Choose a justified next conclusion
What can you write about each series?
The dots show the terms of one series at a time. The nth-term test states that a series diverges if its terms do not approach 0. For each series, you write one conclusion that this test supports.
Show series A and find the limit of its terms. The dots approach 1, not 0, so the series diverges. Series C also diverges, because its dots alternate between 1 and \(-1\), so the limit does not exist.
In series B and D the dots approach 0. This shows neither convergence nor divergence, so the nth-term test does not decide and another test is needed.
The test gives only two conclusions: the series diverges, or the test does not decide.
One question, two answers
The dots now show the terms of a general series. They approach a height \(L\), which the slider moves. The text under the graph gives the conclusion of the nth-term test.
For every \(L\) except 0, the series diverges. It also diverges when the limit does not exist.
At \(L=0\), the conclusion changes to "the test does not decide". No value of \(L\) gives the conclusion that the series converges.
The test can prove only divergence. Convergent and divergent series can both have terms that approach 0, so \(L=0\) leaves the question open.
Notation
Find the limit of the terms.
\[\lim_{n\to\infty}\cy{a_n}\]If the limit is not 0, or does not exist, write that the series diverges by the nth-term test.
\[\begin{aligned}&\lim a_n\ne0\ \text{or does not exist}\\ &\Rightarrow\ \cr{\text{diverges}}\end{aligned}\]If the limit is 0, write that the nth-term test does not decide. Then choose another test that fits the form of the terms.
\[\lim a_n=0\ \Rightarrow\ \cq{\text{no conclusion}}\]Name the test behind every conclusion. The nth-term test never supports the word converges.
When it fails
Change the terms of series A to \(\frac{1}{(n+10)^2}\). The dots now approach 0, so the conclusion changes from diverges to "the test does not decide".
It does not change to converges, because the nth-term test cannot prove convergence. Another test must decide this series.
Example 1: A limit that is not 0
Write a justified conclusion about \(\displaystyle\sum_{n=1}^{\infty}\dfrac{3^n}{3^n+1}\).
To find the limit of the terms, divide the top and the bottom by \(3^n\). The term \(3^{-n}\) approaches 0, so the terms approach 1.
\[\frac{3^n}{3^n+1}=\frac{1}{1+3^{-n}}\to\cy{1}\]The far dots lie on the dashed line at height 1, which is not 0.
Because the limit is not 0, the series diverges by the nth-term test.
\[1\ne0\ \Rightarrow\ \cr{\text{diverges}}\]
Example 2: A limit of 0
Write a justified conclusion about \(\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n+3}\), using only the nth-term test.
The bottom grows without bound, so the terms approach 0, as the upper graph shows. Each mark on the horizontal axis is ten times the one before it.
\[\lim_{n\to\infty}\frac{1}{n+3}=0\]A tempting wrong move: The terms approach 0, so the series converges.
\[\lim a_n=0\ \Rightarrow\ \text{converges}\]The teal dots in the lower graph are the partial sums, the totals of the first \(n\) terms. They continue to rise and reach about 5.66, circled in red, after 1000 terms. This is the harmonic series without its first three terms, so it diverges.
Because the limit is 0, the nth-term test does not decide, so another test must decide the series.
Practice
A student claims that the terms of \(\displaystyle\sum_{n=1}^{\infty}ne^{-n}\) approach 0, so the series converges by the nth-term test. Write a justified conclusion.
Move the slider to the right and follow the orange dot. Then decide which conclusions the nth-term test can support.
Solution
Replace \(n\) by \(x\). The top and the bottom of \(\frac{x}{e^x}\) both grow without bound, so l'Hôpital's rule applies and gives the limit 0.
\[\lim_{x\to\infty}\frac{x}{e^x}=\lim_{x\to\infty}\frac{1}{e^x}=0\]The limit is 0, but the nth-term test cannot show convergence, so the claim of the student is not justified.
A justified conclusion is that the nth-term test does not decide, so the series needs another test.
Questions
Questions 1 and 2 use the figure. Each graph shows the dots for the terms of one series, for \(n=1\) to 30.
Which conclusion about the series in the upper graph is justified?
- It diverges by the nth-term test, because the terms have no limit.
- The nth-term test does not decide, because half of the terms are 0.
- It converges, because no term is larger than 1.
Answer
It diverges by the nth-term test, because the terms have no limit.
Which conclusion about the series in the lower graph is justified, using only the nth-term test?
- The test does not decide, and another test is needed.
- The series converges, because the dots fall toward 0.
- The series diverges, because the dots fall.
Answer
The test does not decide, and another test is needed.
What does the nth-term test show for \(\displaystyle\sum_{n=1}^{\infty}\dfrac{4n^2-n}{n^2+9}\)?
- It diverges, because the terms approach 4.
- It converges to 4.
- The test does not decide, because the limit is 4.
- It converges, because the terms have a limit.
Answer
It diverges, because the terms approach 4.
Summary
If the limit of the terms is not 0, or does not exist, the series diverges by the nth-term test.
If the limit is 0, the test does not decide, because every convergent series has terms that approach 0 but some divergent series do too.
You can use the test first on any series, but a conclusion of convergence always needs another test.
This is one of four skills in topic 10.3 of Which & Why: AP Calculus.