How To Tell If A Series Is Convergent Or Divergent

Diving into the world of infinite series can feel like navigating a vast ocean. One of the fundamental questions that arises is whether a given series converges to a finite value or diverges indefinitely. Understanding the convergence or divergence of a series is crucial in various fields such as calculus, physics, engineering, and computer science. Determining whether a series is convergent or divergent requires a toolbox of tests and techniques, each suited to different types of series.

Unveiling the Essence: Convergence and Divergence Defined

Before we plunge into the tests, let's clarify what convergence and divergence truly mean.

A series is the sum of an infinite sequence of terms. Mathematically, if we have a sequence {a_n}, the series is represented as:

∑[n=1 to ∞] a_n = a_1 + a_2 + a_3 + ...

The partial sum, denoted as S_n, is the sum of the first n terms of the series:

S_n = a_1 + a_2 + a_3 + ... + a_n

• A series converges if the sequence of its partial sums {S_n} approaches a finite limit L as n approaches infinity. In other words:

  lim [n→∞] S_n = L

  If this limit exists and is finite, the series converges, and L is the sum of the series.

• A series diverges if the sequence of its partial sums {S_n} does not approach a finite limit. This can happen in several ways:

  • The partial sums grow without bound (approach infinity).
  • The partial sums oscillate between two or more values.
  • The partial sums behave erratically without settling down.

The Arsenal of Convergence and Divergence Tests

Now, let's explore the various tests that help us determine the fate of a series.

1. The Divergence Test (or the nth-Term Test)

This is often the first test you should apply. It's simple and can quickly identify many divergent series.

Theorem: If lim [n→∞] a_n ≠ 0, then the series ∑[n=1 to ∞] a_n diverges.

Important Note: If lim [n→∞] a_n = 0, the test is inconclusive. The series may converge or diverge; further testing is required. The divergence test only proves divergence.

Example:

Consider the series ∑[n=1 to ∞] n/(n+1).

lim [n→∞] n/(n+1) = 1 ≠ 0. Therefore, the series diverges.

2. The Integral Test

The integral test connects a series to an improper integral. It's particularly useful when the terms of the series resemble a function that is easy to integrate.

Theorem: Let f(x) be a continuous, positive, and decreasing function on the interval [1, ∞). Then, the series ∑[n=1 to ∞] f(n) converges if and only if the improper integral ∫[1 to ∞] f(x) dx converges.

In other words:

• If ∫[1 to ∞] f(x) dx converges, then ∑[n=1 to ∞] f(n) converges.
• If ∫[1 to ∞] f(x) dx diverges, then ∑[n=1 to ∞] f(n) diverges.

Example:

Consider the series ∑[n=1 to ∞] 1/n^2.

Let f(x) = 1/x^2. This function is continuous, positive, and decreasing on [1, ∞).

∫[1 to ∞] (1/x^2) dx = lim [t→∞] ∫[1 to t] (1/x^2) dx = lim [t→∞] [-1/x]_1^t = lim [t→∞] (-1/t + 1) = 1

Since the integral converges to 1, the series ∑[n=1 to ∞] 1/n^2 also converges. (Note: the integral test doesn't tell you what the series converges to, just that it converges).

3. The Comparison Test

The comparison test allows you to compare a given series with another series whose convergence or divergence is known.

Theorem: Suppose ∑a_n and ∑b_n are series with positive terms.

• If ∑b_n converges and a_n ≤ b_n for all n greater than some integer N, then ∑a_n also converges.
• If ∑b_n diverges and a_n ≥ b_n for all n greater than some integer N, then ∑a_n also diverges.

Key Idea: Find a known convergent series that is larger than your series, or a known divergent series that is smaller than your series.

Example:

Consider the series ∑[n=1 to ∞] 1/(n^2 + 1).

We know that ∑[n=1 to ∞] 1/n^2 converges (p-series with p=2 > 1).

Since 1/(n^2 + 1) < 1/n^2 for all n, and ∑[n=1 to ∞] 1/n^2 converges, the series ∑[n=1 to ∞] 1/(n^2 + 1) also converges by the comparison test.

4. The Limit Comparison Test

The limit comparison test is a more powerful version of the comparison test. It avoids the need to establish a direct inequality between the terms.

Theorem: Suppose ∑a_n and ∑b_n are series with positive terms. If

lim [n→∞] (a_n / b_n) = c

where 0 < c < ∞ (c is a finite, positive number), then both series either converge or both series diverge.

Key Idea: If the ratio of the terms approaches a finite, positive number, the two series behave similarly.

Example:

Consider the series ∑[n=1 to ∞] (n + 1)/(n^3 + 2n^2 + 1).

Let's compare it to the series ∑[n=1 to ∞] 1/n^2, which we know converges.

lim [n→∞] ((n + 1)/(n^3 + 2n^2 + 1)) / (1/n^2) = lim [n→∞] (n^3 + n^2) / (n^3 + 2n^2 + 1) = 1

Since the limit is 1 (a finite, positive number), and ∑[n=1 to ∞] 1/n^2 converges, the series ∑[n=1 to ∞] (n + 1)/(n^3 + 2n^2 + 1) also converges by the limit comparison test.

5. The Ratio Test

The ratio test is particularly useful for series involving factorials or exponential terms.

Theorem: Let ∑a_n be a series. Define

L = lim [n→∞] |a_(n+1) / a_n|

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

Key Idea: The ratio test examines the relative size of consecutive terms. If the terms are decreasing rapidly enough (L < 1), the series converges.

Example:

Consider the series ∑[n=1 to ∞] n! / n^n.

L = lim [n→∞] |((n+1)! / (n+1)^(n+1)) / (n! / n^n)| = lim [n→∞] ((n+1)! * n^n) / (n! * (n+1)^(n+1)) = lim [n→∞] (n+1) * n^n / (n+1)^(n+1) = lim [n→∞] n^n / (n+1)^n = lim [n→∞] (n/(n+1))^n = lim [n→∞] (1/(1 + 1/n))^n = 1/e

Since 1/e < 1, the series ∑[n=1 to ∞] n! / n^n converges absolutely by the ratio test.

6. The Root Test

The root test is another test useful for series where the terms involve exponents.

Theorem: Let ∑a_n be a series. Define

L = lim [n→∞] |a_n|^(1/n)

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

Key Idea: The root test examines the nth root of the absolute value of the terms.

Example:

Consider the series ∑[n=1 to ∞] ( (2n + 3) / (3n + 2) )^n

L = lim [n→∞] |( (2n + 3) / (3n + 2) )^n |^(1/n) = lim [n→∞] (2n + 3) / (3n + 2) = 2/3

Since 2/3 < 1, the series ∑[n=1 to ∞] ( (2n + 3) / (3n + 2) )^n converges absolutely by the root test.

7. Alternating Series Test

This test applies specifically to alternating series, where the terms alternate in sign.

Theorem (Alternating Series Test): Consider the alternating series ∑[n=1 to ∞] (-1)^n b_n or ∑[n=1 to ∞] (-1)^(n+1) b_n, where b_n > 0 for all n. If the following two conditions are satisfied:

• b_(n+1) ≤ b_n for all n greater than some integer N (the terms are decreasing in magnitude)
• lim [n→∞] b_n = 0

Then the series converges.

Key Idea: For an alternating series to converge, the terms must decrease in magnitude and approach zero.

Example:

Consider the series ∑[n=1 to ∞] (-1)^(n+1) / n.

• b_n = 1/n, which is positive for all n.
• 1/(n+1) < 1/n, so the terms are decreasing.
• lim [n→∞] 1/n = 0

Therefore, the series ∑[n=1 to ∞] (-1)^(n+1) / n converges by the alternating series test. This is a classic example of a conditionally convergent series (see below).

8. Absolute and Conditional Convergence

These concepts further refine our understanding of convergence.

• A series ∑a_n converges absolutely if the series ∑|a_n| converges. If a series converges absolutely, it also converges.
• A series ∑a_n converges conditionally if the series ∑a_n converges, but the series ∑|a_n| diverges.

Example:

We saw that ∑[n=1 to ∞] (-1)^(n+1) / n converges. However, the series of absolute values, ∑[n=1 to ∞] |(-1)^(n+1) / n| = ∑[n=1 to ∞] 1/n, is the harmonic series, which diverges. Therefore, the series ∑[n=1 to ∞] (-1)^(n+1) / n converges conditionally.

If we had shown that ∑[n=1 to ∞] |a_n| converges using a test like the ratio test or root test, then we would know that ∑a_n converges absolutely.

Strategic Application: A Guide to Choosing the Right Test

Selecting the appropriate test is crucial for efficiently determining convergence or divergence. Here's a general strategy:

1. Divergence Test: Always start with the divergence test. It's quick and can immediately rule out many divergent series.
2. Geometric Series: Recognize geometric series. They have a simple convergence/divergence criterion. A geometric series ∑ ar^(n-1) converges if |r| < 1 and diverges if |r| ≥ 1.
3. p-Series: Recognize p-series. A p-series ∑ 1/n^p converges if p > 1 and diverges if p ≤ 1.
4. Integral Test: Consider the integral test if the terms of the series resemble a function that is easy to integrate and satisfies the conditions of the test (continuous, positive, decreasing).
5. Comparison Tests (Comparison Test and Limit Comparison Test): Use comparison tests when you can relate your series to a known convergent or divergent series (p-series or geometric series are common choices). The limit comparison test is often easier to apply than the direct comparison test.
6. Ratio Test: Use the ratio test when the series involves factorials or exponential terms.
7. Root Test: Use the root test when the terms of the series involve exponents.
8. Alternating Series Test: If the series is alternating, apply the alternating series test.

Common Series and Their Behavior

It's helpful to be familiar with the behavior of some common series:

• Harmonic Series: ∑[n=1 to ∞] 1/n diverges.
• p-Series: ∑[n=1 to ∞] 1/n^p converges if p > 1 and diverges if p ≤ 1.
• Geometric Series: ∑[n=0 to ∞] ar^n converges to a/(1-r) if |r| < 1 and diverges if |r| ≥ 1.
• The series ∑ 1/n! converges (to e).

Examples Illustrating the Tests

Let's solidify our understanding with a few more examples:

Example 1: Determine whether the series ∑[n=1 to ∞] (2^n) / (n^3) converges or diverges.

• Try the Ratio Test:

  L = lim [n→∞] |(2^(n+1) / (n+1)^3) / (2^n / n^3)| = lim [n→∞] (2^(n+1) * n^3) / (2^n * (n+1)^3) = lim [n→∞] 2 * (n^3 / (n+1)^3) = 2

  Since L = 2 > 1, the series diverges by the ratio test.

Example 2: Determine whether the series ∑[n=2 to ∞] 1 / (n ln(n)) converges or diverges.

• Try the Integral Test:

  Let f(x) = 1 / (x ln(x)). This function is continuous, positive, and decreasing on [2, ∞).

  ∫[2 to ∞] (1 / (x ln(x))) dx = lim [t→∞] ∫[2 to t] (1 / (x ln(x))) dx

  Let u = ln(x), then du = (1/x) dx

  = lim [t→∞] ∫[ln(2) to ln(t)] (1/u) du = lim [t→∞] [ln(u)]_[ln(2) to ln(t)] = lim [t→∞] (ln(ln(t)) - ln(ln(2))) = ∞

  Since the integral diverges, the series ∑[n=2 to ∞] 1 / (n ln(n)) also diverges by the integral test.

Example 3: Determine whether the series ∑[n=1 to ∞] (-1)^n (n / (n^2 + 1)) converges or diverges.

• Try the Alternating Series Test:

  Let b_n = n / (n^2 + 1)

  1. Is b_n decreasing? Consider the function f(x) = x / (x^2 + 1). Its derivative is f'(x) = (1 - x^2) / (x^2 + 1)^2. For x > 1, f'(x) < 0, so f(x) is decreasing. Therefore, b_n is decreasing for n > 1.

  2. Does lim [n→∞] b_n = 0?

     lim [n→∞] (n / (n^2 + 1)) = 0

  Since both conditions of the alternating series test are satisfied, the series converges. To check for absolute convergence, we would need to analyze ∑[n=1 to ∞] n / (n^2 + 1), which diverges (by the limit comparison test with 1/n). Therefore, this series converges conditionally.

Conclusion: Mastering the Art of Series Analysis

Determining the convergence or divergence of an infinite series is a fundamental skill in calculus and its applications. By understanding the definitions of convergence and divergence and mastering the various tests available, you can confidently analyze a wide range of series. Remember to start with the simpler tests like the divergence test and then strategically apply more sophisticated tests as needed. With practice and a solid understanding of the underlying principles, you can navigate the world of infinite series with ease. Don't be afraid to experiment with different tests and approaches until you find the one that works best for a given series. Good luck!