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Theorem 2

Given a series

09_infinite_series-338.gif

one can form a new series by listing the terms in a different order, for example

a1 + a3 + a2 + a5 + a4 + ....

Such a series is called a rearrangement of 09_infinite_series-339.gif. The difference between absolute convergence and conditional convergence is shown emphatically by the following pair of theorems.

THEOREM 2

A.    Every rearrangement of an absolutely convergent series is also convergent and has the same sum.

B.     Let 09_infinite_series-340.gif be a conditionally convergent series.

(i) The series has a rearrangement which diverges to ∞. (ii) The series has another rearrangement which diverges to -∞.

(iii) For each real number r, the series has a rearrangement which converges to r.

We shall not prove these theorems. Instead we give a pair of rearrangements of the conditionally convergent series

09_infinite_series-341.gif

one diverging to ∞ and the other converging to - 1. The alternating series

09_infinite_series-342.gif

conditionally converges to a number between ½ and 1.

To get a rearrangement which diverges to ∞, we write down terms in the following order:

  • 1st positive term,
  • 1st negative term,
  • next 2 positive terms,
  • 2nd negative term,
  • next 4 positive terms,
  • 3rd negative term,
  • next 2m positive terms,
  • mth negative term,

We thus obtain the series

09_infinite_series-343.gif

Each block of 2m positive terms adds up to at least ,

09_infinite_series-344.gif

However, all the negative terms except -1/2 and - have absolute value ≤ 1/6. Hence after the mth negative term the partial sum is more than

09_infinite_series-345.gif

Therefore the partial sums, and hence the series, diverge to x.

To get a rearrangement which converges conditionally to -1 we proceed as follows:

Write down negative terms until the partial sum is below -1, then positive terms until the partial sum is above -1, then negative terms until the partial sum is below -1, and so on.

The mth time the partial sum goes above -1, it must be between -1 and - 1 + (1/m). The with time it goes below -1 it must be between -1 and -1 -(1/m). Therefore the series converges to -1.


Last Update: 2006-11-08