# 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):

For infinite series, we can only make this statement when \( |a_i| \le b_n \), otherwise the sum for \( a \) could diverge. 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

#### Proof

Two earlier propositions are:

- Every convergent sequence of real numbers is bounded (6.1.17).
- An increasing 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 absolutely convergent, and in fact

#### Proof

Let \( SA_N := \sum_{n=m}^{N}|a_n| \) be the Nth partial sum of \( \sum_{n=m}^{\infty}|a_n| \) and \( SB_N := \sum_{n=m}^{N}b_n \) be the Nth partial sum of \( \sum_{n=m}^{\infty}b_n \). As \( \sum_{n=m}^{\infty}b_n \) is bounded according to the proposition above, and by the the comparison test for finite series we have: \[ SA_N \le SB_N \le M \text{ for all } N \ge m \text{ and for some bounding real } M \] Through this transitivity, \( (SA_N)_{N=n}^{\infty} \) also meets the conditions for being bounded. Then from the above propositive again, it must be that \( \sum_{n=m}^{\infty}|a_n| \) is convergent. The Absolute Convergence Test informs us that absolute convergence implies conditional convergence, and in addition that \( \left| \sum_{n=m}^{\infty}a_n \right| \le \sum_{n=m}^{\infty}|a_n| \). Thus, with all of the below infinite series converging we can compare them like so:We can also run the Comparison Test in the contrapositive: