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We remark that this theorem is true more generally as long as there exists some integer N such that 0 b n + 1 b n for all n N .

Convergence of alternating series

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

  1. n = 1 ( −1 ) n + 1 / n 2
  2. n = 1 ( −1 ) n + 1 n / ( n + 1 )
  1. Since
    1 ( n + 1 ) 2 < 1 n 2 and 1 n 2 0 ,
    the series converges.
  2. Since n / ( n + 1 ) 0 as n , we cannot apply the alternating series test. Instead, we use the n th term test for divergence. Since
    lim n ( −1 ) n + 1 n n + 1 0 ,
    the series diverges.
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Determine whether the series n = 1 ( −1 ) n + 1 n / 2 n converges or diverges.

The series converges.

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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

n = 1 ( −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 [link] , we see that for any integer N 1 , the remainder R N satisfies

| R N | = | S S N | | S N + 1 S N | = b n + 1 .

Remainders in alternating series

Consider an alternating series of the form

n = 1 ( −1 ) n + 1 b n or n = 1 ( −1 ) n b n

that satisfies the hypotheses of the alternating series test. Let S denote the sum of the series and S N denote the N th partial sum. For any integer N 1 , the remainder R N = S S N satisfies

| R 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 th partial sum S N is in magnitude at most the size of the next term b N + 1 .

Estimating the remainder of an alternating series

Consider the alternating series

n = 1 ( −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 .

From the theorem stated above,

| R 10 | b 11 = 1 11 2 0.008265 .

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Find a bound for R 20 when approximating n = 1 ( −1 ) n + 1 / n by S 20 .


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Absolute and conditional convergence

Consider a series n = 1 a n and the related series n = 1 | a n | . Here we discuss possibilities for the relationship between the convergence of these two series. For example, consider the alternating harmonic series n = 1 ( −1 ) n + 1 / n . The series whose terms are the absolute value of these terms is the harmonic series, since n = 1 | ( −1 ) n + 1 / n | = n = 1 1 / 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 n = 1 ( −1 ) n + 1 / n 2 . The series whose terms are the absolute values of the terms of this series is the series n = 1 1 / n 2 . Since both of these series converge, we say the series n = 1 ( −1 ) n + 1 / n 2 exhibits absolute convergence.


A series n = 1 a n exhibits conditional convergence    if n = 1 | a n | converges. A series n = 1 a n exhibits absolute convergence    if n = 1 a n converges but n = 1 | a n | diverges.

As shown by the alternating harmonic series, a series n = 1 a n may converge, but n = 1 | a n | may diverge. In the following theorem, however, we show that if n = 1 | a n | converges, then n = 1 a n converges.

Questions & Answers

Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
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Introduction about quantum dots in nanotechnology
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nano basically means 10^(-9). nanometer is a unit to measure length.
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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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