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):

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 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

\sum_{n = m}^{N} a_{n} \leq M for all integers N \geq m

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 $| a_{n} | \leq b_{n}$ 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 absolutely convergent, and in fact

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

Proof

Let

S A_{N} := \sum_{n = m}^{N} | a_{n} |

be the Nth partial sum of

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

and

S B_{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:

S A_{N} \leq S B_{N} \leq M for all N \geq m and for some bounding real M

Through this transitivity,

(S A_{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

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

. Thus, with all of the below infinite series converging we can compare them like so:

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

We can also run the Comparison Test in the contrapositive:

\sum_{n = m}^{\infty} a_{n}

is not absolutely convergent then

\sum_{n = m}^{\infty} b_{n}

is not conditionally convergent.

Source

p171

25.11.2019