The Divergence and Integral Tests

Limit Test

The first test for divergence is the limit test. The limit test essentially tells us whether or not the series is a candidate for being convergent. It is as follows:

Limit Test for Convergence

If a series $S=\sum _{n}^{\infty }{s_{n}}$ and if $\lim _{n\rightarrow \infty }{s_{n}}\neq 0$ the series must be divergent. If the limit is zero, the test is inconclusive and further analysis is needed.

This follows because if the summand does not approach zero, when $n$ becomes very large, $s_{n}$ will be close to the non-zero $L$ and the series will start behaving like an arithmetic series; remember that arithmetic series never converge.

However, one should not misuse this test. This is a test for divergence and not convergence. A series fails this test if the limit of the summand is zero, not if it is some non-zero $L$. If the limit is zero, you will need to do other tests to conclude that the series is divergent or convergent.

Example 1

Determine whether or not the series

$\sum _{n=0}^{\infty }{\frac {n+1}{n}}$

is divergent or if the limit test fails.

Solution

Because

$\lim _{n\rightarrow \infty }{\frac {n+1}{n}}=1$

the limit is not zero and so the series is divergent by the limit test.

Example 2

Determine whether or not the series

$\sum _{n=0}^{\infty }{\frac {n^{2}+5n+6}{3n^{2}+1}}$

is divergent or if the limit test fails.

Solution

Because

$\lim _{n\rightarrow \infty }{\frac {n^{2}+5n+6}{3n^{2}+1}}={\frac {1}{3}}$

the limit is not zero and so the series is divergent by the limit test.

Example 3

Determine whether or not the series

$\sum _{n=0}^{\infty }{\frac {n^{2}+5n+6}{n^{3}}}$

is divergent or if the limit test fails.

Solution

Because

$\lim _{n\rightarrow \infty }{\frac {n^{2}+5n+6}{n^{3}}}=0$

the limit is zero and so the test is inconclusive. Further analysis is needed to determine whether or not the series converges or diverges.

Integral Test

The next test for convergence for infinite series is the integral test. The integral test utilizes the fact that an integral is essentially an Riemann Sum—which is itself an infinite sum—over an infinite interval which is useful because integration is relatively straight forward and familiar.

The test is as follows:

Integral Test for Convergence and Divergence

If a series $S=\sum _{n=j}^{\infty }{s_{n}}$, and if the discrete function $s(n)$ is continuous and decreasing on the interval $[j,\infty )$ then if

1. $\int _{j}^{\infty }{s(n)dn}$ is convergent, so is $S$
2. $\int _{j}^{\infty }{s(n)dn}$ is divergent, so is $S$

Let us look at why this this test works. As an example, we'll use the harmonic series $\sum _{n=1}^{\infty }{\frac {1}{n}}$. The harmonic series is well known series that actually happens to be divergent. This series actually passes the limit test, however it does not pass the integral test. If we want to approximate the integral, we can use rectangles like in a Riemann Sum:

Note that this right-handed method will always under approximate the integral (assuming that the function is decreasing on our selected interval). This implies that if the right-handed sum is equal to the actual infinite series, then the integral itself must be greater than the sum. This can help show convergence, because if the integral from the starting point to infinity is convergent, then by the comparison test the original function on that interval must also be convergent. And so we can see that the integral test is actually a "special case" of the comparison test.

But what about divergence? This case is also satisfied—if we use a left-handed approximation instead of a right handed one, we see that we again attain the original series, however there is an important difference:

The key difference, in this case, is that the integral becomes an under approximation for the series, and we can use the new "series" of the integral to show divergence with the comparison test.

This test is useful but is unfortunately only useful on functions that can be integrated and are decreasing in size. The latter may seem like a trivial and unnecessary addition, however consider how this test works; it relies on the fact that the integral of a function decreasing on an interval will always yield an under/over approximation of a series; if the function is not decreasing everywhere on the interval, the integral will not necessarily yield an under/over approximation every time.

Example 1

Use the integral test to determine if the following series is convergent or divergent:

$\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$

Solution

The improper integral from $1$ returns $\lim _{x\rightarrow \infty }{\frac {-1}{n}}-{\frac {-1}{(1)}}$. The limit of the first term is zero, and so because the integral converges the series does too.

Example 2

Determine if the series $\sum _{x=1}^{\infty }{\frac {x^{2}+1}{x}}$ converges or diverges.

Solution

This series does not pass the first requirement that the series is decreasing over the desired interval; $\lim _{x\rightarrow \infty }{\frac {x^{2}+1}{x}}=\infty$. Applying the integral test would still show the series' convergence.

Example 3

Determine if the series $\sum _{n=1}^{\infty }{\frac {1}{(n-3)^{2}+1}}$ converges or diverges.

Solution

However, this series is not decreasing everywhere on the interval. However, it has a relative maximum at $n=1$, after which it does decrease forever. And so we can write this series as $\sum _{n=1}^{2}{\frac {1}{(n-3)^{2}+1}}+\sum _{n=3}^{\infty }{\frac {1}{(n-3)^{2}+1}}$. Integrating the function yields the improper integral $\tan ^{-1}(n-3)$ from $3$ to infinity, which converges, and so the series converges too.

Resources

The Divergence Test

• The Nth Term Divergence Test Video by James Sousa, Math is Power 4U
• The Nth Term Divergence Test (Rational) Video by James Sousa, Math is Power 4U
• The Nth Term Divergence Test (Geometric) Video by James Sousa, Math is Power 4U
• Test for Divergence for Series, Two Examples Video by Patrick JMT
• The nth Term Test, Divergence Test, Zero Test Video by Krista King
• Divergence Test For Series Video by The Organic Chemistry Tutor

The Integral Test

• The Integral Test Video by James Sousa, Math is Power 4U
• Infinite Series - The Integral Test Video by James Sousa, Math is Power 4U
• Integral Test Example Rational Function Video by James Sousa, Math is Power 4U
• Integral Test Example Rational Function Video by James Sousa, Math is Power 4U
• Integral Test Example Radical Function Video by James Sousa, Math is Power 4U
• Integral Test Example Exponential Function Video by James Sousa, Math is Power 4U
• Integral Test Example Involving Arctangent Video by James Sousa, Math is Power 4U
• Integral Test Example Involving Integration by Parts Video by James Sousa, Math is Power 4U

• Integral Test - Basic Idea Video by Patrick JMT
• Integral Test for Series Example 1 Video by Patrick JMT
• Integral Test for Series Example 2 Video by Patrick JMT
• Integral Test to Evaluate Series Ex 1 Video by Patrick JMT
• Integral Test to Evaluate Series Ex 2 Video by Patrick JMT
• Integral Test to Evaluate Series Ex 3 Video by Patrick JMT
• Integral Test to Evaluate Series Ex 4 Video by Patrick JMT

• How to Use the Integral Test Video by Krista King

• Integral Test for Series Video by The Organic Chemistry Tutor

The p-Series Test

• The p-Series Test Video by James Sousa, Math is Power 4U
• Infinite Series - The p-Series Test Video by James Sousa, Math is Power 4U
• The p-Series Test Example 1 Video by James Sousa, Math is Power 4U
• The p-Series Test Example 2 Video by James Sousa, Math is Power 4U

• The p-Series Test Video by Patrick JMT

• The p-Series Test Video by Krista King

• The p-Series Test Video by The Organic Chemistry Tutor

The Remainder Estimate for the Integral Test

• Remainder Estimate for the Integral Test Video by Patrick JMT

• Estimating the Error/Remainder of a Series Video by Krista King

• Remainder Estimate for the Integral Test Video by The Organic Chemistry Tutor