9.6: Ratio and Root Tests
( \newcommand{\kernel}{\mathrm{null}\,}\)
- Use the ratio test to determine absolute convergence of a series.
- Use the root test to determine absolute convergence of a series.
- Describe a strategy for testing the convergence of a given series.
In this section, we prove the last two series convergence tests: the ratio test and the root test. These tests are particularly nice because they do not require us to find a comparable series. The ratio test will be especially useful in the discussion of power series in the next chapter. Throughout this chapter, we have seen that no single convergence test works for all series. Therefore, at the end of this section we discuss a strategy for choosing which convergence test to use for a given series.
Ratio Test
Consider a series ∞∑n=1an. From our earlier discussion and examples, we know that limn→∞an=0 is not a sufficient condition for the series to converge. Not only do we need an→0, but we need an→0 quickly enough. For example, consider the series ∞∑n=11n and the series ∞∑n=11n2. We know that 1n→0 and 1n2→0. However, only the series ∞∑n=11n2 converges. The series ∞∑n=11n diverges because the terms in the sequence {1n} do not approach zero fast enough as n→∞. Here we introduce the ratio test, which provides a way of measuring how fast the terms of a series approach zero.
Let ∞∑n=1an be a series with nonzero terms. Let
ρ=limn→∞|an+1an|.
- If 0≤ρ<1, then ∞∑n=1an converges absolutely.
- If ρ>1 or ρ=∞, then ∞∑n=1an diverges.
- If ρ=1, the test does not provide any information.
Let ∞∑n=1an be a series with nonzero terms.
We begin with the proof of part i. In this case, ρ=limn→∞|an+1an|<1. Since 0≤ρ<1, there exists R such that 0≤ρ<R<1. Let ε=R−ρ>0. By the definition of limit of a sequence, there exists some integer N such that
||an+1an|−ρ|<ε,for alln≥N.
Therefore,
|an+1an|<ρ+ε=R,for alln≥N
and, thus,
|aN+1|<R|aN|
|aN+2|<R|aN+1|<R2|aN|
|aN+3|<R|aN+2|<R2|aN+1|<R3|aN|
|aN+4|<R|aN+3|<R2|aN+2|<R3|aN+1|<R4|aN|
⋮.
Since R<1, the geometric series
R|aN|+R2|aN|+R3|aN|+⋯
converges. Given the inequalities above, we can apply the comparison test and conclude that the series
|aN+1|+|aN+2|+|aN+3|+|aN+4|+⋯
converges. Therefore, since
∞∑n=1|an|=N∑n=1|an|+∞∑n=N+1|an|
where N∑n=1|an| is a finite sum and ∞∑n=N+1|an| converges, we conclude that ∞∑n=1|an| converges.
For part ii.
ρ=limn→∞|an+1an|>1.
Since ρ>1, there exists R such that ρ>R>1. Let ε=ρ−R>0. By the definition of the limit of a sequence, there exists an integer N such that
||an+1an|−ρ|<ε,for alln≥N.
Therefore,
R=ρ−ε<|an+1an|,for alln≥N,
and, thus,
|aN+1|>R|aN|
|aN+2|>R|aN+1|>R2|aN|
|aN+3|>R|aN+2|>R2|aN+1|>R3|aN|
|aN+4|>R|aN+3|>R2|aN+2|>R3|aN+1|>R4|aN|.
Since R>1, the geometric series
R|aN|+R2|aN|+R3|aN|+⋯
diverges. Applying the comparison test, we conclude that the series
|aN+1|+|aN+2|+|aN+3|+⋯
diverges, and therefore the series ∞∑n=1|an| diverges.
For part iii. we show that the test does not provide any information if ρ=1 by considering the p−series ∞∑n=11np. For any real number p,
ρ=limn→∞1/(n+1)p1/np=limn→∞np(n+1)p=1.
However, we know that if p≤1, the p−series ∞∑n=11np diverges, whereas ∞∑n=11np converges if p>1.
□
The ratio test is particularly useful for series whose terms contain factorials or exponential, where the ratio of terms simplifies the expression. The ratio test is convenient because it does not require us to find a comparative series. The drawback is that the test sometimes does not provide any information regarding convergence.
For each of the following series, use the ratio test to determine whether the series converges or diverges.
- ∞∑n=12nn!
- ∞∑n=1nnn!
- ∞∑n=1(−1)n(n!)2(2n)!
Solution
a. From the ratio test, we can see that
ρ=limn→∞2n+1/(n+1)!2n/n!=limn→∞2n+1(n+1)!⋅n!2n.
Since (n+1)!=(n+1)⋅n!,
ρ=limn→∞2n+1=0.
Since ρ<1, the series converges.
b. We can see that
ρ=limn→∞(n+1)n+1/(n+1)!nn/n!=limn→∞(n+1)n+1(n+1)!⋅n!nn=limn→∞(n+1n)n=limn→∞(1+1n)n=e.
Since ρ>1, the series diverges.
c. Since
|(−1)n+1((n+1)!)2/(2(n+1))!(−1)n(n!)2/(2n)!|=(n+1)!(n+1)!(2n+2)!⋅(2n)!n!n!=(n+1)(n+1)(2n+2)(2n+1)
we see that
ρ=limn→∞(n+1)(n+1)(2n+2)(2n+1)=14.
Since ρ<1, the series converges.
Use the ratio test to determine whether the series ∞∑n=1n33n converges or diverges.
- Hint
-
Evaluate limn→∞(n+1)33n+1⋅3nn3.
- Answer
-
The series converges.
Root Test
The approach of the root test is similar to that of the ratio test. Consider a series ∞∑n=1an such that limn→∞n√|an|=ρ for some real number ρ. Then for N sufficiently large, |aN|≈ρN. Therefore, we can approximate ∞∑n=N|an| by writing
|aN|+|aN+1|+|aN+2|+⋯≈ρN+ρN+1+ρN+2+⋯.
The expression on the right-hand side is a geometric series. As in the ratio test, the series ∞∑n=1an converges absolutely if 0≤ρ<1 and the series diverges if ρ≥1. If ρ=1, the test does not provide any information. For example, for any p-series, ∞∑n=11np, we see that
ρ=limn→∞n√|1np|=limn→∞1np/n.
To evaluate this limit, we use the natural logarithm function. Doing so, we see that
lnρ=ln(limn→∞1np/n)=limn→∞ln(1n)p/n=limn→∞pn⋅ln(1n)=limn→∞pln(1/n)n.
Using L’Hôpital’s rule, it follows that lnρ=0, and therefore ρ=1 for all p. However, we know that the p-series only converges if p>1 and diverges if p<1.
Consider the series ∞∑n=1an. Let
ρ=limn→∞n√|an|.
- If 0≤ρ<1, then ∞∑n=1an converges absolutely.
- If ρ>1 or ρ=∞, then ∞∑n=1an diverges.
- If ρ=1, the test does not provide any information.
The root test is useful for series whose terms involve exponentials. In particular, for a series whose terms an satisfy |an|=(bn)n, then n√|an|=bn and we need only evaluate limn→∞bn.
For each of the following series, use the root test to determine whether the series converges or diverges.
- ∞∑n=1(n2+3n)n(4n2+5)n
- ∞∑n=1nn(ln(n))n
Solution
a. To apply the root test, we compute
ρ=limn→∞n√(n2+3n)n(4n2+5)n=limn→∞n2+3n4n2+5=14.
Since ρ<1, the series converges absolutely.
b. We have
ρ=limn→∞n√nn(ln(n))n=limn→∞nlnn=∞by L’Hôpital’s rule.
Since ρ=∞, the series diverges.
Use the root test to determine whether the series ∞∑n=11nn converges or diverges.
- Hint
-
Evaluate limn→∞n√1nn.
- Answer
-
The series converges.
Choosing a Convergence Test
At this point, we have a long list of convergence tests. However, not all tests can be used for all series. When given a series, we must determine which test is the best to use. Here is a strategy for finding the best test to apply.
Consider a series ∞∑n=1an. In the steps below, we outline a strategy for determining whether the series converges.
- Is ∞∑n=1an a familiar series? For example, is it the harmonic series (which diverges) or the alternating harmonic series (which converges)? Is it a p−series or geometric series? If so, check the power p or the ratio r to determine if the series converges.
- Is it an alternating series? Are we interested in absolute convergence or just convergence? If we are just interested in whether the series converges, apply the alternating series test. If we are interested in absolute convergence, proceed to step 3, considering the series of absolute values ∞∑n=1|an|.
- Is the series similar to a p−series or geometric series? If so, try the comparison test or limit comparison test.
- Do the terms in the series contain a factorial or power? If the terms are powers such that an=(bn)n, try the root test first. Otherwise, try the ratio test first.
- Use the divergence test. If this test does not provide any information, try the integral test.
For each of the following series, determine which convergence test is the best to use and explain why. Then determine if the series converges or diverges. If the series is an alternating series, determine whether it converges absolutely, converges conditionally, or diverges.
- ∞∑n=1n2+2nn3+3n2+1
- ∞∑n=1(−1)n+1(3n+1)n!
- ∞∑n=1enn3
- ∞∑n=13n(n+1)n
Solution
a. Step 1. The series is not a p–series or geometric series.
Step 2. The series is not alternating.
Step 3. For large values of n, we approximate the series by the expression
n2+2nn3+3n2+1≈n2n3=1n.
Therefore, it seems reasonable to apply the comparison test or limit comparison test using the series ∞∑n=11/n. Using the limit comparison test, we see that
limn→∞(n2+2n)/(n3+3n2+1)1/n=limn→∞n3+2n2n3+3n2+1=1.
Since the series ∞∑n=11/n
diverges, this series diverges as well.
b. Step 1.The series is not a familiar series.
Step 2. The series is alternating. Since we are interested in absolute convergence, consider the series
∞∑n=13n(n+1)!.
Step 3. The series is not similar to a p-series or geometric series.
Step 4. Since each term contains a factorial, apply the ratio test. We see that
limn→∞(3(n+1))/(n+1)!(3n+1)/n!=limn→∞3n+3(n+1)!⋅n!3n+1=limn→∞3n+3(n+1)(3n+1)=0.
Therefore, this series converges, and we conclude that the original series converges absolutely, and thus converges.
c. Step 1. The series is not a familiar series.
Step 2. It is not an alternating series.
Step 3. There is no obvious series with which to compare this series.
Step 4. There is no factorial. There is a power, but it is not an ideal situation for the root test.
Step 5. To apply the divergence test, we calculate that
limn→∞enn3=∞.
Therefore, by the divergence test, the series diverges.
d. Step 1. This series is not a familiar series.
Step 2. It is not an alternating series.
Step 3. There is no obvious series with which to compare this series.
Step 4. Since each term is a power of n,we can apply the root test. Since
limn→∞n√(3n+1)n=limn→∞3n+1=0,
by the root test, we conclude that the series converges.
For the series ∞∑n=12n3n+n, determine which convergence test is the best to use and explain why.
- Hint
-
The series is similar to the geometric series ∞∑n=1(23)n.
- Answer
-
The comparison test because 2n3n+n<2n3n for all positive integers n. The limit comparison test could also be used.
In Table, we summarize the convergence tests and when each can be applied. Note that while the comparison test, limit comparison test, and integral test require the series ∞∑n=1an to have nonnegative terms, if ∞∑n=1an has negative terms, these tests can be applied to ∞∑n=1|an| to test for absolute convergence.
Series or Test | Conclusions | Comments |
---|---|---|
Divergence Test For any series ∞∑n=1an, evaluate limn→∞an. |
If limn→∞an=0, the test is inconclusive. | This test cannot prove convergence of a series. |
If limn→∞an≠0, the series diverges. | ||
Geometric Series ∞∑n=1arn−1 |
If |r|<1, the series converges to a/(1−r). | Any geometric series can be reindexed to be written in the form a+ar+ar2+⋯, where a is the initial term and r is the ratio. |
If |r|≥1, the series diverges. | ||
p-Series ∞∑n=11np |
If p>1, the series converges. | For p=1, we have the harmonic series ∞∑n=11/n. |
If p≤1, the series diverges. | ||
Comparison Test For ∞∑n=1an with nonnegative terms, compare with a known series ∞∑n=1bn. |
If an≤bn for all n≥N and ∞∑n=1bn converges, then ∞∑n=1an converges. | Typically used for a series similar to a geometric or p-series. It can sometimes be difficult to find an appropriate series. |
If an≥bn for all n≥N and ∞∑n=1bn diverges, then ∞∑n=1an diverges. | ||
Limit Comparison Test For ∞∑n=1an with positive terms, compare with a series ∞∑n=1bn by evaluating L=limn→∞anbn. |
If L is a real number and L≠0, then ∞∑n=1an and ∞∑n=1bn both converge or both diverge. | Typically used for a series similar to a geometric or p-series. Often easier to apply than the comparison test. |
If L=0 and ∞∑n=1bn converges, then ∞∑n=1an converges. | ||
If L=∞ and ∞∑n=1bn diverges, then ∞∑n=1an diverges. | ||
Integral Test If there exists a positive, continuous, decreasing function f such that an=f(n) for all n≥N, evaluate ∫∞Nf(x)dx. |
∫∞Nf(x)dx and ∞∑n=1an both converge or both diverge. | Limited to those series for which the corresponding function f can be easily integrated. |
Alternating Series ∞∑n=1(−1)n+1bn or ∞∑n=1(−1)nbn |
If bn+1≤bn for all n≥1 and bn→0, then the series converges. | Only applies to alternating series. |
Ratio Test For any series ∞∑n=1an with nonzero terms, let ρ=limn→∞|an+1an| |
If 0≤ρ<1, the series converges absolutely. | Often used for series involving factorials or exponentials. |
If ρ>1 or ρ=∞, the series diverges. | ||
If ρ=1, the test is inconclusive. | ||
Root Test For any series ∞∑n=1an, let \displaystyle ρ=\lim_{n→∞}\sqrt[n]{|a_n|}. |
If 0≤ρ<1, the series converges absolutely. | Often used for series where |a_n|=(b_n)^n. |
If ρ>1 or ρ=∞, the series diverges. | ||
If ρ=1, the test is inconclusive. |
Dozens of series exist that converge to π or an algebraic expression containing π. Here we look at several examples and compare their rates of convergence. By rate of convergence, we mean the number of terms necessary for a partial sum to be within a certain amount of the actual value. The series representations of π in the first two examples can be explained using Maclaurin series, which are discussed in the next chapter. The third example relies on material beyond the scope of this text.
1. The series
π=4\sum_{n=1}^∞\frac{(−1)^{n+1}}{2n−1}=4−\frac{4}{3}+\frac{4}{5}−\frac{4}{7}+\frac{4}{9}−⋯ \nonumber
was discovered by Gregory and Leibniz in the late 1600s. This result follows from the Maclaurin series for f(x)=\tan^{−1}x. We will discuss this series in the next chapter.
a. Prove that this series converges.
b. Evaluate the partial sums S_n for n=10,20,50,100.
c. Use the remainder estimate for alternating series to get a bound on the error R_n.
d. What is the smallest value of N that guarantees |R_N|<0.01? Evaluate S_N.
2. The series
π=6\sum^∞_{n=0}\frac{(2n)!}{2^{4n+1}(n!)^2(2n+1)}=6\left(\frac{1}{2}+\frac{1}{2⋅3}\left(\frac{1}{2}\right)^3+\frac{1⋅3}{2⋅4⋅5}⋅\left(\frac{1}{2}\right)^5+\frac{1⋅3⋅5}{2⋅4⋅6⋅7}\left(\frac{1}{2}\right)^7+⋯\right) \nonumber
has been attributed to Newton in the late 1600s. The proof of this result uses the Maclaurin series for f(x)=\sin^{−1}x.
a. Prove that the series converges.
b. Evaluate the partial sums S_n for n=5,10,20.
c. Compare S_n to π for n=5,10,20 and discuss the number of correct decimal places.
3. The series
\frac{1}{π}=\frac{\sqrt{8}}{9801}\sum_{n=0}^∞\frac{(4n)!(1103+26390n)}{(n!)^4396^{4n}} \nonumber
was discovered by Ramanujan in the early 1900s. William Gosper, Jr., used this series to calculate π to an accuracy of more than 17 million digits in the mid-1980s. At the time, that was a world record. Since that time, this series and others by Ramanujan have led mathematicians to find many other series representations for π and 1/π.
a. Prove that this series converges.
b. Evaluate the first term in this series. Compare this number with the value of π from a calculating utility. To how many decimal places do these two numbers agree? What if we add the first two terms in the series?
c. Investigate the life of Srinivasa Ramanujan (1887–1920) and write a brief summary. Ramanujan is one of the most fascinating stories in the history of mathematics. He was basically self-taught, with no formal training in mathematics, yet he contributed in highly original ways to many advanced areas of mathematics.
Key Concepts
- For the ratio test, we consider ρ=\lim_{n→∞}\left|\frac{a_{n+1}}{a_n}\right|. \nonumber If ρ<1, the series \displaystyle \sum_{n=1}^∞a_n converges absolutely. If ρ>1, the series diverges. If ρ=1, the test does not provide any information. This test is useful for series whose terms involve factorials.
- For the root test, we consider ρ=\lim_{n→∞}\sqrt[n]{|a_n|}. \nonumber If ρ<1, the series \displaystyle \sum_{n=1}^∞a_n converges absolutely. If ρ>1, the series diverges. If ρ=1, the test does not provide any information. The root test is useful for series whose terms involve powers.
- For a series that is similar to a geometric series or p−series, consider one of the comparison tests.
Glossary
- ratio test
- for a series \displaystyle \sum^∞_{n=1}a_n with nonzero terms, let \displaystyle ρ=\lim_{n→∞}\left|\frac{a_{n+1}}{a_n}\right|; if 0≤ρ<1, the series converges absolutely; if ρ>1, the series diverges; if ρ=1, the test is inconclusive
- root test
- for a series \displaystyle \sum^∞_{n=1}a_n, let \displaystyle ρ=\lim_{n→∞}\sqrt[n]{|a_n|}; if 0≤ρ<1, the series converges absolutely; if ρ>1, the series diverges; if ρ=1, the test is inconclusive