Evaluate and interpret the ratio limit
Ratios that approach a limit
The bars show the terms of two series. Series A has the terms \(\frac{n^2}{3^n}\), and series B has the terms \(\frac{3^n}{n^2}\). The dots show the ratio of neighboring terms of the series you select. Drag the slider to add terms.
In series A the first ratio is 4/3, which is above 1, but the later ratios fall toward the orange dashed line at 1/3. After the second bar, each bar is smaller than the one before it.
Select series B. Its first ratio is 3/4, which is below 1, but the later ratios rise toward 3. After the second bar, each bar is larger than the one before it.
The first ratios do not decide the result, because only the ratios for large n show how the terms behave far along the series.
Reading the limit against 1
The dots show the ratio of neighboring terms of a series. They approach the orange line at the height \(L\), which is the limit of the ratio. Drag the slider to choose \(L\).
If \(L\) is less than 1, choose a number \(r\) between \(L\) and 1, shown by the teal dashed line. After some \(n\), every dot lies below it, so each absolute value is less than \(r\) times the one before it. The absolute values then shrink faster than the terms of the geometric series with ratio \(r\), which converges because \(r < 1\). The series of absolute values therefore converges, and the series is said to converge absolutely.
If \(L\) is more than 1, each absolute value is eventually larger than the one before it. The terms do not approach 0, so the series diverges.
If \(L\) is 1, the terms change by a factor close to 1. The dots do not show whether the series converges.
Notation
Find the limit of the absolute ratio, which may be a number or infinite.
\[\cy{L} = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|\]If \(L < 1\), the series converges absolutely, so the series itself also converges.
\[L < 1 \ \Rightarrow \ \sum |a_n| \text{ converges}\]If \(L > 1\), or if \(L\) is infinite, the series diverges.
\[L > 1 \ \Rightarrow \ \sum a_n \text{ diverges}\]If \(L = 1\), the test gives no conclusion.
\[L = 1 \ \Rightarrow \ \text{no conclusion}\]The value of \(L\) is a limit of ratios of terms. It is not the sum of the series.
When it fails
The ratios can fail to approach a value. For the terms 1/2, 1, 1/2, 1, ... the ratios alternate between 2 and 1/2, so their limit does not exist.
The ratio test needs a limit, so it gives no conclusion here. The series diverges, but another test shows it: its terms do not approach 0.
Example 1: A limit below 1
Use the ratio test on \(\displaystyle\sum_{n=1}^{\infty} \frac{n^3}{2^n}\).
Form the ratio and cancel the powers of 2.
\[\begin{aligned}\left|\frac{a_{n+1}}{a_n}\right| &= \frac{(n+1)^3}{2^{n+1}} \cdot \frac{2^n}{n^3}\\ &= \frac{1}{2}\left(\frac{n+1}{n}\right)^3\end{aligned}\]As \(n\) grows, \(\frac{n+1}{n} = 1 + \frac{1}{n}\) approaches 1, so its cube approaches 1. The dots approach the orange dashed line at \(\frac12\).
\[L = \frac{1}{2} \cdot 1^3 = \cy{\frac{1}{2}}\]The limit is less than 1, so the series converges absolutely. The first three ratios are above 1, but the test uses only the limit.
Example 2: A limit above 1
Use the ratio test on \(\displaystyle\sum_{n=1}^{\infty} \frac{n!}{4^n}\).
Form the ratio and cancel the factorials and the powers of 4.
\[\begin{aligned}\left|\frac{a_{n+1}}{a_n}\right| &= \frac{(n+1)!}{4^{n+1}} \cdot \frac{4^n}{n!}\\ &= \frac{n+1}{4}\end{aligned}\]A tempting wrong move: The first ratios are \(\frac12\) and \(\frac34\). They are below 1, so the terms shrink and the series converges.
The red dots are the first two ratios, and the test does not use them. The dots reach the dashed line at n = 3 and then rise, so after the fourth bar each bar is larger than the one before it.
The ratio \(\frac{n+1}{4}\) grows without bound, so the limit is infinite.
\[L = \lim_{n\to\infty} \frac{n+1}{4} = \infty\]An infinite limit counts as more than 1, so the terms eventually grow. They do not approach 0, so the series diverges.
Practice
Use the ratio test on \(\displaystyle\sum_{n=1}^{\infty} \frac{(n+1)!}{n^n}\).
Cancel the factorials first. Then move the slider to large \(n\) and compare the value that the dots approach with 1.
Solution
Form the ratio. The factorials cancel to \(n+2\). Write \((n+1)^{n+1}\) as \((n+1)(n+1)^n\), so that \(n^n\) and \((n+1)^n\) form one power.
\[\begin{aligned}\left|\frac{a_{n+1}}{a_n}\right| &= \frac{(n+2)!}{(n+1)^{n+1}} \cdot \frac{n^n}{(n+1)!}\\ &= \frac{n+2}{n+1}\left(\frac{n}{n+1}\right)^n\end{aligned}\]The first part approaches 1. The second part equals \(\left(1 + \frac{1}{n}\right)^{-n}\), which approaches \(\frac{1}{e}\), because \(\left(1 + \frac{1}{n}\right)^n\) approaches \(e\).
\[L = 1 \cdot \frac{1}{e} \approx 0.37\]The limit is less than 1, so the series converges absolutely.
Questions
The graph shows the ratio of neighboring terms of a series as dots, for n from 1 to 24. The dashed line is at the height 1.
What number do the dots approach as n grows?
- 0.6
- 1.2
- 1
- 0
Answer
0.6
The first dot is above the dashed line. What does the ratio test conclude?
- The series converges absolutely, because the dots approach a value below 1.
- The series diverges, because one ratio is above 1.
- The test is inconclusive, because the first ratio is above 1.
- The series converges, and its sum is 0.6.
Answer
The series converges absolutely, because the dots approach a value below 1.
What does the ratio test show for \(\displaystyle\sum (n+2)\left(\frac{5}{8}\right)^n\)?
- It converges absolutely, because \(L = \frac{5}{8}\).
- It converges, and its sum is \(\frac{5}{8}\).
- It diverges, because \(L\) is not 0.
Answer
It converges absolutely, because \(L = \frac{5}{8}\).
Summary
A ratio limit L below 1 means absolute convergence, L above 1 means divergence, and L equal to 1 gives no conclusion.
This works because, far along the series, each term is about L times the one before it, as in a geometric series with ratio L.
The test suits terms that contain factorials, or powers with n in the exponent.
This is one of four skills in topic 10.8 of Which & Why: AP Calculus.