Divergence Test: Determining If a Series Converges or Diverges

Ray is a licensed engineer in the Philippines. He loves to write about mathematics and civil engineering.

What Is a Divergence Test?

The Divergence Test is a method used to determine whether the sum of a series diverges, and if it does, it is impossible to converge. If the series does not diverge, then the test is inconclusive. Take note that the divergence test is not a test for convergence.

We have learned that if a series converges, then the summed sequence's terms must converge to 0. How about when the series converges to a non-zero or does not exist?

Nth Term Test for Divergence Definition

The nth term for Divergence states that if lim_n→∞ a_n does not exist, or if lim_n→∞ (a_n ≠ 0), then the series ∑_n=1^∞ (a_n) is divergent. In other words, if the limit of a_n is not zero or does not exist, then the sum diverges.

Suppose we have a series ∑_n=1^∞ (a_n) where the sequence a_n converges to a non-zero limit. For instance, let us try to test the divergence of the constant a_n=5. The partial sums of the series are 2n (unbounded), so the series doesn’t converge. Generally, any constant sequence a_n = a (a ≠ 0) will diverge.

How to Do a Divergence Test

Remember to use the Divergence Test when it seems like the series terms' limit does not converge to zero. Take note to use the Divergence Test if a glance suggests that a series's limit is not zero or does not exist. Indicated below is how one would know if a series is a convergence or divergence.

1. In performing the Divergence Test, replace the sigma notation with a limit.
2. Find the limit of the series using the different methods. Remember that one may encounter problems that use pure algebraic simplification to solve the limit, and some utilize the L'Hopital's Rule.
3. If the series's limit is not equal to zero or does not exist, then the series is divergent. Always be careful with two of the few mistakes when solving for the Divergence Test:

• When the limit equals zero, the series converges.
• When the limit equals zero, the convergence of the given series cannot be established.

When using the Nth Term Divergence Test and the limit results to zero, the test yields no conclusion, or the series is inconclusive. However, use a different test to determine the convergence or divergence of a series.

Example 1: Using the Test for Divergence

Show that the series ∑_n=1^∞ [n²] / [5n²+4] diverges.

Solution 1

The divergence test asks whether the nth term of the series has a non-zero limit. If the result is a non-zero value, then the series diverges. Using L’Hopital’s Rule, find the limit of the given function.

lim_n→∞ (a_n) = lim_n→∞ (n²) / (5n²+4)

lim_n→∞ (a_n) = lim_n→∞ (n²) / (5n²+4) = lim_n→∞ [(d/dn) n²] / [(d/dn) (5n²+4)]

lim_n→∞ [(d/dn) n²] / [(d/dn) (5n²+4)] = lim_n→∞ [(d/dn) 2n] / [(d/dn) (10n)]

lim_n→∞ [(d/dn) 2n] / [(d/dn) (10n)] = 1/5 ≠ 0

Answer

Since the function’s limit is not equal to zero using L’Hopital’s Rule, the series diverges by the Test for Divergence.

Solution 2

Another solution aside from using L’Hospital’s Rule is rewriting the equation.

lim_n→∞ (a_n) = lim_n→∞ (n²) / (5n²+4)

lim_n→∞ (n²) / (5n²+4) = lim_n→∞ (1) / [5 + (4/n²)]

lim_n→∞ (1) / [5 + (4/n²)] = 1/5 ≠ 0

Answer

Since the limit of the function is not equal to zero, the series diverges by the Test for Divergence.

Example 2: Application of Divergence Test

Using the divergence test, determine whether or not the following series diverges.

∑_x=1^∞ [1] / [√1 + x²]

Solution

The first step in applying the divergence test, replace the sigma notation of the function with a limit.

∑_x=1^∞ [1] / [√1 + x²] = lim_x→∞ [1] / [√1 + x²]

Simplify the algebraic expression. Divide both the numerator and denominator by the highest power of k, which is 1.

lim_x→∞ [1] / [√1 + x²] = lim_x→∞ [1] / [√1 + x²] x [(1/x) / (1/x)]

lim_x→∞ [1] / [√1 + x²] = lim_x→∞ (1/x) / [(1/x) √1 + x²]

lim_x→∞ [1] / [√1 + x²] = lim_x→∞ (1/x) / [(√1/x²) √1 + x²]

lim_x→∞ [1] / [√1 + x²] = lim_x→∞ (1/x) / [√(1/k²) + 1]

lim_x→∞ [1] / [√1 + x²] = 0

Answer

Since the limit results to zero, the divergence test is inconclusive.

Example 3: Determining If a Series Diverges or Converges

Determine whether the series diverges using the Divergence Test for series.

∑_x=1^∞ [x + 1] / [x]

Solution

Apply the Divergence Test and replace the sigma notation with the limit of the function as x approaches ∞.

∑_x=1^∞ [x + 1] / [x] = lim_x→∞ [x + 1] / [x]

Use the L’Hopital’s Rule and differentiate both the numerator and the denominator.

lim_x→∞ [x + 1] / [x] = lim_x→∞ [(d/dx) (x + 1)] / [(d/dx) (x)]

lim_x→∞ [x + 1] / [x] = 1/1

lim_x→∞ [x + 1] / [x] = 1

Answer

The sum of the given series diverges since the limit results in 1.

Example 4: Application of Divergence Test

For each of the following series, apply the divergence test.

a. ∑_n=1^∞(n) / (4n-1)

b. ∑_n=1^∞(1 /n³)

Recommended for You

Hindu Funerals in Varanasi, India

The Witty and Wise Jane Austen: A Mini Biography

Prince John (1905-1919): The British Royal Who Most People Don't Know Existed

Solution

The divergence test merely asks whether the nth term of the series has a non-zero limit. If the result is a non-zero value, then the series diverges. Using L’Hopital’s Rule, find the limit of the functions shown above. Start with the series ∑_n=1^∞(n) / (4n-1).

∑_n=1^∞(n) / (4n-1) = lim_n→∞ (n) / (4n-1)

lim_n→∞ (n) / (4n-1) = ¼

∑_n=1^∞(1 /n³) = lim_n→∞ (1 /n³)

lim_n→∞ (1 /n³) = 1/∞ = 0

Answer

The sum of series ∑_n=1^∞(n) / (4n-1) diverges since the limit of (n) / (4n-1) as n approaches infinity is ¼ ≠ 0. On the other hand, the limit of the series ∑_n=1^∞(1 /n³) is equal to 0 which means the series is inconclusive.

Example 5: Evaluating a Series Using the Divergence Test

Determine if the given series converges or diverges.

∑_n=0^∞(−1)ⁿ

Solution

In solving this series’s divergence or convergence, the general formula for the partial sums is needless. Write down manually the first few partial sums of the series.

S₀ = (−1)⁰

S₀ = 1

S₁ = (−1)⁰ + (−1)¹

S₁ = 1 – 1

S₁ = 0

S₂ = (−1)⁰ + (−1)¹ + (−1)²

S₂ = 1 – 1 + 1

S₂ = 1

S₃ = (−1)⁰ + (−1)¹ + (−1)² + (−1)³

S₃ = 1 – 1 + 1 – 1

S₃ = 0

The solution above shows that the sequence of partial sums is {S_n} = {1,0,1,0,1,0,1, 0…}.

Answer

The sequence ∑_n=0^∞(−1)ⁿ diverges since the lim_n→_∞ does not exist.

Example 6: Applying the Divergence Test

Does the series ∑_n=1^∞[2n² + n³] / [4n³ – 3n] converge?

Solution

Use the L’Hopital’s Rule in testing the divergence of the limit of the series as n approaches ∞. Find the limit of the given function. Observe from the solution below that you have to use L’Hopital’s Rule 2 times to finally get an answer. Remember that you apply L’Hopital’s Rule many times if the previous results are still indeterminate.

∑_n=1^∞[2n² + n³] / [4n³ – 3n] = lim_n→∞ [2n² + n³] / [4n³ – 3n]

lim_n→∞ [2n² + n³] / [4n³ – 3n] = lim_n→∞ [4n + 3n²] / [12n² – 3]

lim_n→∞ [4n + 3n²] / [12n² – 3] = lim_n→∞ [4 + 6n] / [24n]

Apply the laws of limits for the final evaluation of limit. Divide the equation by the highest denominator power which is 24, leaving 4/x + 6 in the limit equation.

lim_n→∞ [4 + 6n] / [24n] = (1/24) lim_n→∞ (4/x) + 6

lim_n→∞ [4 + 6n] / [24n] = (1/24) [lim_n→∞ (4/x) + lim_n→∞ (6)]

lim_n→∞ [4 + 6n] / [24n] = (1/24) [0 + 6]

lim_n→∞ [4 + 6n] / [24n] = 1/4

Answer

Since the limit of the series is not zero, we can safely say that the series does not converge, and therefore diverges.

Example 7: Evaluating If a Series Diverges or Converges Using the Divergence Test

Determine if the series ∑_n=0^∞[5n² – n³] / [3 + 8n³] is convergent or divergent.

Solution

Take the limit of the series given and use the Divergence Test in identifying if the series is divergent or convergent. Divide the given equation by the highest denominator power which is n³.

∑_n=0^∞[5n² – n³] / [3 + 8n³] = lim_n→∞ [5n² – n³] / [3 + 8n³]

lim_n→∞ [5n² – n³] / [3 + 8n³] = lim_n→∞ [5/n – 1] / [(3/n³) + 8]

With the exception of indeterminate form,

lim_n→∞ [5/n – 1] / [(3/n³) + 8] = lim_n→∞ [5/n – 1] / lim_n→∞ [(3/n³) + 8]

lim_n→∞ [5/n – 1] / [(3/n³) + 8] = -1/8

Answer

The limit of the series ∑_n=0^∞[5 – 6n²] / [3 - 8n²] is not zero and so by the Divergence Test the series diverges.

• A Full Guide to the 30-60-90 Triangle (With Formulas and Examples)
  This article is a full guide to solving problems on 30-60-90 triangles. It includes pattern formulas and rules necessary to understand the concept of 30-60-90 triangles. There are also examples provided to show the step-by-step procedure on how to so
• Using the Factor Theorem in Finding the Factors of Polynomials (With Examples)
  This article is a full guide in solving problems involving the use of the factor theorem. It includes the definition, proof, examples and solutions about the factor theorem.
• Solving Related Rates Problems in Calculus
  Learn to solve different kinds of related rates problems in Calculus. This article is a full guide that shows the step-by-step procedure of solving problems involving related/associated rates.
• Same-Side Interior Angles: Theorem, Proof, and Examples
  In this article, you can learn the concept of the Same-Side Interior Angles Theorem in Geometry through solving various examples provided. The article also includes the Converse of the Same-Side Interior Angles Theorem and its proof.
• Limit Laws and Evaluating Limits
  This article will help you learn to evaluate limits by solving various problems in Calculus that require applying the limit laws.
• Power-Reducing Formulas and How to Use Them (With Examples)
  In this article, you can learn how to use the power-reducing formulas in simplifying and evaluating trigonometric functions of different powers.
• The Derivative of a Constant (With Examples)
  This article can help you learn how to find the derivative of constants in various forms.
• Linear Approximation and Differentials in Calculus
  Learn how to find the linear approximation or differentials of a function at a given point. This article also includes formulas, proof, and examples with solutions that can help you fully understand the Linear Approximation topic in Calculus.

This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualized advice from a qualified professional.

© 2021 Ray