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Math and science::Analysis::Tao::07. Series

The comparison test (and bounded series of non-negative numbers)

We wish to extend the comparison test for finite series to apply to infinite series.

For finite series, the comparison test appeared as a simple opening lemma (7.1.4 f):

\[ \text{When } a_i \le b_i \text{ we have } \sum_{i=m}^{n} a_i \le \sum_{i=m}^{n} b_i \]

For infinite series, we can only make this statement when \( |a_i| \le b_n \), otherwise [...]. We set this up and prove it below.

First, a useful (and very simple) proposition.

Sums of non-negative numbers are bounded iff they are convergent.

Let \( \sum_{n=m}^{\infty}a_n \) be a formal series of non-negative real numbers. Then this series is convergent if and only if there is a real number \( M \) such that

\[ \sum_{n=m}^{N}a_n \le M \text{ for all integers } N \ge m \]

Proof

Two earlier propositions are:

  • Every [...] of real numbers is bounded (6.1.17).
  • An [...] sequence which has an upper bound is convergent (6.3.8).

The sequence \( (S_N)_{n=m}^{\infty} \) representing the series \( \sum_{n=m}^{\infty}a_n \) is increasing, by definition. So the above two propositions form each side of the proposition's iff statement. 

Now we can introduce the comparison test.

Comparison test

Let \( \sum_{n=m}^{\infty}a_n \) and \( \sum_{n=m}^{\infty}b_n \) be two formal series of real numbers, and suppose that \( |a_n| \le b_n \) for all \( n \ge m \) (thus all numbers in \( (b_n) \) must be non-negative). Then if \( \sum_{n=m}^{\infty}b_n \) is convergent, then \( \sum_{n=m}^{\infty}a_n \) is [...], and in fact

\[  \left| \sum_{n=m}^{\infty}a_n \right| \le \sum_{n=m}^{\infty}|a_n| \le  \sum_{n=m}^{\infty}b_n \]