9.6: Ratio and Root Tests
 Page ID
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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 \(\displaystyle \sum_{n=1}^∞a_n\). From our earlier discussion and examples, we know that \(\displaystyle \lim_{n→∞}a_n=0\) is not a sufficient condition for the series to converge. Not only do we need \( a_n→0\), but we need \( a_n→0\) quickly enough. For example, consider the series \(\displaystyle \sum_{n=1}^∞\frac{1}{n}\) and the series \(\displaystyle \sum_{n=1}^∞\frac{1}{n^2}\). We know that \( \frac{1}{n}→0\) and \( \frac{1}{n^2}→0\). However, only the series \(\displaystyle \sum_{n=1}^∞ \frac{1}{n^2}\) converges. The series \(\displaystyle \sum_{n=1}^∞\frac{1}{n}\) diverges because the terms in the sequence \( \left\{\frac{1}{n}\right\}\) 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.
Ratio Test
Let \(\displaystyle \sum^∞_{n=1}a_n\) be a series with nonzero terms. Let
\[ρ=\lim_{n→∞} \left\frac{a_{n+1}}{a_n}\right.\]
 If \( 0≤ρ<1,\) then \(\displaystyle \sum^∞_{n=1}a_n\) converges absolutely.
 If \( ρ>1\) or \( ρ=∞\), then \(\displaystyle \sum^∞_{n=1}a_n\) diverges.
 If \( ρ=1,\) the test does not provide any information.
Proof
Let \(\displaystyle \sum_{n=1}^∞a_n\) be a series with nonzero terms.
We begin with the proof of part i. In this case, \( ρ=\lim_{n→∞}∣\frac{a_{n+1}}{a_n}∣<1.\) 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
\[\left\left\frac{a_{n+1}}{a_n}\right−ρ\right<ε,\;\text{for all}\; n≥N.\]
Therefore,
\[\left\frac{a_{n+1}}{a_n}\right<ρ+ε=R, \;\text{for all}\; n≥N\]
and, thus,
\( a_{N+1}<Ra_N\)
\( ∣a_{N+2}∣<R∣a_{N+1}∣<R^2∣a_N∣\)
\( ∣a_{N+3}∣<R∣a_{N+2}∣<R^2∣a_{N+1}∣<R^3∣a_N∣\)
\( ∣a_{N+4}∣<R∣a_{N+3}∣<R^2∣a_{N+2}∣<R^3∣a_{N+1}∣<R^4∣a_N∣\)
\( ⋮.\)
Since \( R<1,\) the geometric series
\[R∣a_N∣+R^2∣a_N∣+R^3∣a_N∣+⋯\]
converges. Given the inequalities above, we can apply the comparison test and conclude that the series
\[a_{N+1}+a_{N+2}+a_{N+3}+a_{N+4}+⋯\]
converges. Therefore, since
\[\sum_{n=1}^∞a_n=\sum_{n=1}^Na_n+\sum_{n=N+1}^∞a_n\]
where \(\displaystyle \sum_{n=1}^Na_n\) is a finite sum and \(\displaystyle \sum_{n=N+1}^∞a_n\) converges, we conclude that \(\displaystyle \sum_{n=1}^∞a_n\) converges.
For part ii.
\[ρ=\lim_{n→∞}\left\frac{a_{n+1}}{a_n}\right>1.\]
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
\[\left\left\frac{a_{n+1}}{a_n}\right−ρ\right<ε, \;\text{for all}\; n≥N.\]
Therefore,
\[R=ρ−ε<\left\frac{a_{n+1}}{a_n}\right, \;\text{for all}\; n≥N,\]
and, thus,
\( a_{N+1}>Ra_N\)
\( ∣a_{N+2}∣>R∣a_{N+1}∣>R^2∣a_N∣\)
\( ∣a_{N+3}∣>R∣a_{N+2}∣>R^2∣a_{N+1}∣>R^3∣a_N∣\)
\( ∣a_{N+4}∣>R∣a_{N+3}∣>R^2∣a_{N+2}∣>R^3∣a_{N+1}∣>R^4∣a_N∣.\)
Since \( R>1,\) the geometric series
\[R∣a_N∣+R^2∣a_N∣+R^3∣a_N∣+⋯\]
diverges. Applying the comparison test, we conclude that the series
\[a_{N+1}+a_{N+2}+a_{N+3}+⋯\]
diverges, and therefore the series \(\displaystyle \sum_{n=1}^∞a_n\) diverges.
For part iii. we show that the test does not provide any information if \( ρ=1\) by considering the \( p−series\) \(\displaystyle \sum^∞_{n=1}\frac{1}{n^p}\). For any real number \( p\),
\[ρ=\lim_{n→∞}\frac{1/(n+1)^p}{1/n^p}=\lim_{n→∞}\frac{n^p}{(n+1)^p}=1.\]
However, we know that if \( p≤1,\) the p−series \(\displaystyle \sum^∞_{n=1}\frac{1}{n^p}\) diverges, whereas \(\displaystyle \sum^∞_{n=1}\frac{1}{n^p}\) 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.
Example \( \PageIndex{1}\): Using the Ratio Test
For each of the following series, use the ratio test to determine whether the series converges or diverges.
 \(\displaystyle \sum^∞_{n=1}\frac{2^n}{n!}\)
 \(\displaystyle \sum^∞_{n=1}\frac{n^n}{n!} \)
 \(\displaystyle \sum_{n=1}^∞\frac{(−1)^n(n!)^2}{(2n)!}\)
Solution
a. From the ratio test, we can see that
\[ ρ=\lim_{n→∞}\frac{2^{n+1}/(n+1)!}{2^n/n!}=\lim_{n→∞}\frac{2^{n+1}}{(n+1)!}⋅\frac{n!}{2^n}.\]
Since \( (n+1)!=(n+1)⋅n!,\)
\[ ρ=\lim_{n→∞}\frac{2}{n+1}=0.\]
Since \( ρ<1,\) the series converges.
b. We can see that
\[ ρ=\lim_{n→∞}\frac{(n+1)^{n+1}/(n+1)!}{n^n/n!}=\lim_{n→∞}\frac{(n+1)^{n+1}}{(n+1)!}⋅\frac{n!}{n^n}=\lim_{n→∞}(\frac{n+1}{n})^n=\lim_{n→∞}(1+\frac{1}{n})^n=e.\]
Since \( ρ>1,\) the series diverges.
c. Since
\[ ∣\frac{(−1)^{n+1}((n+1)!)^2/(2(n+1))!}{(−1)^n(n!)^2/(2n)!}∣=\frac{(n+1)!(n+1)!}{(2n+2)!}⋅\frac{(2n)!}{n!n!}=\frac{(n+1)(n+1)}{(2n+2)(2n+1)}\]
we see that
\[ ρ=\lim_{n→∞}\frac{(n+1)(n+1)}{(2n+2)(2n+1)}=\frac{1}{4}.\]
Since \( ρ<1\), the series converges.
Exercise \(\PageIndex{1}\)
Use the ratio test to determine whether the series \(\displaystyle \sum^∞_{n=1}\frac{n^3}{3^n}\) converges or diverges.
 Hint

Evaluate \(\displaystyle \lim_{n→∞}\frac{(n+1)^3}{3^{n+1}}⋅\frac{3^n}{n^3}.\)
 Answer

The series converges.
Root Test
The approach of the root test is similar to that of the ratio test. Consider a series \(\displaystyle \sum^∞_{n=1}a_n\) such that \(\displaystyle \lim_{n→∞}\sqrt[n]{a_n}=ρ\) for some real number \( ρ\). Then for \( N\) sufficiently large, \( ∣a_N∣≈ρN.\) Therefore, we can approximate \(\displaystyle \sum_{n=N}^∞a_n\) by writing
\[∣a_N∣+∣a_{N+1}∣+∣a_{N+2}∣+⋯≈ρ^N+ρ^{N+1}+ρ^{N+2}+⋯.\]
The expression on the righthand side is a geometric series. As in the ratio test, the series \(\displaystyle \sum^∞_{n=1}a_n\) 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 pseries, \(\displaystyle \sum_{n=1}^∞\frac{1}{n^p}\), we see that
\[ρ=\lim_{n→∞}\sqrt[n]{∣\frac{1}{n^p}∣}=\lim_{n→∞}\frac{1}{n^{p/n}}\].
To evaluate this limit, we use the natural logarithm function. Doing so, we see that
\( \ln ρ=\ln(\lim_{n→∞}\frac{1}{n^{p/n}})=\lim_{n→∞}\ln(\frac{1}{n})^{p/n}=\lim_{n→∞}\frac{p}{n}⋅\ln(\frac{1}{n})=\lim_{n→∞}\frac{p\ln(1/n)}{n}.\)
Using L’Hôpital’s rule, it follows that \( \ln ρ=0\), and therefore \( ρ=1\) for all \( p\). However, we know that the pseries only converges if \( p>1\) and diverges if \( p<1\).
Root Test
Consider the series \(\displaystyle \sum^∞_{n=1}a_n\). Let
\[ρ=\lim_{n→∞}\sqrt[n]{a_n}\].
 If \( 0≤ρ<1,\) then \(\displaystyle \sum^∞_{n=1}a_n\) converges absolutely.
 If \( ρ>1\) or \( ρ=∞\), then \(\displaystyle \sum^∞_{n=1}a_n\) 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 \( a_n\) satisfy \( a_n=(b_n)^n\), then \( \sqrt[n]{a_n}=b_n\) and we need only evaluate \(\displaystyle \lim_{n→∞}b_n\).
Example \( \PageIndex{2}\): Using the Root Test
For each of the following series, use the root test to determine whether the series converges or diverges.
 \(\displaystyle \sum^∞_{n=1}\frac{(n^2+3n)^n}{(4n^2+5)^n}\)
 \(\displaystyle \sum^∞_{n=1}\frac{n^n}{(\ln(n))^n}\)
Solution
a. To apply the root test, we compute
\[ ρ=\lim_{n→∞}\sqrt[n]{(n^2+3n)^n/(4n^2+5)^n}=\lim_{n→∞}\frac{n^2+3n}{4n^2+5}=\frac{1}{4}.\]
Since \( ρ<1,\) the series converges absolutely.
b. We have
\[ ρ=\lim_{n→∞}\sqrt[n]{n^n/(\ln n)^n}=\lim_{n→∞}\frac{n}{\ln n}=∞\quad \text{by L’Hôpital’s rule.}\]
Since \( ρ=∞\), the series diverges.
Exercise \(\PageIndex{2}\)
Use the root test to determine whether the series \(\displaystyle \sum^∞_{n=1}1/n^n\) converges or diverges.
 Hint

Evaluate \(\displaystyle \lim_{n→∞}\sqrt[n]{\frac{1}{n^n}}\).
 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.
ProblemSolving Strategy: Choosing a Convergence Test for a Series
Consider a series \(\displaystyle \sum_{n=1}^∞a_n.\) In the steps below, we outline a strategy for determining whether the series converges.
 Is \(\displaystyle \sum_{n=1}^∞a_n\) 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 \(\displaystyle \sum_{n=1}^∞a_n.\)
 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 \( a_n=(b_n)^n,\) 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.
Visit this website for more information on testing series for convergence, plus general information on sequences and series.
Example \( \PageIndex{3}\): Using Convergence Tests
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.
 \(\displaystyle \sum^∞_{n=1}\frac{n^2+2n}{n^3+3n^2+1}\)
 \(\displaystyle \sum^∞_{n=1}\frac{(−1)^{n+1}(3n+1)}{n!}\)
 \(\displaystyle \sum^∞_{n=1}\frac{e^n}{n^3}\)
 \(\displaystyle \sum^∞_{n=1}\frac{3^n}{(n+1)^n}\)
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
\( \frac{n^2+2n}{n^3+3n^2+1}≈\frac{n^2}{n^3}=\frac{1}{n}.\)
Therefore, it seems reasonable to apply the comparison test or limit comparison test using the series \(\displaystyle \sum_{n=1}^∞1/n\). Using the limit comparison test, we see that
\(\displaystyle \lim_{n→∞}\frac{(n^2+2n)/(n^3+3n^2+1)}{1/n}=\lim_{n→∞}\frac{n^3+2n^2}{n^3+3n^2+1}=1.\)
Since the series \(\displaystyle \sum_{n=1}^∞1/n\)
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
\(\displaystyle \sum_{n=1}^∞\frac{3n}{(n+1)!}.\)
Step 3. The series is not similar to a pseries or geometric series.
Step 4. Since each term contains a factorial, apply the ratio test. We see that
\(\displaystyle \lim_{n→∞}\frac{(3(n+1))/(n+1)!}{(3n+1)/n!}=\lim_{n→∞}\frac{3n+3}{(n+1)!}⋅\frac{n!}{3n+1}=\lim_{n→∞}\frac{3n+3}{(n+1)(3n+1)}=0.\)
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
\(\displaystyle \lim_{n→∞}\frac{e^n}{n^3}=∞.\)
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
\(\displaystyle \lim_{n→∞}\sqrt[n]{(\frac{3}{n+1})^n}=\lim_{n→∞}\frac{3}{n+1}=0,\)
by the root test, we conclude that the series converges.
Exercise \(\PageIndex{3}\)
For the series \(\displaystyle \sum^∞_{n=1}\frac{2^n}{3^n+n}\), determine which convergence test is the best to use and explain why.
 Hint

The series is similar to the geometric series \(\displaystyle \sum^∞_{n=1}\left(\frac{2}{3}\right)^n\).
 Answer

The comparison test because \( \dfrac{2^n}{3^n+n}<\dfrac{2^n}{3^n}\) 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 \(\displaystyle \sum_{n=1}^∞a_n\) to have nonnegative terms, if \(\displaystyle \sum_{n=1}^∞a_n\) has negative terms, these tests can be applied to \(\displaystyle \sum_{n=1}^∞a_n\) to test for absolute convergence.
Series or Test  Conclusions  Comments 

Divergence Test For any series \(\displaystyle \sum^∞_{n=1}a_n\), evaluate \(\displaystyle \lim_{n→∞}a_n\). 
If \(\displaystyle \lim_{n→∞}a_n=0\), the test is inconclusive.  This test cannot prove convergence of a series. 
If \(\displaystyle \lim_{n→∞}a_n≠0\), the series diverges.  
Geometric Series \(\displaystyle \sum^∞_{n=1}ar^{n−1}\) 
If \( r<1\), the series converges to \( a/(1−r)\).  Any geometric series can be reindexed to be written in the form \( a+ar+ar^2+⋯\), where \( a\) is the initial term and r is the ratio. 
If \( r≥1,\) the series diverges.  
pSeries \(\displaystyle \sum^∞_{n=1}\frac{1}{n^p}\) 
If \( p>1\), the series converges.  For \( p=1\), we have the harmonic series \(\displaystyle \sum^∞_{n=1}1/n\). 
If \( p≤1\), the series diverges.  
Comparison Test For \(\displaystyle \sum^∞_{n=1}a_n \) with nonnegative terms, compare with a known series \(\displaystyle \sum^∞_{n=1}b_n\). 
If \( a_n≤b_n\) for all \( n≥N\) and \(\displaystyle \sum^∞_{n=1}b_n\) converges, then \(\displaystyle \sum^∞_{n=1}a_n\) converges.  Typically used for a series similar to a geometric or \( p\)series. It can sometimes be difficult to find an appropriate series. 
If \( a_n≥b_n\) for all \( n≥N\) and \(\displaystyle \sum^∞_{n=1}b_n\) diverges, then \(\displaystyle \sum^∞_{n=1}a_n\) diverges.  
Limit Comparison Test For \(\displaystyle \sum^∞_{n=1}a_n\) with positive terms, compare with a series \(\displaystyle \sum^∞_{n=1}b_n\) by evaluating \( L=\displaystyle \lim_{n→∞}\frac{a_n}{b_n}.\) 
If \( L\) is a real number and \( L≠0\), then \(\displaystyle \sum^∞_{n=1}a_n\) and \(\displaystyle \sum^∞_{n=1}b_n\) 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 \(\displaystyle \sum^∞_{n=1}b_n\) converges, then \(\displaystyle \sum^∞_{n=1}a_n\) converges.  
If \( L=∞\) and \(\displaystyle \sum^∞_{n=1}b_n\) diverges, then \(\displaystyle \sum^∞_{n=1}a_n\) diverges.  
Integral Test If there exists a positive, continuous, decreasing function \( f\) such that \( a_n=f(n)\) for all \( n≥N\), evaluate \( \displaystyle ∫^∞_Nf(x)dx.\) 
\( ∫^∞_Nf(x)dx\) and \(\displaystyle \sum^∞_{n=1}a_n\) both converge or both diverge.  Limited to those series for which the corresponding function f can be easily integrated. 
Alternating Series \(\displaystyle \sum^∞_{n=1}(−1)^{n+1}b_n\) or \(\displaystyle \sum^∞_{n=1}(−1)^nb_n\) 
If \( b_{n+1}≤b_n\) for all \( n≥1\) and \( b_n→0\), then the series converges.  Only applies to alternating series. 
Ratio Test For any 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. 
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 \(\displaystyle \sum^∞_{n=1}a_n\), 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. 
Series Converging to \( π\) and \( 1/π\)
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
\[π=4\sum_{n=1}^∞\frac{(−1)^{n+1}}{2n−1}=4−\frac{4}{3}+\frac{4}{5}−\frac{4}{7}+\frac{4}{9}−⋯\]
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
\[π=6\sum^∞_{n=0}\frac{(2n)!}{2^{4n+1}(n!)^2(2n+1)}=6\left(\frac{1}{2}+\frac{1}{2⋅3}\left(\frac{1}{2}\right)^3+\frac{1⋅3}{2⋅4⋅5}⋅\left(\frac{1}{2}\right)^5+\frac{1⋅3⋅5}{2⋅4⋅6⋅7}\left(\frac{1}{2}\right)^7+⋯\right)\]
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
\[\frac{1}{π}=\frac{\sqrt{8}}{9801}\sum_{n=0}^∞\frac{(4n)!(1103+26390n)}{(n!)^4396^{4n}}\]
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 \( mid1980s\). 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 selftaught, 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
\[ρ=\lim_{n→∞}∣\frac{a_{n+1}}{a_n}∣.\]
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
\[ρ=\lim_{n→∞}\sqrt[n]{a_n}\].
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 \( \displaystyle ρ=\lim_{n→∞}a_{n+1}/a_n\); 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
Contributors
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CCBYSANC 4.0 license. Download for free at http://cnx.org.