# Wednesday, April 3, 2013 Comparison tests, absolute and conditional convergence

## 1Comparison test¶

Comparison test: Let $(x_n), (y_n)$ be sequences of real numbers and suppose $\exists k in \mathbb N$ such that $0 \leq x_n \leq y_n$ for $n \geq k$.
i) if $\sum y_n$ converges then $\sum x_n$ converges
ii) if $\sum x_n$ diverges then $\sum y_n$ diverges

Proof: i) Suppose $\sum y_n$ converges, then for any $\epsilon > 0$ $\exists N \in \mathbb N$ such that if $m > n \geq N$ then $|y_{n+1} + y_{n+2} + \ldots + y_m| < \epsilon$ but $|y_{n+1} + \ldots + y_m| = y_{n+1} + \ldots + y_m$ if $n+1 \geq k$. Thus for $m > n > \sup\{N, k\}$, $|x_{n+1} + x_{n+2} + \ldots + x_m| = x_{n+1} + x_{n+2} + \ldots + x_m \leq y_n+1 + \ldots + y_m < \epsilon$.

Thus $\sum x_n$ converges by Cauchy Criterion. ii) is simply the contrapositive of i).

Establishing the inequality needed for the comparison test can be hard; the following test is often more useful:

## 2Limit Comparison test¶

Limit Comparison test: Let $(x_n), (y_n)$ be sequences of strictly positive numbers and suppose $\lim x_n / y_n = R \in \mathbb R$ (i.e. exists and is finite).
i) if $R \neq 0$, $\sum x_n$ converges $\iff$ $\sum y_n$ converges
ii) if $R = 0$ and $\sum y_n$ converges then $\sum x_n$ converges.

Proof: i) if $R \neq 0$ let $\epsilon = R / 2$. Then there exists $N \in \mathbb N$ such that if $n \geq N$,

$$\left| \frac{x_n}{y_n} - R \right| < \frac{R}{2}$$

i.e.

$$\frac{R}{2} < \frac{x_n}{y_n} < \frac{3R}{2} \quad\text{for } n \geq N$$

i.e. if $n \geq N$, $0 \leq x_n \leq \frac{3R}{2}y_n$. If $\sum y_n$ converges, then $\sum \frac{3R}{2}y_n = \frac{3R}{2}\sum y_n$ also converges $\implies$ $\sum x_n$ converges by comparison test. Similarly, if $n \geq N$, $\frac{R}{2}y_n \leq x_n$ or $0 \leq y_n \leq \frac{2}{R}x_n$. If $\sum x_n$ converges, so does $\sum\frac{2}{R}x_n = \frac{2}{R}\sum x_n$ and thus $\sum y_n$ converges by comparison test

ii) if $R = 0$, let $\epsilon = 1$, then $\exists K \in \mathbb N$ such that if $n \geq k$,

$$\left| \frac{x_n}{y_n} - 0 \right| < 1$$

i.e. $x_n / y_n < 1$ so $0 \leq x_n \leq y_n$ since $x_n, y_n \geq 0$. And thus if $\sum y_n$ converges, $\sum x_n$ will also converge by comparison test.

## 3Absolute and Conditional convergence¶

Definition: The series $\sum a_n$ is said to be absolutely convergent (or to converge absolutely) if $\sum |a_n|$ converges.

Theorem: if $\sum a_n$ converges absolutely, then $\sum a_n$ converges

Proof: For $\epsilon > 0$ given, $\exists N \in \mathbb N$ such that if $m > n \geq N$,

$$\big| |a_{n+1}| + |a_{n+2}| + \ldots + |a_m| \big| < \epsilon$$

thus if $m > n \geq N$ then

$$|a_{n+1} + a_{n+2} + \ldots + a_m| \leq |a_{n+1}| + |a_{n+2}| + \ldots + |a_m| = \big| |a_{n+1}| + |a_{n+2}| + \ldots + |a_m| \big| < \epsilon$$

Definition: If $\sum a_n$ is a series where $\sum|a_n|$ diverges but $\sum a_n$ still converges, $\sum a_n$ is said to conditionally converge.

As an example, take the sequence

$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$$

which is not absolutely convergent as the absolute value gives the series $\sum \frac{1}{n}$ which we know diverges. Let $S_n$ be the sequence of partial sums of $\sum \frac{(-1)^{k+1}}{k}$

\begin{align*} S_{2n} &= \left(1 - \frac{1}{2}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \ldots + \left(\frac{1}{2n-1} - \frac{1}{2n}\right) \geq 0 \ S_{2n+1} &= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots - \frac{1}{2n} + \frac{1}{2n+1} \leq 1 \ &= S_{2n} + \frac{1}{2n+1} \end{align*}

so $0 \leq \underbrace{S_{2n}}_\text{increasing} \leq \underbrace{S_{2n+1}}_\text{decreasing} \leq 1$. $S_{2n+1} = S_{2n} + \underbrace{\frac{1}{2n+1}}_{\to 0}$. So $S_{2n} \to L$ and $S_{2n+1} \to L$ thus $S_k \to L$ and our series converges conditionally.

Page 269 in the book:
2. Show that if a series is conditionally convergent, then the series obtained from adding its positive terms is divergent (negative similarly).

Proof: Let $\sum a_n$ be a conditionally convergent series. Let $\sum a_n^+$ be the series of positive terms. Consider

$$\frac{a_n + |a_n|}{2}$$

if $a_n \leq 0$. Then

$$\frac{a_n + |a_n|}{2} = \frac{a_n + a_n}{2} = a_n$$

If $a_n < 0$, $|a_n| = -a_n$, and

$$\frac{a_n + |a_n|}{2} = 0$$

So if we define

$$a_n^+ = \frac{a_n + |a_n|}{2}$$

then

$$a_n^+ = \begin{cases} a_n & a_n \geq 0\\ 0 & a_n < 0 \end{cases}$$

Thus $\sum a_n^+$ is the series of positive terms. Does $\sum a_n^+$ converge? If so, since $\sum a_n$ converges then $\sum a_n / 2$ converges and

$$\sum\left( \frac{a_n + |a_n|}{2} - \frac{a_n}{2} \right) = \sum \frac{|a_n|}{2} = \frac{1}{2}\sum|a_n|$$

giving us a contradiction. We can do this similarly for the negative terms with

$$a_n^- = \frac{a_n - |a_n|}{2}$$