Ratio and Root Tests

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.

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.

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1. 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. \(\sum _{n=1}^{\infty }\frac{{(-1)}^{n}{(n\text{!})}^{2}}{(2n)\text{!}}\)
   Prebrodi to. Nasvet

   Odkrij odgovor

   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.

2. Use the ratio test to determine whether the series \(\sum _{n=1}^{\infty }\frac{{n}^{3}}{{3}^{n}}\) converges or diverges.

   Prebrodi to. Nasvet

   Odkrij odgovor

   The series converges.

3. 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. \(\sum _{n=2}^{\infty }\frac{{n}^{n}}{{(\text{ln}(n))}^{n}}\)
   Prebrodi to. Nasvet

   Odkrij odgovor

   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.

4. Use the root test to determine whether the series \(\sum _{n=1}^{\infty }1\text{/}{n}^{n}\) converges or diverges.

   Prebrodi to. Nasvet

   Odkrij odgovor

   The series converges.

5. 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. \(\sum _{n=1}^{\infty }\frac{{3}^{n}}{{(n+1)}^{n}}\)
   Prebrodi to. Nasvet

   Odkrij odgovor

   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.

6. 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.

   Prebrodi to. Nasvet

   Odkrij odgovor

   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.

7. \({a}_{n}=1\text{/}n\text{!}\)

   Prebrodi to. Nasvet

   Odkrij odgovor

   \({a}_{n+1}\text{/}{a}_{n}\to 0.\) Converges.

8. \({a}_{n}={10}^{n}\text{/}n\text{!}\)

   Prebrodi to. Nasvet

9. \({a}_{n}={n}^{2}\text{/}{2}^{n}\)

   Prebrodi to. Nasvet

   Odkrij odgovor

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

10. \({a}_{n}={n}^{10}\text{/}{2}^{n}\)

    Prebrodi to. Nasvet

11. \(\sum _{n=1}^{\infty }\frac{{(n\text{!})}^{3}}{(3n)\text{!}}\)

    Prebrodi to. Nasvet

    Odkrij odgovor

    \(\frac{{a}_{n+1}}{{a}_{n}}\to 1\text{/}27<1.\) Converges.

12. \(\sum _{n=1}^{\infty }\frac{{2}^{3n}{(n\text{!})}^{3}}{(3n)\text{!}}\)

    Prebrodi to. Nasvet

13. \(\sum _{n=1}^{\infty }\frac{(2n)\text{!}}{{n}^{2n}}\)

    Prebrodi to. Nasvet

    Odkrij odgovor

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

14. \(\sum _{n=1}^{\infty }\frac{(2n)\text{!}}{{(2n)}^{n}}\)

    Prebrodi to. Nasvet

15. \(\sum _{n=1}^{\infty }\frac{n\text{!}}{{(n\text{/}e)}^{n}}\)

    Prebrodi to. Nasvet

    Odkrij odgovor

    \(\frac{{a}_{n+1}}{{a}_{n}}\to 1.\) Ratio test is inconclusive.

16. \(\sum _{n=1}^{\infty }\frac{(2n)\text{!}}{{(n\text{/}e)}^{2n}}\)

    Prebrodi to. Nasvet

17. \(\sum _{n=1}^{\infty }\frac{{({2}^{n}n\text{!})}^{2}}{{(2n)}^{2n}}\)

    Prebrodi to. Nasvet

    Odkrij odgovor

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

18. \({a}_{k}={(\frac{k-1}{2k+3})}^{k}\)

    Prebrodi to. Nasvet

19. \({a}_{k}={(\frac{2{k}^{2}-1}{{k}^{2}+3})}^{k}\)

    Prebrodi to. Nasvet

    Odkrij odgovor

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

20. \({a}_{n}=\frac{{(\text{ln}\ n)}^{2n}}{{n}^{n}}\)

    Prebrodi to. Nasvet

21. \({a}_{n}=n\text{/}{2}^{n}\)

    Prebrodi to. Nasvet

    Odkrij odgovor

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

22. \({a}_{n}=n\text{/}{e}^{n}\)

    Prebrodi to. Nasvet

23. \({a}_{k}=\frac{{k}^{e}}{{e}^{k}}\)

    Prebrodi to. Nasvet

    Odkrij odgovor

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

24. \({a}_{k}=\frac{{\pi }^{k}}{{k}^{\pi }}\)

    Prebrodi to. Nasvet

25. \({a}_{n}={(\frac{1}{e}+\frac{1}{n})}^{n}\)

    Prebrodi to. Nasvet

    Odkrij odgovor

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

26. \({a}_{k}=\frac{1}{{(1+\text{ln}\ k)}^{k}}\)

    Prebrodi to. Nasvet

27. \({a}_{n}=\frac{{(\text{ln}(1+\text{ln}\ n))}^{n}}{{(\text{ln}\ n)}^{n}}\)

    Prebrodi to. Nasvet

    Odkrij odgovor

    \({a}_{n}^{1\text{/}n}=\frac{(\text{ln}(1+\text{ln}\ n))}{(\text{ln}\ n)}\to 0\) by L’Hôpital’s rule. Converges.

28. \({a}_{k}=\frac{k\text{!}}{1\cdot 3\cdot 5\cdots (2k-1)}\)

    Prebrodi to. Nasvet

29. \({a}_{k}=\frac{2\cdot 4\cdot 6\cdots 2k}{(2k)\text{!}}\)

    Prebrodi to. Nasvet

    Odkrij odgovor

    \(\frac{{a}_{k+1}}{{a}_{k}}=\frac{1}{2k+1}\to 0.\) Converges by ratio test.

30. \({a}_{k}=\frac{1\cdot 4\cdot 7\cdots (3k-2)}{{3}^{k}k\text{!}}\)

    Prebrodi to. Nasvet

31. \({a}_{n}={(1-\frac{1}{n})}^{{n}^{2}}\)

    Prebrodi to. Nasvet

    Odkrij odgovor

    \({({a}_{n})}^{1\text{/}n}\to 1\text{/}e.\) Converges by root test.

32. \({a}_{k}={(\frac{1}{k+1}+\frac{1}{k+2}+\cdots +\frac{1}{2k})}^{k}\) (Hint: Compare \({a}_{k}^{1\text{/}k}\) to \({\int }_{k}^{2k}\frac{dt}{t}.)\)

    Prebrodi to. Nasvet

33. \({a}_{k}={(\frac{1}{k+1}+\frac{1}{k+2}+\cdots +\frac{1}{3k})}^{k}\)

    Prebrodi to. Nasvet

    Odkrij odgovor

    \({a}_{k}^{1\text{/}k}\to \text{ln}(3)>1.\) Diverges by root test.

34. \({a}_{n}={({n}^{1\text{/}n}-1)}^{n}\)

    Prebrodi to. Nasvet

35. \(\sum _{n=1}^{\infty }\frac{{3}^{{n}^{2}}}{{2}^{{n}^{3}}}\)

    Prebrodi to. Nasvet

    Odkrij odgovor

    \(\frac{{a}_{n+1}}{{a}_{n}}=\) \(\frac{{3}^{2n+1}}{{2}^{3{n}^{2}+3n+1}}\to 0.\) Converge.

36. \(\sum _{n=1}^{\infty }\frac{{2}^{{n}^{2}}}{{n}^{n}n\text{!}}\)

    Prebrodi to. Nasvet

37. \({a}_{n}=1\text{/}{x}_{n}^{n}\) where \({x}_{n+1}=\frac{1}{2}{x}_{n}+\frac{1}{{x}_{n}},\) \({x}_{1}=1\) (Hint: Find limit of \(\{{x}_{n}\}.)\)

    Prebrodi to. Nasvet

    Odkrij odgovor

    Converges by root test and limit comparison test since \({x}_{n}\to \sqrt{2}.\)

38. \(\sum _{n=1}^{\infty }\frac{(n+1)}{{n}^{3}+{n}^{2}+n+1}\)

    Prebrodi to. Nasvet

39. \(\sum _{n=1}^{\infty }\frac{{(-1)}^{n+1}(n+1)}{{n}^{3}+3{n}^{2}+3n+1}\)

    Prebrodi to. Nasvet

    Odkrij odgovor

    Converges absolutely by limit comparison with \(p-\text{series,}\) \(p=2.\)

40. \(\sum _{n=1}^{\infty }\frac{{(n+1)}^{2}}{{n}^{3}+{(1.1)}^{n}}\)

    Prebrodi to. Nasvet

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\sum_{k=1}^{n} a_ksummation
Add a_k for k = 1 up to n.\sqrt{x}square root, n-th root
The non-negative number whose square (n-th power) is x.\inftyinfinity
Not a number: "grows without bound" in limits and intervals.\leqless/greater than or equal
Inequalities that allow equality; < and > exclude it.\supsupremum, infimum
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Ratio of a circle's circumference to its diameter, 3.14159…eEuler's number
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Constants of integration fixed by initial conditions.

Kako: Ratio and Root Tests

1. Use the ratio test to determine absolute convergence of a series.
2. Use the root test to determine absolute convergence of a series.
3. Describe a strategy for testing the convergence of a given series.
4. If
5. If
6. If
7. From the ratio test, we can see that
8. We can see that

Vprašanja, ki jih ljudje sprašujejo

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.

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Več v Calculus

LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests