# Lecture 21

## Review

Recall the alternating series test from calculus: "Suppose $(a_n)^\infty_{n=1}$ is a sequence satisfies the following conditions:

1. The sequence is nonnegative. (For all $n\in \mathbb{N}$, $a_n\geq 0$.)
2. The sequence is decreasing. ($a_1\geq a_2\geq a_3\geq \cdots$)
3. $\lim_{n\to\infty}a_n=0$.

Then $\sum_{n=1}^\infty (-1)^{n+1}a_n$ converges."

Exercise: Show that the statement above is false if we remove the second condition.

[Hint: Use the fact that $\sum_{n=1}^\infty \frac{1}{n}$ diverges.]

Let the sequence $a_n$ be defined as $a_n=\frac{1}{n},a_{n+1}=0$ for all $n\in \mathbb{N}$. This sequence satisfies the 1,3 but not the 2. 

And the harmonic series is not convergent.

## New Material

### Other tests for convergence of series

Recall the integration by parts formula:

Let $A(t),a(t),b(t)$ be functions of $t$ and $A'(t)=a(t)$.

Then 
$$
\begin{aligned}
\int_p^q a(t)b(t)\,dt&=\int_p^q b(t)A'(t)\,dt\\
&=\left.b(t)A(t)\right|_p^q-\int_p^q A(t)b'(t)\,dt
\end{aligned}
$$

#### Theorem 3.41 Summation by parts

Let $a_n,b_n$ be sequences.

Let $A(n)=\sum_{k=1}^n a_k$. ($A_{-1}=0$). If $0\leq p\leq q$, then

$$
\sum_{n=p}^q a_nb_n=A_q b_q-A_{p-1}b_p-\sum_{n=p}^{q-1}A_n (b_n-b_{n+1})
$$

Proof:

$$
\begin{aligned}
\sum_{n=p}^q a_nb_n&=\sum_{n=p}^q (A_n-A_{n-1})b_n\\
&=\sum_{n=p}^q A_nb_n-\sum_{n=p}^q A_{n-1}b_n\\
&=\sum_{n=p}^q A_nb_n-\sum_{n=p-1}^{q-1}A_n b_{n+1}\\
&=A_qb_q-A_{p-1}b_p-\sum_{n=p}^{q-1} A_nb_n-\sum_{n=p}^{q-1} A_n b_{n+1}\\
&=A_qb_q-A_{p-1}b_p-\sum_{n=p}^{q-1} A_n (b_n-b_{n+1})
\end{aligned}
$$

QED

#### Theorem 3.42 (Dirichlet's test)

Suppose

(a) the partial sum $A_n$ of $\sum a_n$ form a bounded sequence.  
(b) $b_0\geq b_1\geq b_2\geq \cdots$ (non-increasing)  
(c) $\lim_{n\to\infty}b_n=0$.

Then $\sum a_nb_n$ converges.

Proof:

By Cauchy criterion, it's enough to prove

$\forall \epsilon >0, \exists N\in \mathbb{N}$ such that for all $p\geq q\geq N$,

$$
\left|\sum_{n=p}^q a_nb_n\right|<\epsilon
$$

By the partial sum $A_n$ of $\sum a_n$ form a bounded sequence. Let $\left|A_n\right|\leq M$ for all $n\in \mathbb{N}$.

$$
\begin{aligned}
\left|\sum_{n=p}^q a_nb_n\right|&=\left|A_qb_q-A_{p-1}b_p-\sum_{n=p}^{q-1}A_n (b_n-b_{n+1})\right|\\
&\leq |A_qb_q|+|A_{p-1}b_p|+\sum_{n=p}^{q-1}|A_n (b_n-b_{n+1})|\\
&\leq M|b_q|+M|b_p|+\sum_{n=p}^{q-1}M(b_n-b_{n+1})\\
&=M|b_q|+M|b_p|+M\sum_{n=p}^{q-1}(b_n-b_{n+1})\\
&=M|b_q|+M|b_p|+M(b_p-b_q)\\
&=2M|b_p|
\end{aligned}
$$

Then we let $\epsilon >0$ be given. Since $b_n\to 0$, there exists $N\in \mathbb{N}$ such that for all $n\geq N$, $|b_n|<\frac{\epsilon}{2M}$.

If $q\geq p\geq N$, then

$$
\left|\sum_{n=p}^q a_nb_n\right|\leq 2M|b_p|<\epsilon
$$

So $\sum a_nb_n$ converges.

QED

#### Theorem 3.43 (Alternating series test)

Let $(b_n)^\infty_{n=1}$ be a sequence such that:

(a) $b_1\geq b_2\geq b_3\geq \cdots$ (non-increasing)
(b) $\lim_{n\to\infty}b_n=0$

Then $\sum_{n=1}^\infty (-1)^{n+1}b_n$ converges.

Proof:

Let $a_n=(-1)^{n+1}$

$A_n=\sum_{k=1}^n a_k=1$ if $n$ is odd, $0$ if $n$ is even.

So $|A_n|\leq 1$ for all $n\in \mathbb{N}$.

By Theorem 3.42, $\sum_{n=1}^\infty a_n b_n$ converges.

QED

Example:

Consider the power series $\sum_{n=0}^\infty \frac{z^n}{n}$.

The radius of convergence is $1$.

We claim that the series converges for all $z\in \mathbb{C}$ with $|z|=1$ and $z\neq 1$.

#### Theorem 3.44 Abel's test

Let $(b_n)^\infty_{n=0}$ be a sequence such that:

(a) $b_0\geq b_1\geq b_2\geq \cdots$ (non-increasing)
(b) $\lim_{n\to\infty}b_n=0$

Then if $|z|=1$ and $z\neq 1$, $\sum_{n=0}^\infty b_nz^n$ converges.

Proof:

Fix $z\in \mathbb{C}$ with $|z|=1$ and $z\neq 1$.

Let $a_n=z^n$.

Then $A_n=\sum_{k=0}^n z^k=\frac{1-z^{n+1}}{1-z}$._

$|A_n|\leq \frac{|1-z^{n+1}|}{|1-z|}$ for all $n\in \mathbb{N}$.

By triangle inequality, $|1-z^{n+1}|\leq |1|+|z^{n+1}|=1+|z^{n+1}|$.

And since $|z|=1$, $|z^{n+1}|=|z|^{n+1}=1$.

So $|1-z^{n+1}|\leq 2$.

So $|A_n|\leq \frac{2}{|1-z|}$ for all $n\in \mathbb{N}$.

By Dirichlet's test, $\sum_{n=0}^\infty b_nz^n$ 

QED

### Absolute convergence

The series $\sum_{n=0}^\infty a_n$ is said to **converge absolutely** if $\sum_{n=0}^\infty |a_n|$ converges.

If $\sum_{n=0}^\infty a_n$ converges but does not converge absolutely, then $\sum_{n=0}^\infty a_n$ is said to **converge conditionally**.

_Absolute convergence are nice but conditionally convergent series are not._

#### Theorem 3.45 (Absolute convergence)

If $\sum_{n=0}^\infty a_n$ converges absolutely, then $\sum_{n=0}^\infty a_n$ converges.

Proof:

Use comparison test.

$$
\sum_{n=0}^\infty |a_n|\geq \sum_{n=0}^\infty a_n
$$

QED

Rearrangement of series:

Let $f:\mathbb{N}\to \mathbb{N}$ be a bijection.

If $\sum_{n=0}^\infty a_n$ is a sequence and $b_n=a_{f(n)}$, then $(b_n)^\infty_{n=0}$ is a rearrangement of $(a_n)^\infty_{n=0}$.

If $\sum_{n=0}^\infty a_n$ converges absolutely, then any rearrangement of $\sum_{n=0}^\infty a_n$ converges to the same sum.

Example:

$a_n=\frac{(-1)^{n+1}}{n}$. $b_n=a_{f(n)}$.

|n|1|2|3|4|5|6|7|8|9|
|:-:|:-:|:-:|:-:|:-:|:-:|:-:|:-:|:-:|:-:|
|$f(n)$|1|2|4|3|6|8|5|10|12|
|$b_n$|1|-1/2|-1/4|1/3|-1/6|-1/8|1/5|-1/10|-1/12|

$$
\sum_{n=1}^\infty a_n=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\log 2
$$

$$
\begin{aligned}
\sum_{n=1}^\infty b_n&=1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}-\frac{1}{10}-\frac{1}{12}+\cdots\\
&=\left(1-\frac{1}{2}\right)-\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{6}\right)-\frac{1}{8}+\left(\frac{1}{5}-\frac{1}{10}\right)-\frac{1}{12}+\cdots\\
&=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots\\
&=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\right)\\
&=\frac{1}{2}\log 2
\end{aligned}
$$

You cannot always rearrange series.

But, if $\sum_{n=0}^\infty a_n$ converges absolutely, then you can rearrange the series.

#### Theorem 3.55

Let $(a_n)^\infty_{n=0}$ be a sequence in $\mathbb{C}$ such that $\sum_{n=0}^\infty |a_n|$ converges absolutely.

Then any rearrangement of $\sum_{n=0}^\infty a_n$ converges absolutely to the same sum.

$$
\sum_{n=0}^\infty a_n=\sum_{n=0}^\infty a_{f(n)}
$$

Ideas of proof:

Let $f:\mathbb{N}\to \mathbb{N}$ be a bijection.

and let $b_n=a_{f(n)}$.

Let $s_n=\sum_{k=0}^n a_k,t_n=\sum_{k=0}^n b_k=\sum_{k=0}^n a_{f(k)}$.

$I_n=\{1,2,\cdots,n\}$.

$J_n=\{f(1),f(2),\cdots,f(n)\}$.

$$
\begin{aligned}
s_n-t_n&=\sum_{k=0}^n a_k-\sum_{k=0}^n a_{f(k)}\\
&=\sum_{k\in I_n} a_k-\sum_{k\in J_n} a_k\\
&= \sum_{k\in I_n\setminus J_n} a_k+\sum_{k\in J_n\setminus I_n} a_k\\
&\leq \sum_{k\in I_n\setminus J_n} |a_k|+\sum_{k\in J_n\setminus I_n} |a_k|
\end{aligned}
$$

Key observation:

For every $n\in \mathbb{N}$, there exists a $p$ such that $\{1,2,\cdots,n\}\subset I_n\cap J_n$.

Then $|s_n-t_n|\leq \sum_{k=N+1}^\infty |a_k|$.

QED
