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Alternating Series

Use the alternating series test to test an alternating series for convergence.

The Alternating Series Test

A series whose terms alternate between positive and negative values is an alternating series. For example, the series

\[\sum _{n=1}^{\infty }{(-\frac{1}{2})}^{n}=-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\cdots\]

and

\[\sum _{n=1}^{\infty }\frac{{(-1)}^{n+1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\]

are both alternating series.

Series (1), shown in , is a geometric series. Since \(|r|=|\text{-}1\text{/}2|<1,\) the series converges. Series (2), shown in , is called the alternating harmonic series. We will show that whereas the harmonic series diverges, the alternating harmonic series converges.

To prove this, we look at the sequence of partial sums \(\{{S}_{k}\}\) ().

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

Remainder of an Alternating Series

It is difficult to explicitly calculate the sum of most alternating series, so typically the sum is approximated by using a partial sum. When doing so, we are interested in the amount of error in our approximation. Consider an alternating series

\[\sum _{n=1}^{\infty }{(-1)}^{n+1}{b}_{n}\]

satisfying the hypotheses of the alternating series test. Let \(S\) denote the sum of this series and \(\{{S}_{k}\}\) be the corresponding sequence of partial sums. From , we see that for any integer \(N\ge 1,\) the remainder \({R}_{N}\) satisfies

\[|{R}_{N}|=|S-{S}_{N}|\le |{S}_{N+1}-{S}_{N}|={b}_{n+1}.\]

In other words, if the conditions of the alternating series test apply, then the error in approximating the infinite series by the \(N\text{th}\) partial sum \({S}_{N}\) is in magnitude at most the size of the next term \({b}_{N+1}.\)

Example

Try it.

Consider the alternating series

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

Use the remainder estimate to determine a bound on the error \({R}_{10}\) if we approximate the sum of the series by the partial sum \({S}_{10}.\)

Solution

From the theorem stated above,

\(|{R}_{10}|\le {b}_{11}=\frac{1}{{11}^{2}}\approx 0.008265.\)

Absolute and Conditional Convergence

Consider a series \(\sum _{n=1}^{\infty }{a}_{n}\) and the related series \(\sum _{n=1}^{\infty }|{a}_{n}|.\) Here we discuss possibilities for the relationship between the convergence of these two series. For example, consider the alternating harmonic series \(\sum _{n=1}^{\infty }{(-1)}^{n+1}\text{/}n.\) The series whose terms are the absolute value of these terms is the harmonic series, since \(\sum _{n=1}^{\infty }|{(-1)}^{n+1}\text{/}n|=\sum _{n=1}^{\infty }1\text{/}n.\) Since the alternating harmonic series converges, but the harmonic series diverges, we say the alternating harmonic series exhibits conditional convergence.

By comparison, consider the series \(\sum _{n=1}^{\infty }{(-1)}^{n+1}\text{/}{n}^{2}.\) The series whose terms are the absolute values of the terms of this series is the series \(\sum _{n=1}^{\infty }1\text{/}{n}^{2}.\) Since both of these series converge, we say the series \(\sum _{n=1}^{\infty }{(-1)}^{n+1}\text{/}{n}^{2}\) exhibits absolute convergence.

As shown by the alternating harmonic series, a series \(\sum _{n=1}^{\infty }{a}_{n}\) may converge, but \(\sum _{n=1}^{\infty }|{a}_{n}|\) may diverge. In the following theorem, however, we show that if \(\sum _{n=1}^{\infty }|{a}_{n}|\) converges, then \(\sum _{n=1}^{\infty }{a}_{n}\) converges.

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

Alternating Series

State whether each of the following series converges absolutely, conditionally, or not at all.

In each of the following problems, use the estimate \(|{R}_{N}|\le {b}_{N+1}\) to find a value of \(N\) that guarantees that the sum of the first \(N\) terms of the alternating series \(\sum _{n=1}^{\infty }{(-1)}^{n+1}{b}_{n}\) differs from the infinite sum by at most the given error. Calculate the partial sum \({S}_{N}\) for this \(N.\)

For the following exercises, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.

The following series do not satisfy the hypotheses of the alternating series test as stated.

In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.

The following alternating series converge to given multiples of \(\pi .\) Find the value of \(N\) predicted by the remainder estimate such that the \(N\text{th}\) partial sum of the series accurately approximates the left-hand side to within the given error. Find the minimum \(N\) for which the error bound holds, and give the desired approximate value in each case. Up to \(15\) decimals places, \(\pi =3.141592653589793\ldots .\)

Ora tu Nessuna calcolatrice risolve questa domanda, ma le sue parti sono calcolabili. Provane una qui sotto, o digita la tua.

Pratica (40)

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  1. For each of the following alternating series, determine whether the series converges or diverges.

    1. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}\text{/}{n}^{2}\)
    2. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}n\text{/}(n+1)\)
    Rivela la risposta
    1. Since
      \(\frac{1}{{(n+1)}^{2}}<\frac{1}{{n}^{2}}\ \text{and}\ \frac{1}{{n}^{2}}\to 0,\)
      the series converges.
    2. Since \(n\text{/}(n+1)↛0\) as \(n\to \infty ,\) we cannot apply the alternating series test. Instead, we use the nth term test for divergence. Since
      \(\underset{n\to \infty }{\text{lim}}\frac{{(-1)}^{n+1}n}{n+1}\ne 0,\)
      the series diverges.

  2. Determine whether the series \(\sum _{n=1}^{\infty }{(-1)}^{n+1}n\text{/}{2}^{n}\) converges or diverges.

    Rivela la risposta

    The series converges.

  3. Consider the alternating series

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

    Use the remainder estimate to determine a bound on the error \({R}_{10}\) if we approximate the sum of the series by the partial sum \({S}_{10}.\)

    Rivela la risposta

    From the theorem stated above,

    \(|{R}_{10}|\le {b}_{11}=\frac{1}{{11}^{2}}\approx 0.008265.\)

  4. Find a bound for \({R}_{20}\) when approximating \(\sum _{n=1}^{\infty }{(-1)}^{n+1}\text{/}n\) by \({S}_{20}.\)

    Rivela la risposta

    \(0.04762\)

  5. For each of the following series, determine whether the series converges absolutely, converges conditionally, or diverges.

    1. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}\text{/}(3n+1)\)
    2. \(\sum _{n=1}^{\infty }\text{cos}(n)\text{/}{n}^{2}\)
    Rivela la risposta
    1. We can see that
      \[\sum _{n=1}^{\infty }|\frac{{(-1)}^{n+1}}{3n+1}|=\sum _{n=1}^{\infty }\frac{1}{3n+1}\]
      diverges by using the limit comparison test with the harmonic series.
      \[\underset{n\to \infty }{\text{lim}}\frac{1\text{/}(3n+1)}{1\text{/}n}=\frac{1}{3}\]
      Thus, applying Theorem 5.13, the series cannot converge absolutely. Moreover, because of the alternating series test, we can see that the series converges.
      \[\frac{1}{3(n+1)+1}<\frac{1}{3n+1}\ \text{and}\ \frac{1}{3n+1}\to 0,\]
      We can conclude that \(\sum _{n=1}^{\infty }{(-1)}^{n+1}\text{/}(3n+1)\) converges conditionally.
    2. Noting that \(|\text{cos}\ n|\le 1,\) to determine whether the series converges absolutely, compare
      \[\sum _{n=1}^{\infty }|\frac{\text{cos}\ n}{{n}^{2}}|\]
      with the series \(\sum _{n=1}^{\infty }1\text{/}{n}^{2}.\) Since \(\sum _{n=1}^{\infty }1\text{/}{n}^{2}\) converges, by the comparison test, \(\sum _{n=1}^{\infty }|\text{cos}\ n\text{/}{n}^{2}|\) converges, and therefore \(\sum _{n=1}^{\infty }\text{cos}\ n\text{/}{n}^{2}\) converges absolutely.

  6. Determine whether the series \(\sum _{n=1}^{\infty }{(-1)}^{n+1}n\text{/}(2{n}^{3}+1)\) converges absolutely, converges conditionally, or diverges.

    Rivela la risposta

    The series converges absolutely.

  7. Use the fact that

    \[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots =\text{ln}\ 2\]

    to rearrange the terms in the alternating harmonic series so the sum of the rearranged series is \(3\ \text{ln}(2)\text{/}2.\)

    Rivela la risposta

    Let

    \[\sum _{n=1}^{\infty }{a}_{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\cdots .\]

    Since \(\sum _{n=1}^{\infty }{a}_{n}=\text{ln}(2),\) by the algebraic properties of convergent series,

    \[\sum _{n=1}^{\infty }\frac{1}{2}{a}_{n}=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\cdots =\frac{1}{2}\sum _{n=1}^{\infty }{a}_{n}=\frac{\text{ln}\ 2}{2}.\]

    Now introduce the series \(\sum _{n=1}^{\infty }{b}_{n}\) such that for all \(n\ge 1,\) \({b}_{2n-1}=0\) and \({b}_{2n}={a}_{n}\text{/}2.\) Then

    \[\sum _{n=1}^{\infty }{b}_{n}=0+\frac{1}{2}+0-\frac{1}{4}+0+\frac{1}{6}+0-\frac{1}{8}+\cdots =\frac{\text{ln}\ 2}{2}.\]

    Then using the algebraic limit properties of convergent series, since \(\sum _{n=1}^{\infty }{a}_{n}\) and \(\sum _{n=1}^{\infty }{b}_{n}\) converge, the series \(\sum _{n=1}^{\infty }({a}_{n}+{b}_{n})\) converges and

    \[\sum _{n=1}^{\infty }({a}_{n}+{b}_{n})=\sum _{n=1}^{\infty }{a}_{n}+\sum _{n=1}^{\infty }{b}_{n}=\text{ln}\ 2+\frac{\text{ln}\ 2}{2}=\frac{3\ \text{ln}\ 2}{2}.\]

    Now adding the corresponding terms, \({a}_{n}\) and \({b}_{n},\) we see that

    \[\begin{array}{ll}\sum _{n=1}^{\infty }({a}_{n}+{b}_{n}) & =(1+0)+(-\frac{1}{2}+\frac{1}{2})+(\frac{1}{3}+0)+(-\frac{1}{4}-\frac{1}{4})+(\frac{1}{5}+0)+(-\frac{1}{6}+\frac{1}{6}) \\ & \ \\ & \ +(\frac{1}{7}+0)+(-\frac{1}{8}-\frac{1}{8})+\cdots \\ & =1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots .\end{array}\]

    We notice that the series on the right side of the equal sign is a rearrangement of the alternating harmonic series. Since \(\sum _{n=1}^{\infty }({a}_{n}+{b}_{n})=3\ \text{ln}(2)\text{/}2,\) we conclude that

    \[1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots =\frac{3\ \text{ln}(2)}{2}.\]

    Therefore, we have found a rearrangement of the alternating harmonic series having the desired property.

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

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

    Rivela la risposta

    Does not converge by divergence test. Terms do not tend to zero.

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

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

    Rivela la risposta

    Converges conditionally by alternating series test, since \(\sqrt{n+3}\text{/}n\) is decreasing. Does not converge absolutely by comparison with p-series, \(p=1\text{/}2.\)

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

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

    Rivela la risposta

    Converges absolutely by limit comparison to \({3}^{n}\text{/}{4}^{n},\) for example.

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

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

    Rivela la risposta

    Diverges by divergence test since \(\underset{n\to \infty }{\text{lim}}|{a}_{n}|=e.\)

  16. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}{\text{sin}}^{2}n\)

  17. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}{\text{cos}}^{2}n\)

    Rivela la risposta

    Does not converge. Terms do not tend to zero.

  18. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}{\text{sin}}^{2}(1\text{/}n)\)

  19. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}{\text{cos}}^{2}(1\text{/}n)\)

    Rivela la risposta

    \(\underset{n\to \infty }{\text{lim}}{\text{cos}}^{2}(1\text{/}n)=1.\) Diverges by divergence test.

  20. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}\text{ln}(1\text{/}n)\)

  21. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}\text{ln}(1+\frac{1}{n})\)

    Rivela la risposta

    Converges by alternating series test.

  22. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}\frac{{n}^{2}}{1+{n}^{4}}\)

  23. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}\frac{{n}^{e}}{1+{n}^{\pi }}\)

    Rivela la risposta

    Converges conditionally by alternating series test. Does not converge absolutely by limit comparison with p-series, \(p=\pi -e\)

  24. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}{2}^{1\text{/}n}\)

  25. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}{n}^{1\text{/}n}\)

    Rivela la risposta

    Diverges; terms do not tend to zero.

  26. \(\sum _{n=1}^{\infty }{(-1)}^{n}(1-{n}^{1\text{/}n})\) (Hint: \({n}^{1\text{/}n}\approx 1+\text{ln}(n)\text{/}n\) for large \(n.)\)

  27. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}n(1-\text{cos}(\frac{1}{n}))\) (Hint: \(\text{cos}(1\text{/}n)\approx 1-1\text{/}{n}^{2}\) for large \(n.)\)

    Rivela la risposta

    Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.

  28. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}(\sqrt{n+1}-\sqrt{n})\) (Hint: Rationalize the numerator.)

  29. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})\) (Hint: Find common denominator then rationalize numerator.)

    Rivela la risposta

    Converges absolutely by limit comparison with p-series, \(p=3\text{/}2,\) after applying the hint.

  30. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}(\text{ln}(n+1)-\text{ln}\ n)\)

  31. \(\sum _{n=1}^{\infty }{(-1)}^{n+1}n({\text{tan}}^{-1}(n+1)-{\text{tan}}^{-1}n)\) (Hint: Use Mean Value Theorem.)

    Rivela la risposta

    Converges by alternating series test since \(n({\text{tan}}^{-1}(n+1)\text{-}{\text{tan}}^{-1}n)\) is decreasing to zero for large \(n.\) Does not converge absolutely by limit comparison with harmonic series after applying hint.

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

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

    Rivela la risposta

    Converges absolutely, since \({a}_{n}=\frac{1}{n}-\frac{1}{n+1}\) are terms of a telescoping series.

  34. \(\sum _{n=1}^{\infty }\frac{\text{cos}(n\pi )}{n}\)

  35. \(\sum _{n=1}^{\infty }\frac{\text{cos}(n\pi )}{{n}^{1\text{/}n}}\)

    Rivela la risposta

    Terms do not tend to zero. Series diverges by divergence test.

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

  37. \(\sum _{n=1}^{\infty }\text{sin}(n\pi \text{/}2)\text{sin}(1\text{/}n)\)

    Rivela la risposta

    Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.

  38. [T] \({b}_{n}=1\text{/}n,\) error \(<{10}^{-5}\)

  39. [T] \({b}_{n}=1\text{/}\text{ln}(n),\) \(n\ge 2,\) error \(<{10}^{-1}\)

    Rivela la risposta

    \(\text{ln}(N+1)>10,\) \(N+1>{e}^{10},\) \(N\ge 22026;\) \({S}_{22026}=-0.9743\ldots\)

  40. [T] \({b}_{n}=1\text{/}\sqrt{n},\) error \(<{10}^{-3}\)

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Toccare qualsiasi simbolo per la definizione completa, un'immagine, e che cosa ogni lettera in esso significa.

Come: Alternating Series

  1. Use the alternating series test to test an alternating series for convergence.
  2. Estimate the sum of an alternating series.
  3. Explain the meaning of absolute convergence and conditional convergence.
  4. Since
  5. Since
  6. We can see that
  7. Noting that
  8. For an alternating series

Domande che la gente chiede

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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