Convergence Tests

In mathematics, convergence tests are essential tools used to determine whether an infinite series converges (approaches a finite value) or diverges (grows without bound or oscillates).

An infinite series is an expression of the form \sum_{n=1}^{\infty} a_n​

where a_n​ represents the terms of the series.

• Convergence: As more terms are added, the sum approaches a finite limit.
• Divergence: The sum grows indefinitely or oscillates without settling to a single value.

Since it’s impossible to compute infinitely many terms directly, convergence tests provide criteria and techniques to decide whether a series converges or diverges. Different tests are suited for different types of series.

Types of Convergence Tests

Some of the common tests to check the 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 Test

Integral Test is used to determine the convergence or divergence of an infinite series by comparing it to an improper integral. Consider a series \sum_{n=1}^{\infty}a_n​, where a_n = f(n) and f(x) is a continuous, positive, and decreasing function for x ≥ 1. The Integral Test states that:

• If it \int_{1}^{\infty} f(x) \, dxconverges, then the series \sum_{n=1}^{\infty} a_nconverges.
• If \int_{1}^{\infty} f(x) \, dxdiverges, then the series \sum_{n=1}^{\infty} a_ndiverges.

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

Solution:

Let f(x) = 1/x^p​.

Check if f(x) is continuous, positive, and decreasing for x ≥ 1:

f(x) = 1/x^p​​ 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: \int_{1}^{\infty} \frac{1}{x^p} \, dx

Evaluate the integral: \int_{1}^{\infty} \frac{1}{x^p} \, dx = \left[ \frac{x^{1-p}}{1-p} \right]_{1}^{\infty}​

If p > 1, the integral converges because: \left. \frac{x^{1-p}}{1-p} \right|_{1}^{\infty} = \frac{1}{1-p} \lim_{x \to \infty} \frac{1}{x^{p-1}} = 0.

Thus, \sum_{n=1}^{\infty} \frac{1}{n^p} converges.

If p ≤ 1, the integral diverges because: \left. \frac{x^{1-p}}{1-p} \right|_{1}^{\infty} \text{ diverges for } p \leq 1​

Thus, \sum_{n=1}^{\infty} \frac{1}{n^p} diverges.

Ratio Test

Ratio Test provides a straightforward way to determine the convergence of an infinite series by examining the ratio of successive terms. Consider a series \sum_{n=1}^{\infty} a_n​.

To apply the Ratio Test, we define the ratio L as follows:

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

The Ratio Test then gives us the following criteria:

• If L < 1, the series \sum_{n=1}^{\infty} a_nconverges absolutely.
• If L > 1 or L = ∞, the series \sum_{n=1}^{\infty} a_n 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 \sum_{n=1}^{\infty} \frac{n!}{n^n}.

Solution:

Identify a_n = \frac{n!}{n^n}.

Calculate the ratio of successive terms: \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} \right|​

Simplify the expression: \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|​

Further simplification: \left| \frac{a_{n+1}}{a_n} \right| = \left| \left( \frac{n}{n+1} \right)^n \cdot \frac{1}{n+1} \right|​

Take the limit as n → ∞: L = \lim_{n \to \infty} \left| \left( \frac{n}{n+1} \right)^n \cdot \frac{1}{n+1} \right|

Recognize that \left( \frac{n}{n+1} \right)^n \approx e^{-1} as n → ∞: L = \lim_{n \to \infty} \left( e^{-1} \cdot \frac{1}{n+1} \right) = 0

Since L = 0 < 1, the series \sum_{n=1}^{\infty} \frac{n!}{n^n}​ converges absolutely.

Comparison Test

Consider two series \sum_{n=1}^{\infty} a_n \text{ and } \sum_{n=1}^{\infty} b_n​, where a_n, b_n > 0 for all n.

Direct Comparison Test

1. If 0 ≤ a_n ≤ b_n for all n and \sum_{n=1}^{\infty} b_n​ converges, then \sum_{n=1}^{\infty} a_n ​an​ also converges.
2. If 0 ≤ b_n ≤ a_n for all n and \sum_{n=1}^{\infty} b_n​ diverges, then \sum_{n=1}^{\infty} a_n​ also diverges.

Limit Comparison Test

If a_n > 0 and b_n > 0 for all n and the limit

L = \lim_{n \to \infty} \frac{a_n}{b_n}

Exists and is a positive finite number (i.e., 0 < L < ∞), then either both series \sum_{n=1}^{\infty} a_n \text{ and } \sum_{n=1}^{\infty} b_n​ converge or both diverge.

Example (Direct Comparison Test): Consider the series \sum_{n=1}^{\infty} \frac{1}{n^2 + 1}.

Solution:

Choose a comparison series \sum_{n=1}^{\infty} b_n​ such that 0 ≤ a_n ≤ b_n.

Let b_n = 1/n²​.

Note that \frac{1}{n^2 + 1} \leq \frac{1}{n^2} for all n ≥ 1.

Since​, \sum_{n=1}^{\infty} \frac{1}{n^2} is a p-series with p = 2 > 1, it converges. By the Direct Comparison Test, \sum_{n=1}^{\infty} \frac{1}{n^2 + 1}​ also converges.

Example (Limit Comparison Test): Consider the series \sum_{n=1}^{\infty} \frac{n^2}{n^3 + 1}​.

Solution:

Choose a comparison series \sum_{n=1}^{\infty} b_n.

Let b_n = 1/n​.

Compute the limit: \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

Since the limit is a positive finite number, and \sum_{n=1}^{\infty} \frac{1}{n} (harmonic series) diverges, the series \sum_{n=1}^{\infty} \frac{n^2}{n^3 + 1} 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 \sum_{n=1}^{\infty} a_n​, define L as: L = \lim_{n \to \infty} \sqrt[n]{|a_n|}

The conclusions of the Root Test are as follows:

• If L < 1, the series \sum_{n=1}^{\infty} a_n ​an​ converges absolutely.
• If L > 1, the series \sum_{n=1}^{\infty} a_n ​an​ diverges.
• If L = 1, the Root Test is inconclusive, and no conclusion can be drawn.

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

Solution:

Calculate the n^th root of the absolute value of the terms:

\sqrt[n]{\left| \left( \frac{3}{4} \right)^n \right|} = \sqrt[n]{\left( \frac{3}{4} \right)^n} = \frac{3}{4}

Take the limit as n approaches infinity:

L = \lim_{n \to \infty} \sqrt[n]{\left| \left( \frac{3}{4} \right)^n \right|} = \frac{3}{4}

Since L = 3/4 < 1, the series \sum_{n=1}^{\infty} \left( \frac{3}{4} \right)^n 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: \sum_{n=1}^{\infty} (-1)^{n-1} b_n​

The series \sum_{n=1}^{\infty} (-1)^{n-1} b_n converges if the following two conditions are fulfilled :

1. b_n​ is decreasing: b_{n + 1} ≤ b_n for all n ≥ N (for some index N).
2. b_n​ tends to zero: lim_⁡n→∞b_n = 0.

Example: Consider the alternating harmonic series: \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots

Solution:

Check if b_n = 1/n​ is decreasing: b_{n+1} = \frac{1}{n+1} \leq \frac{1}{n} = b_n

Check if b_n converges to 0: \lim_{n \to \infty} \frac{1}{n} = 0

Since both conditions are met, the series \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} converges.

Raabe’s Test

Raabe’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 \sum_{n=1}^{\infty}a_n​. To apply Raabe’s Test, we define the limit R as follows:

R = \lim_{n \to \infty} n \left(1 - \left| \frac{a_{n+1}}{a_n} \right|\right)

• 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 \sum_{n=1}^{\infty} \frac{n!}{n^n}.

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.

Related Article:

• Conditional Convergence
• Absolutely Convergence
• Absolute and Conditional Convergence
• Radius of Convergence 

Applications of Covergence Test

• Mathematics: To test the covergence of power like series like Taylor series.
• Engineering: To ensure solution of differntial equation (series solutions) are valid.
• Finance: Used in annuties and present value (geometric series).
• Comptution: Used to approximate function like e^x, sinx with small error.

Practice Problems on Convergence Tests

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

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

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

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

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

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

Problem 7: Consider the series\sum_{n=1}^{\infty} \frac{1}{n} \quad.

Problem 8: Consider the series \sum_{n=1}^{\infty} \frac{1}{n^4}.

Problem 9: Consider the series \sum_{n=1}^{\infty} \frac{n^2}{2^n}.

Problem 10: Consider the series \sum_{n=1}^{\infty} \frac{n^3}{3^n}.