The divergence test

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If an infinite sum converges, then its terms must tend to zero.

In order to determine if a series $\sum _{k=1}^{\infty } a_k$ converges, we took the following approach.

Consider the associated sequence $\{s_n\}$ of partial sums, where $s_n=\sum _{k=1}^n a_k$ .
Try to find an explicit formula for the term . If you can find such a formula, analyze .
- If the limit exists, $\sum _{k=k_0} a_k$ converges, and if we can determine that $\lim _{n \to \infty } s_n =L$ , then $\sum _{k=k_0} a_k=L$ .
- If $\lim _{n \to \infty } s_n$ does not exist, then $\sum _{k=k_0} a_k$ diverges.
If an explicit formula for $s_n$ cannot be found, further analysis is needed.

In the previous section, we studied two types of series where we could find an explicit formula for $s_n$ , but unfortunately, this is not always easy or possible. Fortunately, it is not always necessary to do this in order to determine whether $\lim _{n \to \infty } s_n$ exists. Consider the example below.

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

Let’s think about what we are trying to do here heuristically. If we set $a_k = \frac {k+2}{2k+1}$ , we can write out the first several terms in the sequence $\{a_n\}_{n=1}$ .

$\frac {3}{5}, \frac {4}{7}, \frac {5}{9}, \frac {6}{11},\frac {7}{13}, \ldots$ We can observe also that $\lim _{n \to \infty } a_n = \answer [given]{\frac {1}{2}}$ , so eventually, the terms in this list become as close to $\frac {1}{2}$ as we want. Conceptually, we can interpret that when trying to compute the series, we will have to add infinitely many numbers together that are very close to $\frac {1}{2}$ . Such an attempt cannot produce a finite result, so we expect the series to diverge.

Of course, this is not a formal proof or an acceptable mathematical argument, but it is good intuition. In order to formalize the argument, recall that we have to set $s_n = \sum _{k=1}^n a_k$ and study whether $\lim _{n \to \infty } s_n$ exists. While we do not have an explicit formula for $s_n$ , we do have a recursive formula, which we recall below.

Since $a_n = \frac {n+2}{2n+1}$ , we can write

$s_n = s_{n-1}+\frac {n+2}{2n+1}.$

Now, if $\lim _{n \to \infty } s_n$ exists and is equal to $L$ , we have that $\lim _{n \to \infty } s_{n-1}=L$ as well, so taking the limit of both sides of the above equation gives

$\begin{align*} \lim_{n \to \infty} s_n &= \lim_{n \to \infty} s_{n-1}+ \lim_{n \to \infty} \frac{n+2}{2n+1} \\ L &= L +\frac{1}{2}. \end{align*}$

This statement is blatantly false, so our underlying assumption that $\lim _{n \to \infty } s_n$ exists is false as well. Since $\lim _{n \to \infty } s_n$ therefore does not exist, $\sum _{k=1}^{\infty } \frac {k+2}{2k+1}$ must diverge.

As it turns out, the above argument can be used to make a very important observation; if $\{a_n\}$ is a sequence for which $\sum _{k=k_0}^{\infty } a_k$ converges, then $\lim _{n \to \infty } a_n =0$ . This result is fundamentally important, so we capture it in a theorem.

Suppose that $\{a_n\}_{n=n_0}$ is a sequence and the series $\sum _{n=n_0}^{\infty } a_n$ converges. Then, $\lim _{n \to \infty } a_n =0$ .

Note that the convergence of the series $\sum _{n=n_0}^{\infty } a_n$ tells us something about the sequence $\{a_n\}$ .

The divergence test

Divergence test Let $\{a_n\}_{n=n_0}$ be a sequence and consider the series $\sum _{k=k_0}^\infty a_k$ . If $\lim _{n\to \infty }a_n \neq 0$ , then $\sum _{n=n_0}^\infty a_n$ diverges.

Stated in plain English, the above test ensures that if the terms in a sequence do not tend to zero, then we cannot add all of the terms in that sequence together.

The result of this theorem can be established using a similar argument as in the previous example. Note that there is no explicit reference to the sequence of partial sums in the actual statement of the test. A more complete statement of the test would be:

Let $\{a_n\}_{n=n_0}$ be a sequence and consider the series $\sum _{k=k_0}^\infty a_k$ . If $\lim _{n\to \infty }a_n \neq 0$ , then we can show that $\lim _{n \to \infty } s_n$ does not exist and hence $\sum _{n=n_0}^\infty a_n$ diverges.

This test gives us a quick way to determine if some series diverge.

Determine if the series $\sum _{k=1}^{\infty } \cos \left (\frac {k^2+7^k}{k!} \right )$ converges or diverges.

Here, the sequence whose terms are being summed is given by the formula $a_n =\cos \left (\frac {n^2+7^n}{n!} \right )$ . Let’s try to apply the divergence test. Notice that

$\lim _{n \to \infty } \frac {n^2+7^n}{n!} = \answer [given]{0}$ by growth rates, so the limit of the sequence $\{a_n\}$ is

$\lim _{n \to \infty } \cos \left (\frac {n^2+7^n}{n!} \right ) = \cos (0) = 1 \neq 0.$

Hence, the series $\sum _{k=1}^{\infty } \cos \left (\frac {k^2+7^k}{k!} \right )$ diverges by the divergence test.

Determine if the series $\sum _{k=1}^{\infty } \sin (k)$ converges or diverges.

Here, the sequence whose terms are being summed is given by the formula $a_n =\sin (n)$ . Notice that these terms fluctuate in sign, so maybe when we try adding them all together, we obtain something finite. Let’s try to apply the divergence test. Notice that $\lim _{n \to \infty } \sin (n)$ does not exist; it certainly is not 0. Hence, the series $\sum _{k=1}^{\infty } \sin (k)$ diverges by the divergence test.

In the last example, perhaps the fact that the terms in $a_n = \sin (n)$ fluctuate in sign will ensure that the series cannot be infinite. To think about this, let’s turn to the sequence of partial sums. To gain a bit of visual perspective about what is happening, note that the $n$ -th term in the sequence of partial sums here is

$s_n = \sum _{k=1}^n \sin (k) = \sin (1)+\sin (2)+\ldots +\sin (n).$

Plotting several such terms reveals that the terms sequence of partial sums $\{s_n\}$ seem to fluctuate.

While we will not show it here, the sequence $\{s_n\}$ is bounded; the reason that $\lim _{n \to \infty } s_n$ does not exist is due to the fact that the terms fluctuate (meaning that the sequence is never eventually monotonic).

Implications of the divergence test

Let’s summarize the important points from the previous discussion.

If $\sum _{k=k_0}^\infty a_k$ converges, then $\lim _{n \to \infty } a_n =0$ .
If $\lim _{n \to \infty } a_n \neq 0$ (including the case where the limit does not exist), then $\sum _{k=k_0}^{\infty } a_k$ diverges.

While divergence test was straightforward to apply in the previous examples, there is a major point to address about what it does not say.

The divergence test can never be used to conclude that a series converges. The theorem does not state that if $\lim _{n\to \infty } a_n = 0$ then $\sum _{n=1}^\infty a_n$ converges.

We’ve actually seen an example of this in action.

Recall that in a previous section, we showed that the series $\sum _{k=1}^\infty \ln \left (\frac {k+1}{k}\right )$ is actually telescoping. By recognizing that $\ln \left (\frac {k+1}{k}\right ) = \ln (k+1)-\ln (k)$ , we showed that there is an explicit formula for the $n$ -th term in the sequence of partial sums given by $s_n = \ln (n+1)$ . We concluded that $\sum _{k=1}^\infty \ln \left (\frac {k+1}{k}\right )$ diverges since $\lim _{n \to \infty } s_n = \infty$ .

Note now that the expression in the sum (i.e. the sequence whose terms we are attempting to sum) is $a_n = \ln \left (\frac {n+1}{n}\right )$ , and that since

$\lim _{n \to \infty } \frac {n+1}{n} = \answer [given]{1}$ we have

$\lim _{n \to \infty } \ln \left (\frac {n+1}{n}\right ) = \ln \left (\answer [given]{1}\right ) = 0.$

Thus, we have an example of a sequence whose limit is zero for which the sum of its terms diverges; that is, we have an example where $\lim _{n \to \infty } a_n =0$ but $\sum _{k=1}^{\infty } a_k$ diverges.

The standard example of a sequence for which $\lim _{n \to \infty } a_n =0$ but for which $\sum _{k=1}^{\infty } a_k$ diverges is the harmonic series, $\sum _{k=1}^{\infty } \frac {1}{k}$ . We’ll show later on that this series diverges.

Said another way:

If $\sum _{k=k_0}^{\infty } a_k$ diverges, it’s still possible that $\lim _{n \to \infty } a_n =0$ .

To elaborate a little more, we can say that a series $\sum _{k=k_0} a_k$ “passes the divergence test” if $\lim _{n \to \infty } a_n=0$ . Which of the following series pass the divergence test?

$\sum _{k=3}^\infty \frac {1}{\ln { k }}$ $\sum _{k=0}^\infty \sin (k)$ $\sum _{k=0}^\infty \frac {\sin (k)}{k^2}$ $\sum _{k=5}^\infty \frac {k+7}{k+6}$ $\sum _{k=0}^\infty \frac {2k}{k - 5}$

Restating this point again (because it is very important): passing the divergence test means that a series has a chance to converge. The divergence test cannot tell us whether a series converges.

There are many questions that require that you now have a firm grasp on the concepts presented thus far. We summarize the important points made thus far, then give many examples that require you to synthesize them.

There are two fundamental questions we can ask of any sequence.
- Do the numbers in the list approach a finite value?
- Can I sum all of the numbers in the list and obtain a finite result?
These questions can be asked of a given sequence $\{a_n\}$ and can also be asked about $\{s_n\}$ or any sequence constructed from it.
Given a sequence , we construct the sequence of partial sum whose -th term is given by the formula .
- The symbols $\sum _{k=k_0}^{\infty } a_k$ and $\lim _{n \to \infty } s_n$ are the same.
- By definition $\sum _{k=k_0} a_k$ converges if $\lim _{n \to \infty } s_n$ exists and in this case, the value of each is the same.
- By definition $\sum _{k=k_0} a_k$ diverges if $\lim _{n \to \infty } s_n$ does not exist (which includes if the limit is infinte).
If the limit of a sequence is not zero, the sum of its terms diverges.
If a series converges, the limit of the sequence whose terms is being summed is zero.
If the limit of a sequence is zero, more information is needed to determine whether the sum of its terms converges or diverges.

To answer the following questions, make sure that you understand exactly what is given in the statement of the question first, then try to synthesize the material above.

Suppose that $\{a_n\}_{n=1}$ is a sequence and let $s_n = \sum _{k=1}^{\infty } a_k$ . Suppose that it is also known that $s_n = \frac {2n+6}{n+4}$ .

Which statement below captures the most we can say about $\sum _{k=1}^{\infty } a_k$ ?

$\sum _{k=1}^{\infty } a_k$ converges but we cannot determine its value without more information. $\sum _{k=1}^{\infty } a_k$ converges to $0$ . $\sum _{k=1}^{\infty } a_k$ converges to $2$ . $\sum _{k=1}^{\infty } a_k$ diverges. More information is needed to determine if $\sum _{k=1}^{\infty } a_k$ converges.

The information given in the problem is an explicit formula for the terms in the sequence of partial sums $s_n$ , not $a_n$ . To answer this question, we need to know how $s_n$ relates to finding $\sum _{k=1}^{\infty } a_k$ . Since $\lim _{n \to \infty } s_n$ and $\sum _{k=1}^{\infty }$ are analogous representations of the same idea and $\lim _{n \to \infty } s_n = \answer [given]{2}$ , we have $\sum _{k=1}^{\infty } a_k$ converges to $2$ .

Which statement below captures the most we can say about $\sum _{k=1}^{\infty } s_k$ ?

$\sum _{k=1}^{\infty } s_k$ converges but we cannot determine its value without more information. $\sum _{k=1}^{\infty } s_k$ converges to $0$ . $\sum _{k=1}^{\infty } s_k$ converges to $2$ . $\sum _{k=1}^{\infty } s_k$ diverges. More information is needed to determine if $\sum _{k=1}^{\infty } s_k$ converges.

It may seem daunting to think about what $\sum _{k=1}^{\infty } s_k$ is in general, but note here that the information given in the problem is an explicit formula for $s_n$ . $\{s_n\}_{n=1}$ is a sequence in its own right, and we can ask whether we can sum its terms. Here, we can immediately write down the series in question.

$\sum _{k=1}^{\infty } s_k =\sum _{k=1}^{\infty } \frac {2k+6}{k+4}$ Since $\lim _{n \to \infty } s_n = 2\neq 0$ , we have $\sum _{k=1}^{\infty } s_k$ diverges by the divergence test.

Note that in the previous questions, $\lim _{n \to \infty } s_n$ was used in two different ways. For the first question, $\lim _{n \to \infty } s_n$ is used to answer a question about $\sum _{k=1}^{\infty } a_k$ by using the definition of convergence. In the second question, we are asked to think of $\{s_n\}$ as a sequence in its own right whose terms can be summed. We can use the divergence test to answer this question.

Suppose that $\{a_n\}_{n=1}$ is a sequence and let $s_n = \sum _{k=1}^{\infty } a_k$ . Suppose that it is known that $\sum _{k=1}^{\infty } s_k =3$ . What can be said about $\sum _{k=1}^{\infty } a_k$ ?

$\sum _{k=1}^{\infty } a_k$ converges but we cannot determine its value without more information. $\sum _{k=1}^{\infty } a_k$ converges to $0$ . $\sum _{k=1}^{\infty } a_k$ converges to $3$ . $\sum _{k=1}^{\infty } a_k$ diverges. More information is needed to determine if $\sum _{k=1}^{\infty } a_k$ converges.

The information given in the problem is that $\sum _{k=1}^{\infty } s_k$ is a convergent series. We need to relate this to $\sum _{k=1}^{\infty } a_k$ . The relationship between the sequence of partial sums and $\sum _{k=1}^{\infty } a_k$ is that $\lim _{n \to \infty } s_n$ and $\sum _{k=1}^{\infty } a_k$ are analogous, so we need $\lim _{n \to \infty } s_n$ . Thankfully, we have a fact for that; since $\sum _{k=1}^{\infty } s_k$ is convergent, the limit of the inner sequence whose terms we add must be zero; that is $\lim _{n \to \infty } s_n=0$ . Hence, $\sum _{k=1}^{\infty } a_k=0$ .

Suppose $\{a_n\}_{n \geq 1}$ is a sequence and $\sum ^{\infty }_{n= 1} a_n=5$ . Let $s_n = \sum ^n_{k=1} a_k$ . Select all statements that must be true:

$\lim _{n \to \infty } a_n = 5$ $\lim _{n \to \infty } a_n = 0$ $\lim _{n \to \infty } s_n = 0$ $\lim _{n \to \infty } s_n = 5$ $\sum ^{\infty }_{k=1} s_k$ must diverge. $\sum ^{\infty }_{k=1} (a_k+1) = 5+1=6$ The divergence test tells us $\sum ^{\infty }_{n= 1} a_n$ converges to $L$ .

Let’s look at each of these statements.

For the first four choices, notice that since $\sum _{k=1}^{\infty } a_k =5$ , we have two immediate consequences. First, by definition $\lim _{n \to \infty } s_n = 5$ . Secondly, $\sum _{k=1}^{\infty } a_k$ converges, so $\lim _{n \to \infty } a_n = 0$ .
Now that we know $\{s_n\}$ is a sequence that does not tend to zero, the divergence test tells us we cannot sum its terms; i.e. since $\lim _{n \to \infty } s_n \neq 0$ , $\sum _{k=1}^{\infty }$ must diverge.
First, notice that there is a huge difference in the series $\sum ^{\infty }_{n=1} (a_n+1)$ and $\sum ^{\infty }_{n=1} a_n+1$ , where the latter sum is to be interpreted as $\left (\sum ^{\infty }_{n=1} a_n\right )+1$ . Since $\lim _{n \to \infty } a_n =0, \lim _{n \to \infty } (a_n+1) \neq 0$ , so $\sum ^{\infty }_{k=1} (a_k+1)$ diverges by divergence test.
The last choice is never true; we can never determine that a series converges by the divergence test.