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In mathematics, convergence tests are essential tools used to determine whether an infinite series converges or diverges.
Convergence Tests is a compilation of many tests which help us find the convergence of infinite series. Infinite series are sums of infinitely many terms, and understanding their behavior is crucial in various fields of science and engineering. There are several convergence tests, each designed to handle different types of series.
One of the fundamental tests is the nth Term Test, which states that if the limit of the nth term of a series does not approach zero, the series diverges. However, if the limit does approach zero, the test is inconclusive, and other tests must be used.
In this article, we will discuss various test for checking convergence of series.
What are Convergence Tests?Convergence tests are mathematical tools used to determine whether an infinite series converges or diverges.
An infinite series is an expression of the form [Tex]\sum_{n=1}^{\infty} a_n[/Tex], where an represents the terms of the series. Convergence of a series implies that as more terms are added, the series approaches a finite value. Conversely, divergence implies that the series grows without bound or oscillates as more terms are added.
Types of Convergence TestsSome of the common test to check convergence of any series are:
- Integral Test
- Ratio Test
- Comparison Test
- Root Test (Cauchy’s Root Test)
- Alternating Series Test (Leibniz’s Test)
- Raabe’s Test
Integral TestIntegral Test is used to determine the convergence or divergence of an infinite series by comparing it to an improper integral. Consider a series [Tex]\sum_{n=1}^{\infty}a_n[/Tex], where an = f(n) and f(x) is a continuous, positive, and decreasing function for x ≥ 1. The Integral Test states that:
- If [Tex]\int_{1}^{\infty} f(x) \, dx[/Tex] converges, then the series [Tex]\sum_{n=1}^{\infty} a_n[/Tex] converges.
- If [Tex]\int_{1}^{\infty} f(x) \, dx[/Tex] diverges, then the series [Tex]\sum_{n=1}^{\infty} a_n[/Tex] diverges.
Example: Consider the series [Tex]\sum_{n=1}^{\infty} \frac{1}{n^p}[/Tex].
Solution:
Let f(x) = 1/xp.
Check if f(x) is continuous, positive, and decreasing for x ≥ 1:
f(x) = 1/xp is continuous for x ≥ 1 when p > 0.
f(x) > 0 for x > 0 when p > 0.
f(x) is decreasing for x ≥ 1 when p > 0.
Now, apply the Integral Test: [Tex]\int_{1}^{\infty} \frac{1}{x^p} \, dx[/Tex]
Evaluate the integral: [Tex] \int_{1}^{\infty} \frac{1}{x^p} \, dx = \left[ \frac{x^{1-p}}{1-p} \right]_{1}^{\infty}[/Tex]
If p > 1, the integral converges because: [Tex]\left. \frac{x^{1-p}}{1-p} \right|_{1}^{\infty} = \frac{1}{1-p} \lim_{x \to \infty} \frac{1}{x^{p-1}} = 0[/Tex].
Thus, [Tex]\sum_{n=1}^{\infty} \frac{1}{n^p}[/Tex] converges.
If p ≤ 1, the integral diverges because: [Tex]\left. \frac{x^{1-p}}{1-p} \right|_{1}^{\infty} \text{ diverges for } p \leq 1[/Tex] diverges for p ≤ 1
Thus, [Tex] \sum_{n=1}^{\infty} \frac{1}{n^p}[/Tex] diverges.
Ratio TestRatio Test provides a straightforward way to determine the convergence of an infinite series by examining the ratio of successive terms. Consider a series [Tex]\sum_{n=1}^{\infty} a_n[/Tex].
To apply the Ratio Test, we define the ratio L as follows:
[Tex]L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|[/Tex]
The Ratio Test then gives us the following criteria:
- If L < 1, the series [Tex]\sum_{n=1}^{\infty} a_n[/Tex] converges absolutely.
- If L > 1 or L = ∞, the series [Tex] \sum_{n=1}^{\infty} a_n [/Tex] diverges.
- If L = 1, the test is inconclusive, and the series may converge or diverge; other methods must be used to determine the behavior of the series.
Example: Consider the series [Tex]\sum_{n=1}^{\infty} \frac{n!}{n^n}[/Tex].
Solution:
Identify [Tex]a_n = \frac{n!}{n^n}[/Tex].
Calculate the ratio of successive terms: [Tex] \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} \right|[/Tex]
Simplify the expression: [Tex]\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(n+1) \cdot n!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} \right| = \left| \frac{n^n}{(n+1)^n} \cdot \frac{1}{n+1} \right|[/Tex]
Further simplification: [Tex]\left| \frac{a_{n+1}}{a_n} \right| = \left| \left( \frac{n}{n+1} \right)^n \cdot \frac{1}{n+1} \right|[/Tex]
Take the limit as n → ∞: [Tex]L = \lim_{n \to \infty} \left| \left( \frac{n}{n+1} \right)^n \cdot \frac{1}{n+1} \right|[/Tex]
Recognize that [Tex] \left( \frac{n}{n+1} \right)^n \approx e^{-1}[/Tex] as n → ∞: [Tex]L = \lim_{n \to \infty} \left( e^{-1} \cdot \frac{1}{n+1} \right) = 0[/Tex]
Since L = 0 < 1, the series [Tex]\sum_{n=1}^{\infty} \frac{n!}{n^n}[/Tex] converges absolutely.
Comparison TestConsider two series [Tex] \sum_{n=1}^{\infty} a_n \text{ and } \sum_{n=1}^{\infty} b_n[/Tex], where an, bn > 0 for all n.
Direct Comparison Test
- If 0 ≤ an ≤ bn for all n and [Tex]\sum_{n=1}^{\infty} b_n[/Tex] converges, then [Tex]\sum_{n=1}^{\infty} a_n[/Tex] an also converges.
- If 0 ≤ bn ≤ an for all n and [Tex]\sum_{n=1}^{\infty} b_n[/Tex] diverges, then [Tex]\sum_{n=1}^{\infty} a_n[/Tex] also diverges.
Limit Comparison Test
If an > 0 and bn > 0 for all n and the limit
[Tex]L = \lim_{n \to \infty} \frac{a_n}{b_n}[/Tex]
Exists and is a positive finite number (i.e., 0 < L < ∞), then either both series [Tex]\sum_{n=1}^{\infty} a_n \text{ and } \sum_{n=1}^{\infty} b_n[/Tex] converge or both diverge.
Example (Direct Comparison Test): Consider the series [Tex]\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}[/Tex].
Solution:
Choose a comparison series [Tex]\sum_{n=1}^{\infty} b_n[/Tex] such that 0 ≤ an ≤ bn.
Let bn = 1/n2.
Note that [Tex]\frac{1}{n^2 + 1} \leq \frac{1}{n^2}[/Tex] for all n ≥ 1.
Since, [Tex] \sum_{n=1}^{\infty} \frac{1}{n^2}[/Tex] is a p-series with p = 2 > 1, it converges. By the Direct Comparison Test, [Tex]\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}[/Tex] also converges.
Example (Limit Comparison Test): Consider the series [Tex]\sum_{n=1}^{\infty} \frac{n^2}{n^3 + 1}[/Tex].
Solution:
Choose a comparison series [Tex]\sum_{n=1}^{\infty} b_n[/Tex].
Let bn = 1/n.
Compute the limit: [Tex]\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{n^2}{n^3 + 1}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} = \lim_{n \to \infty} \frac{n^3}{n^3(1 + \frac{1}{n^3})} = \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n^3}} = 1[/Tex]
Since the limit is a positive finite number, and [Tex]\sum_{n=1}^{\infty} \frac{1}{n}[/Tex] (harmonic series) diverges, the series [Tex] \sum_{n=1}^{\infty} \frac{n^2}{n^3 + 1}[/Tex] also diverges by the Limit Comparison Test.
Root Test (Cauchy’s Root Test)Root Test, also known as Cauchy’s Root Test, is a method used to determine the convergence or divergence of an infinite series by examining the n-th root of the absolute value of the terms of the series.
For a series [Tex]\sum_{n=1}^{\infty} a_n[/Tex], define L as: [Tex]L = \lim_{n \to \infty} \sqrt[n]{|a_n|}[/Tex]
The conclusions of the Root Test are as follows:
- If L < 1, the series [Tex]\sum_{n=1}^{\infty} a_n[/Tex] an converges absolutely.
- If L > 1, the series [Tex]\sum_{n=1}^{\infty} a_n [/Tex]an diverges.
- If L = 1, the Root Test is inconclusive, and no conclusion can be drawn.
Example: Consider the series [Tex]\sum_{n=1}^{\infty} \left( \frac{3}{4} \right)^n[/Tex].
Solution:
Calculate the nth root of the absolute value of the terms:
[Tex]\sqrt[n]{\left| \left( \frac{3}{4} \right)^n \right|} = \sqrt[n]{\left( \frac{3}{4} \right)^n} = \frac{3}{4}[/Tex]
Take the limit as n approaches infinity:
[Tex]L = \lim_{n \to \infty} \sqrt[n]{\left| \left( \frac{3}{4} \right)^n \right|} = \frac{3}{4}[/Tex]
Since L = 3/4 < 1, the series [Tex]\sum_{n=1}^{\infty} \left( \frac{3}{4} \right)^n[/Tex] converges absolutely.
Alternating Series Test (Leibniz’s Test) Alternating Series Test, also known as Leibniz’s Test, is used to determine the convergence of an alternating series. An alternating series is one whose terms alternate in sign. Consider an alternating series of the form: [Tex]\sum_{n=1}^{\infty} (-1)^{n-1} b_n[/Tex]
The series [Tex]\sum_{n=1}^{\infty} (-1)^{n-1} b_n[/Tex] converges if the following two conditions are met:
- bn is decreasing: bn + 1 ≤ bn for all n ≥ N (for some index N).
- bn tends to zero: limn→∞bn = 0.
Example: Consider the alternating harmonic series: [Tex]\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} = 1 – \frac{1}{2} + \frac{1}{3} – \frac{1}{4} + \frac{1}{5} – \cdots[/Tex]
Solution:
Check if bn = 1/n is decreasing: [Tex]b_{n+1} = \frac{1}{n+1} \leq \frac{1}{n} = b_n[/Tex]
Check if bn converges to 0: [Tex]\lim_{n \to \infty} \frac{1}{n} = 0[/Tex]
Since both conditions are met, the series [Tex]\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} [/Tex]converges.
Raabe’s TestRaabe’s Test provides a more refined approach to analyzing the convergence of an infinite series by examining the behavior of the ratio of successive terms, similar to the Ratio Test but with an additional factor that can make it more sensitive to convergence. Consider a series [Tex]\sum_{n=1}^{\infty}a_n[/Tex]. To apply Raabe’s Test, we define the limit RRR as follows:
[Tex]R = \lim_{n \to \infty} n \left(1 – \left| \frac{a_{n+1}}{a_n} \right|\right)[/Tex]
The test then provides the following criteria:
- If R > 1, the series \sum_{n=1}^{\infty} a_n converges absolutely.
- If R < 1, the series \sum_{n=1}^{\infty} a_n diverges.
- If R = 1, the test is inconclusive, and the series may converge or diverge; other methods must be used to determine the behavior of the series.
Example: Consider the series [Tex]\sum_{n=1}^{\infty} \frac{n!}{n^n}[/Tex].
Solution:
Calculate the ratio of successive terms:
\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} \right| = \left| \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} \right| = \left| \frac{(n+1)n^n}{(n+1)^{n+1}} \right| = \left| \frac{n^n}{(n+1)^n} \cdot \frac{1}{n+1} \right|
Simplify the expression:
\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{n^n}{(n+1)^n} \cdot \frac{1}{n+1} \right| = \left| \left(\frac{n}{n+1}\right)^n \cdot \frac{1}{n+1} \right|
Take the limit of \left(1 – \left| \frac{a_{n+1}}{a_n} \right|\right):
R = \lim_{n \to \infty} n \left( 1 – \left( \frac{n}{n+1} \right)^n \cdot \frac{1}{n+1} \right)
As n becomes very large, \left( \frac{n}{n+1} \right)^n approaches 1/e, so:
R = \lim_{n \to \infty} n \left( 1 – \frac{1}{e(n+1)} \right) \approx \lim_{n \to \infty} n \left( 1 – \frac{1}{en} \right) = \lim_{n \to \infty} n – \frac{n}{en} = \lim_{n \to \infty} n – \frac{1}{e} = \infty
Since R = ∞, which is greater than 1, the series \sum_{n=1}^{\infty} \frac{n!}{n^n} converges absolutely.
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Practice Problems on Convergence TestsProblem 1: Consider the geometric series [Tex] \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n[/Tex].
Problem 2: Consider the alternating series [Tex]\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}[/Tex].
Problem 3: Consider the series [Tex]\sum_{n=1}^{\infty} \frac{n!}{3^n}[/Tex].
Problem 4: Consider the series [Tex]\sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n[/Tex].
Problem 5: Consider the series [Tex]\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}[/Tex].
Problem 6: Consider the series [Tex]\sum_{n=1}^{\infty} \frac{1}{n^2}[/Tex].
FAQs on Convergence TestsDefine convergence tests.Convergence tests are mathematical tools used to determine whether a given infinite series converges or diverges. These tests help in understanding the behavior of series and are essential in calculus and analysis.
What is the Ratio Test?The Ratio Test is a convergence test where the ratio of consecutive terms is examined. If [Tex]\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1[/Tex], the series∑an converges absolutely. If the limit is greater than 1, the series diverges. If the limit equals 1, the test is inconclusive.
How does the Root Test work?The Root Test examines the nth root of the absolute value of the nth term. If [Tex]\lim_{n \to \infty} \sqrt[n]{|a_n|} < 1[/Tex], the series ∑an converges absolutely. If the limit is greater than 1, the series diverges. If the limit equals 1, the test is inconclusive.
What is the Comparison Test?The Comparison Test compares the series ∑an with a known series ∑bn. If 0 ≤ an ≤ bn for all n and ∑bn converges, then ∑an also converges. If an ≥ bn ≥ 0 for all n and ∑bn diverges, then ∑an also diverges.
What is the Alternating Series Test?The Alternating Series Test, also known as the Leibniz Test, is used for series whose terms alternate in sign. If the absolute value of the terms ∣an∣ decreases monotonically to 0, then the alternating series [Tex] \sum (-1)^n a_n[/Tex] converges.
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