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Absolute convergence implies convergence

If n = 1 | a n | converges, then n = 1 a n converges.

Proof

Suppose that n = 1 | a n | converges. We show this by using the fact that a n = | a n | or a n = | a n | and therefore | a n | + a n = 2 | a n | or | a n | + a n = 0 . Therefore, 0 | a n | + a n 2 | a n | . Consequently, by the comparison test, since 2 n = 1 | a n | converges, the series

n = 1 ( | a n | + a n )

converges. By using the algebraic properties for convergent series, we conclude that

n = 1 a n = n = 1 ( | a n | + a n ) n = 1 | a n |

converges.

Absolute versus conditional convergence

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

  1. n = 1 ( −1 ) n + 1 / ( 3 n + 1 )
  2. n = 1 cos ( n ) / n 2
  1. We can see that
    n = 1 | ( −1 ) n + 1 3 n + 1 | = n = 1 1 3 n + 1

    diverges by using the limit comparison test with the harmonic series. In fact,
    lim n 1 / ( 3 n + 1 ) 1 / n = 1 3 .

    Therefore, the series does not converge absolutely. However, since
    1 3 ( n + 1 ) + 1 < 1 3 n + 1 and 1 3 n + 1 0 ,

    the series converges. We can conclude that n = 1 ( −1 ) n + 1 / ( 3 n + 1 ) converges conditionally.
  2. Noting that | cos n | 1 , to determine whether the series converges absolutely, compare
    n = 1 | cos n n 2 |

    with the series n = 1 1 / n 2 . Since n = 1 1 / n 2 converges, by the comparison test, n = 1 | cos n / n 2 | converges, and therefore n = 1 cos n / n 2 converges absolutely.
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Determine whether the series n = 1 ( −1 ) n + 1 n / ( 2 n 3 + 1 ) converges absolutely, converges conditionally, or diverges.

The series converges absolutely.

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To see the difference between absolute and conditional convergence, look at what happens when we rearrange the terms of the alternating harmonic series n = 1 ( −1 ) n + 1 / n . We show that we can rearrange the terms so that the new series diverges. Certainly if we rearrange the terms of a finite sum, the sum does not change. When we work with an infinite sum, however, interesting things can happen.

Begin by adding enough of the positive terms to produce a sum that is larger than some real number M > 0 . For example, let M = 10 , and find an integer k such that

1 + 1 3 + 1 5 + + 1 2 k 1 > 10 .

(We can do this because the series n = 1 1 / ( 2 n 1 ) diverges to infinity.) Then subtract 1 / 2 . Then add more positive terms until the sum reaches 100. That is, find another integer j > k such that

1 + 1 3 + + 1 2 k 1 1 2 + 1 2 k + 1 + + 1 2 j + 1 > 100 .

Then subtract 1 / 4 . Continuing in this way, we have found a way of rearranging the terms in the alternating harmonic series so that the sequence of partial sums for the rearranged series is unbounded and therefore diverges.

The terms in the alternating harmonic series can also be rearranged so that the new series converges to a different value. In [link] , we show how to rearrange the terms to create a new series that converges to 3 ln ( 2 ) / 2 . We point out that the alternating harmonic series can be rearranged to create a series that converges to any real number r ; however, the proof of that fact is beyond the scope of this text.

In general, any series n = 1 a n that converges conditionally can be rearranged so that the new series diverges or converges to a different real number. A series that converges absolutely does not have this property. For any series n = 1 a n that converges absolutely, the value of n = 1 a n is the same for any rearrangement of the terms. This result is known as the Riemann Rearrangement Theorem, which is beyond the scope of this book.

Practice Key Terms 4

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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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