute convergence guarantees the convergence of the series regardless of the order of terms. This is not true for conditionally convergent series, where rearranging terms can lead to different sums or even divergence.
Can you provide an example of an absolutely convergent series?The geometric series [Tex]\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n[/Tex] is absolutely convergent because the series of absolute values [Tex]\sum_{n=0}^{\infty} \left|\left(\frac{1}{2}\right)^n\right| = \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n[/Tex] converges to 2.
Is every convergent series also absolutely convergent?No, not every convergent series is absolutely convergent. A series can be conditionally convergent, meaning it converges without the series of its absolute values converging.
How can you test for absolute convergence?One common test for absolute convergence is the comparison test. If ∑∣an∣ is convergent, and there exists another series ∑bn such that ∣an∣ ≤ ∣bn∣ for all n, then ∑bn is also convergent.
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