Use a nonzero or missing term limit
Bars that never stop rising
The partial sum \(S_n\) of a series is the sum of its first \(n\) terms. The series converges if its partial sums approach one number, and it diverges otherwise.
The upper graph shows the terms \(a_n=\frac{3n}{n+2}\) as dots. In the lower graph, teal bar \(n\) has height \(S_n\), so each bar is higher than the bar before it by the term \(a_n\).
Every term is at least 1, because \(3n\ge n+2\) when \(n\ge1\), so each bar is at least 1 higher than the bar before it.
The orange ceiling is a height that you choose, and the bars that pass it turn red. With the ceiling at 20, the bars turn red from \(n=10\) on, because \(S_{10}\) is about 20.4.
Raise the ceiling, and the bars pass it later. A running total that passes every ceiling does not approach a number, so this series diverges.
Bars that level off have steps that shrink
Suppose the heights of the bars approach a number \(S\), and draw an orange band around \(S\).
Narrow the band with the slider. Far to the right, every bar lies inside the band and is teal, while the bars outside it are faint.
A term is the step from one bar to the next. Far to the right, both bars lie inside the band, so the step is shorter than the band is wide. The orange strip in the upper graph has that height, and the far dots lie below it.
The band can be as narrow as you like, so the terms approach 0. If a series converges, its terms approach 0.
Turn this around. If the terms do not approach 0, the bars cannot approach a number, so the series diverges.
Notation
The step from bar \(n-1\) to bar \(n\) is the term \(a_n\), because \(S_n\) is \(S_{n-1}\) plus one more term.
\[\cb{S_n}-\cb{S_{n-1}}=\cy{a_n}\]If the series converges to \(S\), then \(S_n\) and \(S_{n-1}\) both approach \(S\), so their difference approaches 0.
\[\cy{a_n}=S_n-S_{n-1}\to S-S=0\]If the limit of the terms is not 0, or does not exist, the series diverges. This is the nth-term test for divergence. In the figure the terms now approach \(\frac12\), so the bars do not level off.
\[\begin{aligned}&\lim_{n\to\infty}a_n\ne0\ \text{or does not exist}\\ &\Rightarrow\ \cr{\sum a_n\ \text{diverges}}\end{aligned}\]The sign of the limit does not matter. The test uses the terms of the series, not the partial sums.
When it fails
Replace the terms by \(\frac1n\). The terms approach 0, but the bars still rise at every step. After 100 terms the total is about 5.19, and it continues to rise.
The test can show divergence only. When the terms approach 0, it does not decide, and you need another test.
Example 1: Terms that approach a number above 0
Use the nth-term test on \(\displaystyle\sum_{n=1}^{\infty}\dfrac{2n-1}{5n+3}\).
Find the limit of the terms. Divide the top and the bottom by \(n\).
\[\frac{2n-1}{5n+3}=\frac{2-1/n}{5+3/n}\to\cy{\tfrac25}\]The limit is \(\frac25\), which is not 0. Each far bar is about \(\frac25\) higher than the bar before it, so the bars rise without end.
By the nth-term test, the series diverges.
\[\tfrac25\ne0\ \Rightarrow\ \cr{\text{diverges}}\]
Example 2: Terms with no limit
Decide whether \(\displaystyle\sum_{n=1}^{\infty}\sin\dfrac{n\pi}{2}\) converges.
The first terms repeat in a cycle of four.
\[1,\ 0,\ -1,\ 0,\ 1,\ 0,\ -1,\ 0,\ \dots\]A tempting wrong move: The partial sums are 1, 1, 0, 0, 1, 1, 0, 0, and so on. They stay between 0 and 1, so the series converges.
\[0\le S_n\le1\]The partial sums stay in the red band but move between 0 and 1 forever, because every second term, circled in red, has size 1. A series converges only if its partial sums approach one height.
The terms take the values 1, 0, \(-1\), 0 again and again, so their limit does not exist. By the nth-term test, the series diverges.
Practice
Use the nth-term test on \(\displaystyle\sum_{n=1}^{\infty}\dfrac{-n}{3n+7}\).
Find the limit of the terms. A negative limit is still not 0, and the far bars then fall by about that much at each step.
Solution
Divide the top and the bottom by \(n\) and take the limit.
\[\frac{-n}{3n+7}=\frac{-1}{3+7/n}\to-\tfrac13\]The limit is \(-\frac13\), which is not 0. Each far bar lies about \(\frac13\) below the bar before it, so the bars fall without end.
By the nth-term test, the series diverges.
Questions
Questions 1 and 2 use the figure. The upper graph shows the terms \(a_n=\frac{n+1}{2n+3}\) as dots. The lower graph shows the partial sums \(S_n\) as bars, for \(n=1\) to 24.
The dots in the upper graph approach height 0.5. What does this show about the series?
- It diverges, because the terms approach 0.5 and not 0.
- It converges, because the dots approach a height.
- It converges, because all the terms are positive.
- The test does not decide.
Answer
It diverges, because the terms approach 0.5 and not 0.
Every term now has the opposite sign, so the dots approach height \(-0.5\). What does the nth-term test show?
- The series diverges, because the terms approach \(-0.5\) and not 0.
- The series converges to \(-0.5\).
- The test does not decide, because the limit is negative.
Answer
The series diverges, because the terms approach \(-0.5\) and not 0.
Which series does the nth-term test show to diverge?
- \(\displaystyle\sum_{n=1}^{\infty}\dfrac{2n^2}{n^2+9}\)
- \(\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n+3}\)
- \(\displaystyle\sum_{n=1}^{\infty}\dfrac{5}{n^2}\)
- \(\displaystyle\sum_{n=1}^{\infty}\left(\dfrac13\right)^n\)
Answer
\(\displaystyle\sum_{n=1}^{\infty}\dfrac{2n^2}{n^2+9}\)
Summary
If the limit of the terms is not 0, or does not exist, the series diverges by the nth-term test.
This works because bars that approach a number have steps that shrink to 0, and the steps are the terms.
You use it as the first test on any series, and only to show divergence.
This is one of four skills in topic 10.3 of Which & Why: AP Calculus.