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):

When a_{i} \leq b_{i} we have \sum_{i = m}^{n} a_{i} \leq \sum_{i = m}^{n} b_{i}

For infinite series, we can only make this statement when $| a_{i} | \leq b_{n}$ , 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} \leq M for all integers N \geq 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 $| a_{n} | \leq b_{n}$ for all $n \geq m$ (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

| \sum_{n = m}^{\infty} a_{n} | \leq \sum_{n = m}^{\infty} | a_{n} | \leq \sum_{n = m}^{\infty} b_{n}

25.11.2019