Theory

Theory 1

Consider these series:

$$\begin{array}{cc}1& +\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\cdots =\infty \\ & \\ -1& -\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-\frac{1}{7}-\cdots =-\infty \\ & \\ 1& -\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\cdots =\ln 2\\ & \\ 1& +\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\cdots =?\end{array}$$

The absolute values of terms are the same between these series, only the signs of terms change.

The first is a positive series because there are no negative terms.

The second series is the negation of a positive series – the study of such series is equivalent to that of positive series, just add a negative sign everywhere. These signs can be factored out of the series. (For example $\sum -\frac{1}{n}=-\sum \frac{1}{n}$.)

The third series is an alternating series because the signs alternate in a strict pattern, every other sign being negative.

The fourth series is not alternating, nor is it positive, nor negative: it has a mysterious or unknown pattern of signs.

A series with any negative signs present, call it ${\sum }_{n=1}^{\infty }{a}_n$, converges absolutely when the positive series of absolute values of terms, namely ${\sum }_{n=1}^{\infty }|a_n|$, converges.

THEOREM: Absolute implies ordinary

If a series ${\sum }_{n=1}^{\infty }{a}_n$ converges absolutely, then it also converges as it stands.

A series might converge due to the presence of negative terms and yet not converge absolutely:

A series ${\sum }_{n=1}^{\infty }{a}_n$ is said to be converge conditionally when the series converges as it stands, but the series produced by inserting absolute values, namely ${\sum }_{n=1}^{\infty }|a_n|$, diverges.

The alternating harmonic series above, $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots =\ln 2$, is therefore conditionally convergent. Let us see why it converges. We can group the terms to create new sequences of pairs, each pair being a positive term. This can be done in two ways. The first creates an increasing sequence, the second a decreasing sequence:

$$\begin{array}{cc}\text{increasing from }0\text{:}& (1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{5}-\frac{1}{6})+(\frac{1}{7}-\frac{1}{8})+\cdots \\ & \\ \text{decreasing from }1\text{:}& 1-(\frac{1}{2}-\frac{1}{3})-(\frac{1}{4}-\frac{1}{5})-(\frac{1}{6}-\frac{1}{7})-\cdots \end{array}$$

Suppose $S_N$ gives the sequence of partial sums of the original series. Then $S_{2N}$ gives the first sequence of pairs, namely $S_2$, $S_4$, $S_6$, $\dots$ . And $S_{2N-1}$ gives the second sequence of pairs, namely $S_1$, $S_3$, $S_5$, $\dots$ .

The second sequence shows that $S_N$ is bounded above by $1$, so $S_{2N}$ is monotone increasing and bounded above, so it converges. Similarly $S_{2N-1}$ is monotone decreasing and bounded below, so it converges too, and of course they must converge to the same thing.

The fact that the terms were decreasing in magnitude was an essential ingredient of the argument for convergence. This fact ensured that the parenthetical pairs were positive numbers.

Alternating Series Test (AST) - “Leibniz Test”

Applicability: Alternating series only: ${\sum }_{n=1}^{\infty }{(-1)}^{n-1}{a}_n$ with $a_n>0$

Test Statement: If:

1. $a_n\to 0$ as $n\to \infty$ (i.e. it passes the SDT: if this fails, conclude diverges)
2. $a_n$ are decreasing, so $a_1>a_2>a_3>a_4>\cdots >0$

Then:

$$\sum _{n=1}^{\infty }{(-1)}^{n-1}{a}_n\text{converges}$$

“Next Term Bound” rule for error of the partial sums:

$$|S-S_N|<a_{N+1}$$

Extra - Alternating Series Test: Theory

Just as for the alternating harmonic series, we can form positive paired-up series because the terms are decreasing:

$$\begin{array}{c}(a_1-a_2)+(a_3-a_4)+(a_5-a_6)+\cdots \\ \\ a_1-(a_2-a_3)-(a_4-a_5)-(a_6-a_7)-\cdots \end{array}$$

The first sequence $S_{2N}$ is monotone increasing from $0$, and the second $S_{2N-1}$ is decreasing from $a_1$. The first is therefore also bounded above by $a_1$. So it converges. Similarly, the second converges. Their difference at any point is $S_{2N}-S_{2N-1}$ which is equal to $-a_{2N}$, and this goes to zero. So the two sequences must converge to the same thing.

By considering these paired-up sequences and the effect of adding each new term one after the other, we obtain the following order relations:

$$0<S_2<S_4<S_6<\cdots <S<\cdots <S_5<S_3<S_1=a_1$$

Thus, for any even $2N$ and any odd $2M-1$:

$$S_{2N}<S<S_{2M-1}$$

Now set $M=N$ and subtract $S_{2N-1}$ from both sides:

$$\begin{array}{cc}{S}_{2N}-S_{2N-1}& <S-S_{2N-1}<0\\ & \\ \gg \gg -a_{2N}& <S-S_{2N-1}<0\end{array}$$

Now set $M=N+1$ and subtract $S_{2N}$ from both sides:

$$\begin{array}{cc}0<S-S_{2N}& <S_{2N+1}-S_{2N}\\ & \\ \gg \gg 0<S-S_{2N}& <a_{2N+1}\end{array}$$

This covers both even cases ($n=2N$) and odd cases ($n=2N-1$). In either case, we have:

$$|S-S_n|<a_{n+1}$$