# 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.

- In performing the Divergence Test, replace the sigma notation with a limit.
- 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.
- 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}^{∞} [n^{2}] / [5n^{2}+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}_{→}_{∞} (n^{2}) / (5n^{2}+4)

lim_{n}_{→}_{∞ }(a_{n}) = lim_{n}_{→}_{∞} (n^{2}) / (5n^{2}+4) = lim_{n}_{→}_{∞ }[(d/dn) n^{2}] / [(d/dn) (5n^{2}+4)]

lim_{n}_{→}_{∞ }[(d/dn) n^{2}] / [(d/dn) (5n^{2}+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}_{→}_{∞} (n^{2}) / (5n^{2}+4)

lim_{n}_{→}_{∞} (n^{2}) / (5n^{2}+4) = lim_{n}_{→}_{∞} (1) / [5 + (4/n^{2})]

lim_{n}_{→}_{∞} (1) / [5 + (4/n^{2})] = 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 + x^{2}]

**Solution**

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

∑_{x}_{=}_{1}^{∞} [1] / [√1 + x^{2}] = lim_{x}_{→}_{∞} [1] / [√1 + x^{2}]

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

lim_{x}_{→}_{∞} [1] / [√1 + x^{2}] = lim_{x}_{→}_{∞} [1] / [√1 + x^{2}] x [(1/x) / (1/x)]

lim_{x}_{→}_{∞} [1] / [√1 + x^{2}] = lim_{x}_{→}_{∞} (1/x) / [(1/x) √1 + x^{2}]

lim_{x}_{→}_{∞} [1] / [√1 + x^{2}] = lim_{x}_{→}_{∞} (1/x) / [(√1/x^{2}) √1 + x^{2}]

lim_{x}_{→}_{∞} [1] / [√1 + x^{2}] = lim_{x}_{→}_{∞ }(1/x) / [√(1/k^{2}) + 1]

lim_{x}_{→}_{∞} [1] / [√1 + x^{2}] = 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 /n^{3})

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**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 /n^{3}) = lim_{n}_{→}_{∞ }(1 /n^{3})

lim_{n}_{→}_{∞ }(1 /n^{3}) = 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 /n^{3}) 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.

S_{0} = (−1)^{0}

S_{0 }= 1

S_{1} = (−1)^{0} + (−1)^{1}

S_{1} = 1 – 1

S_{1} = 0

S_{2} = (−1)^{0} + (−1)^{1} + (−1)^{2}

S_{2} = 1 – 1 + 1

S_{2} = 1

S_{3} = (−1)^{0} + (−1)^{1} + (−1)^{2} + (−1)^{3}

S_{3} = 1 – 1 + 1 – 1

S_{3} = 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}^{∞}[2n^{2} + n^{3}] / [4n^{3} – 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}^{∞}[2n^{2} + n^{3}] / [4n^{3} – 3n] = lim_{n}_{→}_{∞} [2n^{2} + n^{3}] / [4n^{3} – 3n]

lim_{n}_{→}_{∞} [2n^{2} + n^{3}] / [4n^{3} – 3n] = lim_{n}_{→}_{∞} [4n + 3n^{2}] / [12n^{2} – 3]

lim_{n}_{→}_{∞} [4n + 3n^{2}] / [12n^{2} – 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}^{∞}[5n^{2} – n^{3}] / [3 + 8n^{3}] 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 n^{3}.

∑_{n=0}^{∞}[5n^{2} – n^{3}] / [3 + 8n^{3}] = lim_{n}_{→}_{∞} [5n^{2} – n^{3}] / [3 + 8n^{3}]

lim_{n}_{→}_{∞} [5n^{2} – n^{3}] / [3 + 8n^{3}] = lim_{n}_{→}_{∞} [5/n – 1] / [(3/n^{3}) + 8]

With the exception of indeterminate form,

lim_{n}_{→}_{∞} [5/n – 1] / [(3/n^{3}) + 8] = lim_{n}_{→}_{∞} [5/n – 1] / lim_{n}_{→}_{∞} [(3/n^{3}) + 8]

lim_{n}_{→}_{∞} [5/n – 1] / [(3/n^{3}) + 8] = -1/8

**Answer**

The limit of the series ∑_{n=0}^{∞}[5 – 6n^{2}] / [3 - 8n^{2}] is not zero and so by the Divergence Test the series diverges.

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