When you do the limbo, the question is "how low can you go?" In series, it is "how close can you get?". In other words, how close can you get your series to an actual number, or does it not converge at all?
In this article, we cover convergence tests for series.
Calculus Convergence Tests
There are many different kinds of convergence tests for series. In calculus, you look at the ones that are both relatively simple to apply, and the ones that get used frequently. Some tests will have a result that tells you when a series converges and when it diverges. Some are specifically good for checking for divergence. Here you will see some of the ones that involve comparing one series to a second series.
Series Convergence Tests
Suppose you want to know if the series \[\sum_{n=1}^{\infty}a_n\] converges or diverges. If you know something about a different series, sometimes you can compare the one you have to the one you know something about.
Tests like that are called Comparison Tests. Here you will see two of the more common ones, the Direct Comparison test and the Limit Comparison Test, and in the next section of this article are examples showing how to use them.
We start first with the Direct Comparison Test.
Direct Comparison Test
Let \[\sum_{n=1}^{\infty} a_n\] be a series with non-negative terms.
1. The series \[\sum_{n=1}^{\infty} a_n\] converges if there is a convergent series\[\sum_{n=1}^{\infty} c_n\] with \(a_n\leq c_n\) for all \(n>N\) for some \(N\in\mathbb{N}.\)
2. The series \[\sum_{n=1}^{\infty} a_n\] diverges if there is a divergent series
\[\sum_{n=1}^{\infty} d_n\] of non-negative terms with \(a_n\geq d_n\) for all \(n>N\) for some \(N\in\mathbb{N}\).
In the statement of the Direct Comparison Test, it is required that the series
\[\sum_{n=1}^{\infty} a_n\]
has non-negative terms. It turns out that is very important.
For example, if your series is
\[\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\frac{(-1)^n}{n},\]
which is the alternating harmonic series, and you tried using the Direct Comparison Test with the negative harmonic series
\[\sum_{n=1}^{\infty}d_n=\sum_{n=1}^{\infty}\frac{-1}{n}\]
which diverges (see P-series for more information on the harmonic series), you would find that \(a_n\geq d_n\), which would lead you to conclude that the alternating harmonic series diverges because the harmonic series does. In fact, the alternating harmonic series converges (see Alternating Series for details), while the negative harmonic series does not. So it is very important to make sure that the series you are working with have the correct properties before applying the Direct Comparison Test.
We move now to the Limit Comparison Test.
Limit Comparison Test
Suppose \(a_n>0\) and \(b_n>0\) for all \(n>N\) for some \(N\in \mathbb{N}.\)
1. If \[\lim\limits_{n\to \infty}\frac{a_n}{b_n}=c\] where \(0<c<\infty\), then either both series diverge or both series converge.
2. If \[\lim\limits_{n\to \infty}\frac{a_n}{b_n}=0,\] and \[\sum_{n=1}^{\infty} b_n\] converges, then \[\sum_{n=1}^{\infty}a_n\] converges.
3. If \[\lim\limits_{n\to \infty}\frac{a_n}{b_n}=\infty\] and \[\sum_{n=1}^{\infty} b_n\] diverges, then \[\sum_{n=1} ^{\infty} a_n\] diverges.
With the Direct Comparison Test, you needed that your series has non-negative terms. The Limit Comparison Test is more strict in that it requires your series has positive terms. So the Limit Comparison Test can't be used on Alternating series either.
Can you always apply the Limit Comparison Test to a series with positive terms?
Let's take a look at two series, the Harmonic series and the P-series with \(p=2\). You already know that the Harmonic series diverges, and when \(p=2\) the P-series converges.
If you try to use the first part of the Limit Comparison Test,
\[\lim\limits_{n\to \infty} \frac{\frac{1}{n}}{\frac{1}{n^2}}= \lim\limits_{n\to \infty} \frac{n^2}{n}=\infty,\]
and
\[\lim\limits_{n\to \infty} \frac{\frac{1}{n^2}}{\frac{1}{n}}=\lim\limits_{n\to \infty} \frac{1}{n}=0,\]
so you can't use the first part of the Limit Comparison Test.
What if you try to apply the second part of the Limit Comparison Test? When the limit is zero, the Harmonic series is taking the place of the series
\[\sum_{n=1}^{\infty} b_n.\]
But this series diverges, and for the second part of the Limit Comparison Test you need it to converge. That means you can't use the second part of the Limit Comparison Test.
What about the third part of the Limit Comparison Test? In that case the limit is infinity, so the P-series with \(p=2\) is taking the place of the series
\[\sum_{n=1}^{\infty} b_n.\]
But you need this series to diverge, and you know it actually converges, so you can't apply this part of the Limit Comparison Test either.
So in fact just because two series have positive terms doesn't mean the Limit Comparison Test will help you figure out convergence.
Convergence Test Examples
Let's take a look at some examples of using the Direct Comparison Test and the Limit Comparison Test.
In cases where you can't apply either kind of comparison test, you might be able to use the Root Test or the Ratio Test. For more detail on both of these kinds of tests, see Root Test and Ratio Test.
If possible, decide if the series
\[\sum_{n=1}^{\infty}\frac{1}{3^n+n}\]
converges or diverges
Solution
First, note that each term of the series is given by
\[a_n=\frac{1}{3^n+n},\]
and that \(a_n>0\) for all \(n\in\mathbb{N}\). That means it is safe to try and apply either the Direct Comparison Test or the Limit Comparison Test.
If that extra \(n\) wasn't in the denominator this would be a Geometric series, so that would be a good candidate to see if the Direct Comparison Test can be applied. Try
\[\sum_{n=1}^{\infty} c_n=\sum_{n=1}^{\infty}\frac{1}{3^n}.\]
You know this series converges since it is a Geometric series with the common ratio being less than \(1\). Checking the terms of both series, since \(3^n+n>3\), you have that
\[a_n=\frac{1}{3^n+n}<\frac{1}{3^n}=c_n.\]
That means using the first part of the Direct Comparison Test, you know that
\[\sum_{n=1}^{\infty} a_n\]
converges because
\[\sum_{n=1}^{\infty} c_n\]
converges.
If possible, decide if the series
\[\sum_{n=1}^{\infty}\frac{1}{3^n-n}\]
converges or diverges.
Solution
This one is almost the same as the previous example, except for the subtraction in the denominator instead of addition. Here
\[a_n=\frac{1}{3^n-n}.\]
You might try to do the Direct Comparison Test with the same Geometric series
\[\sum_{n=1}^{\infty} c_n=\sum_{n=1}^{\infty}\frac{1}{3^n},\]
but \(3^n-n<3^n\), which means \(a_n>c_n\), and the inequality goes the opposite direction to what you want. So for this problem, you will need to try the Limit Comparison Test.
When using the first part of the Limit Comparison Test, you can actually switch the roles of the series, because if you know one converges and the limit exists and is positive, then both converge. If you try and take the limit one way and it doesn't work, try switching the roles of the series. So trying the limit the first way,
\[\begin{align} \lim\limits_{n\to \infty} \frac{a_n}{c_n}&=\lim\limits_{n\to \infty}\dfrac{\dfrac{1}{3^n-n}}{\dfrac{1}{3^n}}\\ &=\lim\limits_{n\to \infty} \frac{3^n}{3^n-n}, \end{align}\]
which doesn't look like much fun to evaluate. What about if you switched the roles of the series? Then you would look at
\[\begin{align}\lim\limits_{n\to \infty}\frac{c_n}{a_n}&=\lim\limits_{n\to \infty}\frac{\dfrac{1}{3^n}}{\dfrac{1}{3^n-n}}\\ &=\lim\limits_{n\to \infty}\frac{3^n-n}{3^n}\\&=\lim\limits_{n\to \infty} \left[1-\frac{n}{3^n}\right].\end{align}\]
This limit looks much more approachable. You will need to use L’Hôpital’s Rule on the second part of the limit to evaluate it, and you will get
\[\begin{align}\lim\limits_{n\to \infty} \frac{c_n}{a_n}&=\lim\limits_{n\to \infty} \left[1-\frac{n}{3^n}\right]\\&=1-\lim\limits_{n\to \infty}\frac{1}{3^n\ln{(3)}}\\&=1-0\\&=1.\end{align}\]
Since the limit exists and is positive, by the first part of the Limit Comparison Test if one series converges then they both do. Since you know the Geometric series converges, by the Limit Comparison Test so does
\[\sum_{n=1}^{\infty}\frac{1}{3^n-n}.\]
If possible, decide if the series
\[\sum_{n=1}^{\infty}\frac{\ln{n}}{n}\]
converges or diverges.
Solution
This is sort of like the Harmonic series, except there is a natural log in the numerator. Since \(n\geq 1\), you know that
\[a_n=\frac{\ln{n}}{n}\geq 0,\]
so the series in question has non-negative terms. Since it is like the Harmonic series, that is a good one to try comparing it to first. You know that \(\ln{(n)}>1\) for \(n>3\), so
\[\frac{\ln{(n)}}{n}>\frac{1}{n}\quad \text{for}\quad n>3.\]
Remember that \(\ln{e}=1\), and \(e\) is less than 3, so taking \(n>3\) means \(\ln{(n)}>1\).
Taking \(N=3\) in the statement of the Direct Comparison Test, and
\[\sum_{n=1}^{\infty}d_n=\sum_{n=1}^{\infty}\frac{1}{n},\]
since the Harmonic series diverges by the Direct Comparison Test so does
\[\sum_{n=1}^{\infty}\frac{\ln{(n)}}{n}.\]
Let's take a look at another example.
If possible, decide if the series
\[\sum_{n=1}^{\infty}\frac{1}{n3^n}\]
converges or diverges.
Solution
In this example,
\[a_n=\frac{1}{n3^n}\]
which is certainly positive. The question is whether you use the Limit Comparison Test with the Geometric series \[\sum_{n=1}^{\infty}c_n=\sum_{n=1}^{\infty}\frac{1}{3^n}\]
or the Harmonic series
\[\sum_{n=1}^{\infty}d_n=\sum_{n=1}^{\infty}\frac{1}{n}\]
If you can use the Harmonic series and the Limit Comparison Test then you can show that the original series diverges. On the other hand, if you can use the Geometric series and the Limit Comparison Test that will show that the original series converges. If you don't have a good intuition of which one to try, the answer is to try one, and if it doesn't help then to use the other one.
Let's try the Harmonic series first. Then
\[\begin{align} \lim\limits_{n\to \infty}\frac{a_n}{d_n}&=\lim\limits_{n\to \infty} \frac{\frac{1}{n3^n}}{\frac{1}{n}}\\&=\lim\limits_{n\to \infty} \frac{1}{n3^n}.n\\&=\lim\limits_{n\to \infty}\frac{1}{3^n}\\&=0.\end{align}\]
That doesn't help, because if the limit is zero then you need that the second series converges, and you already know that the Harmonic series diverges. So the Harmonic series isn't useful with the Limit Comparison Test to show that the series you care about diverges.
Instead try the Geometric Series. Taking that limit,
\[\begin{align}\lim\limits_{n\to \infty}\frac{a_n}{c_n}&=\lim\limits_{n\to \infty}\frac{\frac{1}{n3^n}}{\frac{1}{3^n}}\\&=\lim\limits_{n\to \infty}\frac{1}{n3^n}.3^n\\&=\lim\limits_{n\to \infty}\frac{1}{n}\\&=0.\end{align}\]
This is actually helpful, because you know that the Geometric series converges. That means the second part of the Limit Comparison Test gives you that
\[\sum_{n=1}^{\infty}\frac{1}{n3^n}\]
converges since the Geometric series does.
Integral Convergence Test
You can figure out whether or not a series converges or diverges if you can find an integral to compare it to. For an explanation and details of how to do this, along with examples, see Integral Test.
Sequence Convergence Tests
While knowing when a sequence converges or diverges can help you in looking at series, here series convergence is discussed. For sequence convergence tests, see Limit of a Sequence.
Convergence Tests - Key takeaways
Direct Comparison Test
Let
\[\sum_{n=1}^{\infty}a_n\] be a series with non-negative terms.
1. The series \[\sum_{n=1}^{\infty}a_n\] converges if there is a convergent series \[\sum_{n=1}^{\infty} c_n\] with \(a_n\leq c_n\) for all \(n>N\) for some \(N\in\mathbb{N}.\)
2. The series \[\sum_{n=1}^{\infty}a_n\] diverges if there is a divergent series \[\sum_{n=1}^{\infty}d_n\] of non-negative terms with \(a_n\geq d_n\) for all \(n>N\) for some \(N\in \mathbb{N}\).
Limit Comparison Test
Suppose \(a_n>0\) and \(b_n>0\) for all \(n>N\) for some \(N\in\mathbb{N}\).
1. If \[\lim\limits_{n\to \infty} \frac{a_n}{b_n}=c\] where \(0<c<\infty\) then either both series diverge or both series converge.
2. If \[\lim\limits_{n\to \infty}\frac{a_n}{b_n}=0 \] and \[\sum_{n=1}^{\infty}b_n\] converges, then \[\sum_{n=1}^{\infty}a_n\]
converges.
3. If \[\lim\limits_{n\to \infty}\frac{a_n}{b_n}=\infty\] and \[\sum_{n=1}^{\infty}b_n\] diverges, then \[\sum_{n=1}^{\infty}a_n\]
diverges.
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