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.

İndi siz Heç bir hesablama proqramı bu ədədi həll edə bilməz, lakin onun hissələri hesablana bilər. Aşağıdakılardan birini sınayın ya da özünüz yazın.

⌨ Növ ✎ Çək

Riyaziyyat probleminiz

π Çöz

Tədris (40)

Hər bir cavabı əvvəlcə kağız üzərində sınayın. Cavabı yoxlamaq üçün göstərin; təsdiqlənmiş cavablar hər addım üçün həlledicidə açıla bilər.

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

   Cavabı göstər

   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.

   İşlə İpucu

   Cavabı göstər

   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. \(\sum _{n=2}^{\infty }\frac{{n}^{n}}{{(\text{ln}(n))}^{n}}\)
   İşlə İpucu

   Cavabı göstər

   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.

   İşlə İpucu

   Cavabı göstər

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

   Cavabı göstər

   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.

   İşlə İpucu

   Cavabı göstər

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

   İşlə İpucu

   Cavabı göstər

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

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

   İşlə İpucu

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

   İşlə İpucu

   Cavabı göstər

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

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

    İşlə İpucu

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

    İşlə İpucu

    Cavabı göstər

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

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

    İşlə İpucu

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

    İşlə İpucu

    Cavabı göstər

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

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

    İşlə İpucu

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

    İşlə İpucu

    Cavabı göstər

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

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

    İşlə İpucu

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

    İşlə İpucu

    Cavabı göstər

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

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

    İşlə İpucu

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

    İşlə İpucu

    Cavabı göstər

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

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

    İşlə İpucu

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

    İşlə İpucu

    Cavabı göstər

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

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

    İşlə İpucu

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

    İşlə İpucu

    Cavabı göstər

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

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

    İşlə İpucu

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

    İşlə İpucu

    Cavabı göstər

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

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

    İşlə İpucu

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

    İşlə İpucu

    Cavabı göstər

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

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

    İşlə İpucu

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

    İşlə İpucu

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

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

    İşlə İpucu

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

    İşlə İpucu

    Cavabı göstər

    \({({a}_{n})}^{1\text{/}n}\to 1\text{/}e.\) 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 \({\int }_{k}^{2k}\frac{dt}{t}.)\)

    İşlə İpucu

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

    İşlə İpucu

    Cavabı göstər

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

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

    İşlə İpucu

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

    İşlə İpucu

    Cavabı göstər

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

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

    İşlə İpucu

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 \(\{{x}_{n}\}.)\)

    İşlə İpucu

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

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

    İşlə İpucu

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

    İşlə İpucu

    Cavabı göstər

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

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

    İşlə İpucu

Özün işlə

Pulsuz hesab hər dərsdə qeydlər əlavə edir, bitirdiyiniz işlərin qeydini saxlayır, həll etdiyiniz problemləri bir yerdə saxlayır və bu səhifə haqqında soruşa biləcəyiniz müəllimi də əlavə edir. Riyaziyyat özü daxil olub-olmadığınıza baxmayaraq hər kəsə açıqdır.

Qeyd Et Giriş

Burada istifadə edilən simvollar

Tam təsviri, şəkli və hər hərfin nə demək olduğunu görmək üçün hər hansı simvolu vurun.

\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
Least upper bound, greatest lower bound.\pipi
Ratio of a circle's circumference to its diameter, 3.14159…eEuler's number
2.71828…, the base whose exponential is its own derivative.\sinsine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.\log_b xlogarithm, 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)derivative
Instantaneous rate of change; slope of the graph.\int f(x)integral
Antiderivative (indefinite) or signed area from a to b (definite).y'first and second derivative of y
Prime notation for derivatives with respect to x (or t).Carbitrary constants
Constants of integration fixed by initial conditions.

Növbəti: 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

İnsanlar soruşurlar

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.

Bu səhifənin bəzi hissələri OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Bu, bizim üçün çox böyük bir uğurdur, biz bunu çox istəyirik. Onun mənbəyi kommersiya məqsədləri üçün lisenziyalı olduğundan, bu dərs pulsuzdur, heç nə satmır və eyni lisenziyaya əsasən paylaşılır.

Daha çox 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