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Ratio and Root Tests

Use the ratio test to determine absolute convergence of a series.

Ratio Test

Consider a series \(\sum _{n=1}^{\infty }{a}_{n}.\) From our earlier discussion and examples, we know that \(\underset{n\to \infty }{\text{lim}}{a}_{n}=0\) is not a sufficient condition for the series to converge. Not only do we need \({a}_{n}\to 0,\) but we need \({a}_{n}\to 0\) quickly enough. For example, consider the series \(\sum _{n=1}^{\infty }1\text{/}n\) and the series \(\sum _{n=1}^{\infty }1\text{/}{n}^{2}.\) We know that \(1\text{/}n\to 0\) and \(1\text{/}{n}^{2}\to 0.\) However, only the series \(\sum _{n=1}^{\infty }1\text{/}{n}^{2}\) converges. The series \(\sum _{n=1}^{\infty }1\text{/}n\) diverges because the terms in the sequence \(\{1\text{/}n\}\) do not approach zero fast enough as \(n\to \infty .\) Here we introduce the ratio test, which provides a way of measuring how fast the terms of a series approach zero.

Condensed — the full section is in OpenStax Calculus Volume 2.

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.

Choosing a Convergence Test for a Series

Consider a series \(\sum _{n=1}^{\infty }{a}_{n}.\) In the steps below, we outline a strategy for determining whether the series converges.

Is \(\sum _{n=1}^{\infty }{a}_{n}\) a familiar series? For example, is it the harmonic series (which diverges) or the alternating harmonic series (which converges)? Is it a \(p-\text{series}\) 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 \(\sum _{n=1}^{\infty }|{a}_{n}|.\)
Is the series similar to a \(p-\text{series}\) 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.

In , 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 \(\sum _{n=1}^{\infty }{a}_{n}\) to have nonnegative terms, if \(\sum _{n=1}^{\infty }{a}_{n}\) has negative terms, these tests can be applied to \(\sum _{n=1}^{\infty }|{a}_{n}|\) to test for absolute convergence.

Condensed — the full section is in OpenStax Calculus Volume 2.

Key Concepts

For the ratio test, we consider
\[\rho =\underset{n\to \infty }{\text{lim}}|\frac{{a}_{n+1}}{{a}_{n}}|.\]
If \(\rho <1,\) the series \(\sum _{n=1}^{\infty }{a}_{n}\) converges absolutely. If \(\rho >1,\) the series diverges. If \(\rho =1,\) the test does not provide any information. This test is useful for series whose terms involve factorials.
For the root test, we consider
\[\rho =\underset{n\to \infty }{\text{lim}}\sqrt[n]{|{a}_{n}|}.\]
If \(\rho <1,\) the series \(\sum _{n=1}^{\infty }{a}_{n}\) converges absolutely. If \(\rho >1,\) the series diverges. If \(\rho =1,\) 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 \(p-\text{series,}\) consider one of the comparison tests.

Ratio and Root Tests

Use the ratio test to determine whether \(\sum _{n=1}^{\infty }{a}_{n}\) converges, where \({a}_{n}\) is given in the following problems. State if the ratio test is inconclusive.

Use the root test to determine whether \(\sum _{n=1}^{\infty }{a}_{n}\) converges, where \({a}_{n}\) is as follows.

For this exercise, let n start at 2.

In the following exercises, use either the ratio test or the root test as appropriate to determine whether the series \(\sum _{k=1}^{\infty }{a}_{k}\) with given terms \({a}_{k}\) converges, or state if the test is inconclusive.

Use the ratio test to determine whether \(\sum _{n=1}^{\infty }{a}_{n}\) converges, or state if the ratio test is inconclusive.

Use the root and limit comparison tests to determine whether \(\sum _{n=1}^{\infty }{a}_{n}\) converges.

In the following exercises, use an appropriate test to determine whether the series converges.

Condensed — the full section is in OpenStax Calculus Volume 2.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

For each of the following series, use the ratio test to determine whether the series converges or diverges.
1. \(\sum _{n=1}^{\infty }\frac{{2}^{n}}{n\text{!}}\)
2. \(\sum _{n=1}^{\infty }\frac{{n}^{n}}{n\text{!}}\)
3. \(\sum _{n=1}^{\infty }\frac{{(-1)}^{n}{(n\text{!})}^{2}}{(2n)\text{!}}\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
1. From the ratio test, we can see that
  \[\rho =\underset{n\to \infty }{\text{lim}}\frac{{2}^{n+1}\text{/}(n+1)\text{!}}{{2}^{n}\text{/}n\text{!}}=\underset{n\to \infty }{\text{lim}}\frac{{2}^{n+1}}{(n+1)\text{!}}\cdot \frac{n\text{!}}{{2}^{n}}.\]
  Since \((n+1)\text{!}=(n+1)\cdot n\text{!},\)
  \[\rho =\underset{n\to \infty }{\text{lim}}\frac{2}{n+1}=0.\]
  Since \(\rho <1,\) the series converges.
2. We can see that
  \[\begin{array}{ll}\rho & =\underset{n\to \infty }{\text{lim}}\frac{{(n+1)}^{n+1}\text{/}(n+1)\text{!}}{{n}^{n}\text{/}n\text{!}} \\ & =\underset{n\to \infty }{\text{lim}}\frac{{(n+1)}^{n+1}}{(n+1)\text{!}}\cdot \frac{n\text{!}}{{n}^{n}} \\ & =\underset{n\to \infty }{\text{lim}}{(\frac{n+1}{n})}^{n}=\underset{n\to \infty }{\text{lim}}{(1+\frac{1}{n})}^{n}=e.\end{array}\]
  Since \(\rho >1,\) the series diverges.
3. Since
  \[\begin{array}{ll}|\frac{{(-1)}^{n+1}{((n+1)\text{!})}^{2}\text{/}(2(n+1))\text{!}}{{(-1)}^{n}{(n\text{!})}^{2}\text{/}(2n)\text{!}}| & =\frac{(n+1)\text{!}(n+1)\text{!}}{(2n+2)\text{!}}\cdot \frac{(2n)\text{!}}{n\text{!}n\text{!}} \\ & =\frac{(n+1)(n+1)}{(2n+2)(2n+1)}\end{array}\]
  we see that
  \[\rho =\underset{n\to \infty }{\text{lim}}\frac{(n+1)(n+1)}{(2n+2)(2n+1)}=\frac{1}{4}.\]
  Since \(\rho <1,\) the series converges.
Use the ratio test to determine whether the series \(\sum _{n=1}^{\infty }\frac{{n}^{3}}{{3}^{n}}\) converges or diverges.

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

The series converges.
For each of the following series, use the root test to determine whether the series converges or diverges.
1. \(\sum _{n=1}^{\infty }\frac{{({n}^{2}+3n)}^{n}}{{(4{n}^{2}+5)}^{n}}\)
2. \(\sum _{n=2}^{\infty }\frac{{n}^{n}}{{(\text{ln}(n))}^{n}}\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
1. To apply the root test, we compute
  \[\rho =\underset{n\to \infty }{\text{lim}}\sqrt[n]{{({n}^{2}+3n)}^{n}\text{/}{(4{n}^{2}+5)}^{n}}=\underset{n\to \infty }{\text{lim}}\frac{{n}^{2}+3n}{4{n}^{2}+5}=\frac{1}{4}.\]
  Since \(\rho <1,\) the series converges absolutely.
2. We have
  \[\rho =\underset{n\to \infty }{\text{lim}}\sqrt[n]{{n}^{n}\text{/}{(\text{ln}\ n)}^{n}}=\underset{n\to \infty }{\text{lim}}\frac{n}{\text{ln}\ n}=\infty \ \text{by L’Hôpital’s rule}.\]
  Since \(\rho =\infty ,\) the series diverges.
Use the root test to determine whether the series \(\sum _{n=1}^{\infty }1\text{/}{n}^{n}\) converges or diverges.

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

The series converges.
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.
1. \(\sum _{n=1}^{\infty }\frac{{n}^{2}+2n}{{n}^{3}+3{n}^{2}+1}\)
2. \(\sum _{n=1}^{\infty }\frac{{(-1)}^{n+1}(3n+1)}{n\text{!}}\)
3. \(\sum _{n=1}^{\infty }\frac{{e}^{n}}{{n}^{3}}\)
4. \(\sum _{n=1}^{\infty }\frac{{3}^{n}}{{(n+1)}^{n}}\)
ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
1. Step 1. The series is not a \(p-\text{series}\) or geometric series.
  Step 2. The series is not alternating.
  Step 3. For large values of \(n,\) we approximate the series by the expression
  \[\frac{{n}^{2}+2n}{{n}^{3}+3{n}^{2}+1}\approx \frac{{n}^{2}}{{n}^{3}}=\frac{1}{n}.\]
  Therefore, it seems reasonable to apply the comparison test or limit comparison test using the series \(\sum _{n=1}^{\infty }1\text{/}n.\) Using the limit comparison test, we see that
  \[\underset{n\to \infty }{\text{lim}}\frac{({n}^{2}+2n)\text{/}({n}^{3}+3{n}^{2}+1)}{1\text{/}n}=\underset{n\to \infty }{\text{lim}}\frac{{n}^{3}+2{n}^{2}}{{n}^{3}+3{n}^{2}+1}=1.\]
  Since the series \(\sum _{n=1}^{\infty }1\text{/}n\) diverges, this series diverges as well.
2. 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
  \[\sum _{n=1}^{\infty }\frac{3n}{(n+1)\text{!}}.\]
  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
  \[\underset{n\to \infty }{\text{lim}}\frac{(3(n+1))\text{/}(n+1)\text{!}}{(3n+1)\text{/}n\text{!}}=\underset{n\to \infty }{\text{lim}}\frac{3n+3}{(n+1)\text{!}}\cdot \frac{n\text{!}}{3n+1}=\underset{n\to \infty }{\text{lim}}\frac{3n+3}{(n+1)(3n+1)}=0.\]
  Therefore, this series converges, and we conclude that the original series converges absolutely, and thus converges.
3. 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
  \[\underset{n\to \infty }{\text{lim}}\frac{{e}^{n}}{{n}^{3}}=\infty .\]
  Therefore, by the divergence test, the series diverges.
4. 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
  \[\underset{n\to \infty }{\text{lim}}\sqrt[n]{{(\frac{3}{n+1})}^{n}}=\underset{n\to \infty }{\text{lim}}\frac{3}{n+1}=0,\]
  by the root test, we conclude that the series converges.
For the series \(\sum _{n=1}^{\infty }\frac{{2}^{n}}{{3}^{n}+n},\) determine which convergence test is the best to use and explain why.

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

The comparison test because \({2}^{n}\text{/}({3}^{n}+n)<{2}^{n}\text{/}{3}^{n}\) for all positive integers \(n.\) The limit comparison test could also be used.
\({a}_{n}=1\text{/}n\text{!}\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

\({a}_{n+1}\text{/}{a}_{n}\to 0.\) Converges.
\({a}_{n}={10}^{n}\text{/}n\text{!}\)
\({a}_{n}={n}^{2}\text{/}{2}^{n}\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

\(\frac{{a}_{n+1}}{{a}_{n}}=\frac{1}{2}{(\frac{n+1}{n})}^{2}\to 1\text{/}2<1.\) Converges.
\({a}_{n}={n}^{10}\text{/}{2}^{n}\)
\(\sum _{n=1}^{\infty }\frac{{(n\text{!})}^{3}}{(3n)\text{!}}\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

\(\frac{{a}_{n+1}}{{a}_{n}}\to 1\text{/}27<1.\) Converges.
\(\sum _{n=1}^{\infty }\frac{{2}^{3n}{(n\text{!})}^{3}}{(3n)\text{!}}\)
\(\sum _{n=1}^{\infty }\frac{(2n)\text{!}}{{n}^{2n}}\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

\(\frac{{a}_{n+1}}{{a}_{n}}\to 4\text{/}{e}^{2}<1.\) Converges.
\(\sum _{n=1}^{\infty }\frac{(2n)\text{!}}{{(2n)}^{n}}\)
\(\sum _{n=1}^{\infty }\frac{n\text{!}}{{(n\text{/}e)}^{n}}\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

\(\frac{{a}_{n+1}}{{a}_{n}}\to 1.\) Ratio test is inconclusive.
\(\sum _{n=1}^{\infty }\frac{(2n)\text{!}}{{(n\text{/}e)}^{2n}}\)
\(\sum _{n=1}^{\infty }\frac{{({2}^{n}n\text{!})}^{2}}{{(2n)}^{2n}}\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

\(\frac{{a}_{n}}{{a}_{n+1}}\to 1\text{/}{e}^{2}.\) Converges.
\({a}_{k}={(\frac{k-1}{2k+3})}^{k}\)
\({a}_{k}={(\frac{2{k}^{2}-1}{{k}^{2}+3})}^{k}\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

\({({a}_{k})}^{1\text{/}k}\to 2>1.\) Diverges.
\({a}_{n}=\frac{{(\text{ln}\ n)}^{2n}}{{n}^{n}}\)
\({a}_{n}=n\text{/}{2}^{n}\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

\({({a}_{n})}^{1\text{/}n}\to 1\text{/}2<1.\) Converges.
\({a}_{n}=n\text{/}{e}^{n}\)
\({a}_{k}=\frac{{k}^{e}}{{e}^{k}}\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

\({({a}_{k})}^{1\text{/}k}\to 1\text{/}e<1.\) Converges.
\({a}_{k}=\frac{{\pi }^{k}}{{k}^{\pi }}\)
\({a}_{n}={(\frac{1}{e}+\frac{1}{n})}^{n}\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

\({a}_{n}^{1\text{/}n}=\frac{1}{e}+\frac{1}{n}\to \frac{1}{e}<1.\) Converges.
\({a}_{k}=\frac{1}{{(1+\text{ln}\ k)}^{k}}\)
\({a}_{n}=\frac{{(\text{ln}(1+\text{ln}\ n))}^{n}}{{(\text{ln}\ n)}^{n}}\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

\({a}_{n}^{1\text{/}n}=\frac{(\text{ln}(1+\text{ln}\ n))}{(\text{ln}\ n)}\to 0\) by L’Hôpital’s rule. Converges.
\({a}_{k}=\frac{k\text{!}}{1\cdot 3\cdot 5\text{\cdots }(2k-1)}\)
\({a}_{k}=\frac{2\cdot 4\cdot 6\text{\cdots }2k}{(2k)\text{!}}\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

\(\frac{{a}_{k+1}}{{a}_{k}}=\frac{1}{2k+1}\to 0.\) Converges by ratio test.
\({a}_{k}=\frac{1\cdot 4\cdot 7\text{\cdots }(3k-2)}{{3}^{k}k\text{!}}\)
\({a}_{n}={(1-\frac{1}{n})}^{{n}^{2}}\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

\({({a}_{n})}^{1\text{/}n}\to 1\text{/}e.\) Converges by root test.
\({a}_{k}={(\frac{1}{k+1}+\frac{1}{k+2}+\text{\cdots }+\frac{1}{2k})}^{k}\) (Hint: Compare \({a}_{k}^{1\text{/}k}\) to \({\int }_{k}^{2k}\frac{dt}{t}.)\)
\({a}_{k}={(\frac{1}{k+1}+\frac{1}{k+2}+\text{\cdots }+\frac{1}{3k})}^{k}\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

\({a}_{k}^{1\text{/}k}\to \text{ln}(3)>1.\) Diverges by root test.
\({a}_{n}={({n}^{1\text{/}n}-1)}^{n}\)
\(\sum _{n=1}^{\infty }\frac{{3}^{{n}^{2}}}{{2}^{{n}^{3}}}\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

\(\frac{{a}_{n+1}}{{a}_{n}}=\) \(\frac{{3}^{2n+1}}{{2}^{3{n}^{2}+3n+1}}\to 0.\) Converge.
\(\sum _{n=1}^{\infty }\frac{{2}^{{n}^{2}}}{{n}^{n}n\text{!}}\)
\({a}_{n}=1\text{/}{x}_{n}^{n}\) where \({x}_{n+1}=\frac{1}{2}{x}_{n}+\frac{1}{{x}_{n}},\) \({x}_{1}=1\) (Hint: Find limit of \(\{{x}_{n}\}.)\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

Converges by root test and limit comparison test since \({x}_{n}\to \sqrt{2}.\)
\(\sum _{n=1}^{\infty }\frac{(n+1)}{{n}^{3}+{n}^{2}+n+1}\)
\(\sum _{n=1}^{\infty }\frac{{(-1)}^{n+1}(n+1)}{{n}^{3}+3{n}^{2}+3n+1}\)

ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

Converges absolutely by limit comparison with \(p-\text{series,}\) \(p=2.\)
\(\sum _{n=1}^{\infty }\frac{{(n+1)}^{2}}{{n}^{3}+{(1.1)}^{n}}\)

Symbols used here

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Ratio and Root Tests

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.
If
If
If
From the ratio test, we can see that
We can see that

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

ನಿಮ್ಮದೇ ಆದದ್ದನ್ನು ಪ್ರಯತ್ನಿಸಿ

Parts of this page are adapted from OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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