(ST1) Test for Divergence#

In this lesson we are going to see how to:

  • use the Test for Divergence to determine if a series is divergent.

Review Videos#

Motivation#

Usually computing the sum of a series \(\sum a_n\) is very difficult. The best we can usually do is to show either:

  • the series is convergent in which case the sum exists or

  • the series is divergent and the sum does not exist.

Since we are interested in knowing whether the sum exists or not and this is usually extremely difficult to do directly (with partial sums) we instead develop a bunch of tests to help us test whether a series is convergent or divergent.

The first test we are going to look at, the Test for Divergence, is a quick way to show that a series is divergent.

The Test#

Test for Divergence

Given a series \(\sum a_n\), calculate the limit \(\displaystyle \lim_{n\to\infty} a_n\)

  • The series is divergent if \(\displaystyle \lim_{n\to\infty} a_n \neq 0\) or the limit does not exist.

  • The test is inconclusive if \(\displaystyle \lim_{n\to\infty} a_n =0 \).

What do we mean when we say the test is inconclusive? Well, we’re trying to determine whether a series \(\sum a_n\) is convergent or divergent, so if the test is inconclusive, then that means we still do not know have an answer either way.

Intuitive Proof#

Suppose \(\displaystyle \lim_{n\to\infty} a_n = L > 0\) (the case where \(L<0\) is similar).

Then \(a_n > L/2\) for all but a finite number of the terms in the sequence. So when we calculate the series \(\sum a_n\) we are essentially trying to add up an infinite number of terms that are all greater than \(\tfrac{L}{2}\) and so the sequence of partial sums gets larger and larger and goes to infinity. Therefore the series diverges.

Warning#

While this test is normally easy to use, it’s also easy to use incorrectly. A common mistake is to claim that since \(\displaystyle \lim_{n\to\infty} a_n =0 \), the series \(\sum a_n\) is convergent. However this is incorrect. Its important to remember that the test is inconclusive when the the limit is \(0\).

Warning

Note that the Test for Divergence cannot be used to show a series is convergent. We can only use this to show that a series is divergent.

If \(\displaystyle \lim_{n\to\infty} a_n = 0\), we can say the original sequence \(\{ a_n\}\) converges to \(0\), but this does not actually tell us anything about what the series \(\sum a_n\) is doing.

Example 1#

Determine if the following series is convergent or divergent.

\[ \sum_{n=1}^{\infty} \dfrac{n^2}{3n^2+1} \]
Solution (Click to see the steps.)

This is not a geometric series nor is it a \(p\)-series, so we attempt to use the Test for Divergence to see what is happening. We start by calculating:

\[\begin{split} \lim_{n\to\infty} a_n & = \lim_{n\to\infty} \dfrac{n^2}{3n^2+1} \\[10pt] & = \lim_{n\to\infty} \dfrac{n^2}{3n^2+1} \cdot \dfrac{\tfrac{1}{n^2}}{\tfrac{1}{n^2}}\\[10pt] & = \lim_{n\to\infty} \dfrac{1}{3+\tfrac{1}{n^2}} \\[10pt] & = \dfrac{1}{3+0} =\dfrac{1}{3}\\[10pt] \end{split}\]

Since, \(\displaystyle \lim_{n\to\infty} a_n = \tfrac{1}{3} \neq 0\), the Test for Divergence tells us this series diverges.

Example 2#

Determine if the following series is convergent or divergent.

\[ \sum_{n=1}^{\infty} \dfrac{1}{n} \]
Solution (Click to see the steps.)

This is a \(p\) series with \(p=1\) and is therefore divergent.

However, notice that if we had tried to use the Test for Divergence, we would have found that the test was inconconlusive since:

\[ \lim_{n\to\infty} a_n = \lim_{n\to\infty} \dfrac{1}{n} = 0 \]

Example 3#

Determine if the following series is convergent or divergent.

\[ \sum_{n=1}^{\infty} \dfrac{1}{n^2} \]
Solution (Click to see the steps.)

This is a \(p\) series with \(p=2\), which is greater than 1. Therefore, the series is convergent.

However, notice that if we had tried to use the Test for Divergence, we would have found that the test was inconconlusive since:

\[ \lim_{n\to\infty} a_n = \lim_{n\to\infty} \dfrac{1}{n^2} = 0 \]