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Infinite Series - Divergence Test |
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Divergence Test |
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If \( \displaystyle{\lim_{n \to \infty}{a_n} \neq 0} \), then \( \displaystyle{\sum_{n=1}^{\infty}{a_n}}\)
diverges. |
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Alternate Form
The Divergence Test can also be stated in this alternate form.
If \( \displaystyle{\sum_{n=1}^{\infty}{a_n}}\) converges, then
\( \displaystyle{\lim_{n \to \infty}{a_n} = 0} \).
Using Mathematical Logic this can be written
\(\displaystyle{ \sum_{n=1}^{\infty}{a_n} \text{ converges } ~ \to ~ \lim_{n \to \infty}{a_n} = 0 } \), which is the contrapositive of the form above.
This alternate form is not very useful for computation. So we use the theorem above for all discussion and practice problems here at 17calculus.
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The Divergence Test is also called the nth-Term Test. |
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How To Use The Divergence Test |
To use the divergence test, just take the limit \( \displaystyle{\lim_{n \to \infty}{a_n} } \). If this limit turns out to be non-zero, the series diverges and you are done. If the limit is equal to zero, then the test is inconclusive and says nothing about the series. It may converge or it may diverge. You need to use another test to determine convergence or divergence.
The Divergence Test seems to be pretty straight-forward. However, there is a trap that many students fall into.
Basically, this test says that, for a series
\( \displaystyle{ \sum_{n=1}^{\infty}{a_n}} \), if
\( \displaystyle{ \lim_{n \to \infty}{a_n} \neq 0 }, \) then the series diverges. Pretty simple, eh?
Trap - - There is a tendency by almost every student to use this theorem to prove convergence. The statement says nothing about convergence. Let us give you a couple of examples that demonstrate what we mean. Let's compare the convergence or divergence of these two very similar series. (If you don't know about p-series yet, just take our word for the convergence/divergence conclusions. You will understand this soon enough.)
A.
| \(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n}}}\) |
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\( \displaystyle{\lim_{n\to\infty}{\frac{1}{n}} = 0} \) |
diverges |
p-series with \(p=1\) |
B. |
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2}}}\) |
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\( \displaystyle{\lim_{n\to\infty}{\frac{1}{n^2}} = 0} \) |
converges |
p-series with \(p=2\) |
It is important to notice that in both cases the limit of the terms goes to zero. However, series A diverges while series B converges. [ See the Infinite Series - Integral Test page for a video showing a proof of the convergence/divergence of these two series. ]
So you can see that just because the limit goes to zero, this does not guarantee the series will converge. The only time you can apply this theorem is when the limit does not go to zero. This guarantees divergence. When the limit does go to zero, you still don't know if the series converges or diverges. You need to use another test to determine convergence.
Below is the nth-term test row from the infinite series table. Notice that the convergence line says that this test cannot be used. Think about this thoroughly and completely so that you get your head around it.
Divergence Test
\(\displaystyle{ \sum_{n=1}^{\infty}{a_n} }\) |
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convergence: |
this test cannot be used |
divergence: |
\(\displaystyle{ \lim_{n \to \infty}{ a_n \neq 0} }\) |
| Here is a quick video that explains this test. [ 1min-26secs ] |
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Instructions - - Unless otherwise instructed, determine whether these series converge or diverge.
Practice A01 |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{ 2 } } }\) |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{ 2 } } }\) diverges by the nth-Term Test. |
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Let's try the nth-Term Test first.
\(\displaystyle{ \lim_{n \to \infty}{n/2} = \frac{\infty}{2} = \infty }\)
Since the limit is not zero, the series diverges by the nth-Term Test.
Practice A01 Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{ 2 } } }\) diverges by the nth-Term Test. |
Practice A02 |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5n}{n^3} } }\) |
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The series converges by the p-series test. |
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Starting with the nth-term test, we need to determine the limit
\(\displaystyle{ \lim_{n \to \infty}{ \frac{5n}{n^3} } }\).
There are two main ways to do this.
1. The first, and easiest way, is to use algebra.
2. The second way is to use L'Hôpital's Rule.
We will use algebra by multiplying the numerator and denominator by \(1/n\).
\(\displaystyle{
\begin{array}{rcl}
\lim_{n \to \infty}{ \frac{5n}{n^3} } & = & \lim_{n \to \infty}{ \frac{5n}{n^3} \frac{1/n}{1/n} } \\
& = & \lim_{n \to \infty}{ \frac{5}{n^2} } = 0
\end{array}
}\)
Since the limit is zero, we cannot say anything about whether this series converges or diverges using only the nth-term test.
By canceling the \(n\) in the numerator with one in the denominator ( which we do since the canceling operation does not change the domain ), we have \(a_n = 5/n^2\). This is a p-series with \(p=2>1\). So this series converges by the p-series test.
Note that even though the nth-term test cannot be used to determine convergence or divergence, we didn't just stop working the problem. The problem statement said to determine convergence or divergence. So stopping after applying the nth-term test would not be the completion of the problem. We actually needed to determine convergence or divergence. So once the first test failed, we moved on to another test until one of the tests determined convergence or divergence.
Practice A02 Final Answer |
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The series converges by the p-series test. |
Practice A03 |
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\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2+3n+1}{2n^2+4}} }\) |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2+3n+1}{2n^2+4}} }\) diverges by the nth-Term Test. |
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Practice A03 Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2+3n+1}{2n^2+4}} }\) diverges by the nth-Term Test. |
Practice A04 |
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\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n^2+2}{n^2+1}}}\) |
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The series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{n^2+2}{n^2+1}}}\) diverges by the nth-Term Test. |
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Practice A04 Final Answer |
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The series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{n^2+2}{n^2+1}}}\) diverges by the nth-Term Test. |
Practice B01 |
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\(\displaystyle{\sum_{n=2}^{\infty}{\frac{(-1)^n n}{(\ln(n))^2}}}\) |
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The series \(\displaystyle{\sum_{n=2}^{\infty}{\frac{(-1)^n n}{(\ln(n))^2}}}\) diverges by the nth-Term Test. |
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Since this is an alternating series, we will start with the Alternating Series Test. (If you haven't studied the Alternating Series Test yet, don't worry. You will still understand most of this solution.)
Condition 1: We need to show \(\displaystyle{ \lim_{n \to \infty}{a_n} = 0 }\).
Since \(\displaystyle{ \sum_{n=2}^{\infty}{(-1)^n a_n} = \sum_{n=2}^{\infty}{ \frac{(-1)^n n}{ (\ln( n ))^2 } } }\) we know that
\(\displaystyle{ a_n = \frac{n}{ (\ln( n ))^2 } }\)
and so we can write
\(\displaystyle{\lim_{n \to \infty}{a_n} = \lim_{n \to \infty}{\frac{n}{(\ln( n ))^2}} }\).
Plugging in infinity directly yields \(\infty / \infty\), which is indeterminate.
So we can use L'Hôpital's Rule.
\(\displaystyle{
\begin{array}{rcl}
\lim_{n \to \infty}{a_n} & = & \lim_{n \to \infty}{\frac{n}{(\ln( n ))^2}} \\\\
& = & \lim_{n \to \infty}{\frac{1}{(2 \ln( n ))(1/n)}} \\\\
& = & \lim_{n \to \infty}{\frac{n}{2 \ln( n )}} \\\\
& = & \lim_{n \to \infty}{\frac{1}{2/n}} \\\\
& = & \lim_{n \to \infty}{\frac{n}{2}} = \infty
\end{array}
}\)
In the work above, we used L'Hôpital's Rule twice. This is actually the nth-Term Test that tells us that the series diverges.
Note: You may be tempted to say that the series diverges by the Alternating Series Test. However, the Alternating Series Test can be used only for convergence. So, it is not correct to say that we used it for divergence. See the infinite series table to confirm this.
Practice B01 Final Answer |
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The series \(\displaystyle{\sum_{n=2}^{\infty}{\frac{(-1)^n n}{(\ln(n))^2}}}\) diverges by the nth-Term Test. |
Practice B02 |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \ln(\cos n) } }\) |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \ln(\cos n) } }\) diverges by the nth-Term Test. |
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Practice B02 Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \ln(\cos n) } }\) diverges by the nth-Term Test. |
Practice C01 |
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\(\displaystyle{\sum_{n=1}^{\infty}{\sin(n\pi/2)}}\) |
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The series \(\displaystyle{\sum_{n=1}^{\infty}{\sin(n\pi/2)}}\) diverges by the nth-Term Test. |
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In order to determine what the limit \(\displaystyle{\lim_{n\to\infty}{\sin(\pi n/2)}}\) does, you need to think about the sine function. Basically, the sine function takes angles and maps them to the range \(-1 \to 1\). So as \(n \to \infty\) the sine function oscillates between -1 and 1. It never settles at a specific value and, more importantly for us here, it never settles or approaches zero.
Since \(\displaystyle{\lim_{n\to\infty}{\sin(\pi n/2)}\neq 0}\), this series diverges by the nth-Term Test.
Practice C01 Final Answer |
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The series \(\displaystyle{\sum_{n=1}^{\infty}{\sin(n\pi/2)}}\) diverges by the nth-Term Test. |
