271 lines
6.6 KiB
Markdown
271 lines
6.6 KiB
Markdown
# Lecture 20
|
|
|
|
## Review
|
|
|
|
Using the binomial theorem, prove that
|
|
|
|
$$
|
|
\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{n!}\geq \left(1+\frac{1}{n}\right)^n
|
|
$$
|
|
|
|
> Binomial theorem: $$(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$$ $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$
|
|
|
|
Proof:
|
|
$$
|
|
\begin{aligned}
|
|
\left(1+\frac{1}{n}\right)^n &= \sum_{k=0}^{n} \binom{n}{k} \left(1\right)^{n-k} \left(\frac{1}{n}\right)^k \\
|
|
&= \sum_{k=0}^{n} \binom{n}{k} \frac{1}{n^k} \\
|
|
&= \sum_{k=0}^{n} \frac{1}{k!} \prod_{j=1}^{k} \frac{n-j+1}{n} \\
|
|
\end{aligned}
|
|
$$
|
|
|
|
Since $j\geq 1$, $\frac{n-j+1}{n} \leq1$.
|
|
|
|
$$
|
|
\begin{aligned}
|
|
&= \sum_{k=0}^{n} \frac{1}{k!} \prod_{j=1}^{k} \frac{n-j+1}{n} \\
|
|
&\geq \sum_{k=0}^{n} \frac{1}{k!} \\
|
|
\end{aligned}
|
|
$$
|
|
|
|
## New material
|
|
|
|
### Series
|
|
|
|
#### Definition 3.30
|
|
|
|
$$
|
|
e=\sum_{n=0}^{\infty} \frac{1}{n!}
|
|
$$
|
|
|
|
#### Lemma 3.30
|
|
|
|
$\sum_{n=0}^{\infty} \frac{1}{n!}$ converges.
|
|
|
|
Proof:
|
|
|
|
If $n\geq 2$,
|
|
|
|
$$
|
|
\begin{aligned}
|
|
\frac{1}{n!} &= \frac{1}{n} \cdot \frac{1}{(n-1)!} \\
|
|
&\leq \frac{1}{2} \cdot \frac{1}{2} \cdot \dots \cdot \frac{1}{2} \\
|
|
&= \frac{1}{2^{n-1}}
|
|
\end{aligned}
|
|
$$
|
|
|
|
$$
|
|
\frac{1}{n!} \leq \frac{1}{2^{n-1}}
|
|
$$
|
|
|
|
So $\sum_{n=0}^{\infty} \frac{1}{n!}$ converges.
|
|
|
|
#### Theorem 3.31
|
|
|
|
$$
|
|
\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e
|
|
$$
|
|
|
|
Proof:
|
|
|
|
Let $s_n = \sum_{k=0}^{n} \frac{1}{k!}$, let $t_n = \left(1+\frac{1}{n}\right)^n$.
|
|
|
|
Goal: $\lim_{n\to\infty} s_n = \lim_{n\to\infty} t_n$. we already proved $\lim_{n\to\infty} s_n$ exists. But we don't know yet if $\lim_{n\to\infty} t_n$ exists.
|
|
|
|
By warmup exercise, $\forall n\geq 0, t_n \leq s_n$.
|
|
|
|
So if $\limsup_{n\to\infty} t_n \leq \limsup_{n\to\infty} s_n$, then $\lim_{n\to\infty} t_n$ exists and $\lim_{n\to\infty} t_n = \lim_{n\to\infty} s_n$.
|
|
|
|
Now we will show $\limsup_{n\to\infty} t_n \geq e$.
|
|
|
|
Ideas: (special case of the argument)
|
|
|
|
If $n\geq 2$, then
|
|
|
|
$$
|
|
\begin{aligned}
|
|
t_n &= \sum_{k=0}^{n} \binom{n}{k} \left(\frac{1}{n}\right)^k \\
|
|
&\geq \binom{n}{0} + \binom{n}{1}\frac{1}{n} + \binom{n}{2}\left(\frac{1}{n}\right)^2 + \cdots + \binom{n}{n}\left(\frac{1}{n}\right)^n \\
|
|
&= 1 + \frac{n}{n} + \frac{n(n-1)}{2n^2} + \cdots + \frac{1}{n^n} \\
|
|
\end{aligned}
|
|
$$
|
|
|
|
Let $n\to\infty$, then
|
|
|
|
$$
|
|
\liminf_{n\to\infty} t_n \geq 1 + 1 + \frac{1}{2} + \frac{1}{3} + \cdots
|
|
$$
|
|
|
|
Fix $m\geq 2$, for any $n\geq m$,
|
|
|
|
$$
|
|
t_n \geq \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!}\frac{n}{n}\frac{n-1}{n}\cdots+\frac{1}{m!}\frac{n}{n}\frac{n-1}{n}\cdots\frac{n-m+1}{n}
|
|
$$
|
|
|
|
Let $n\to\infty$, then
|
|
|
|
$$
|
|
\liminf_{n\to\infty} t_n \geq \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \cdots + \frac{1}{m!}=s_m
|
|
$$
|
|
|
|
So $\liminf_{n\to\infty} t_n \geq \lim_{n\to\infty} s_n = e$.
|
|
|
|
Therefore, $e\leq \liminf_{n\to\infty} t_n\leq \limsup_{n\to\infty} t_n\leq e$.
|
|
|
|
So $\lim_{n\to\infty} t_n$ exists and $\lim_{n\to\infty} t_n = e$.
|
|
|
|
QED
|
|
|
|
#### Theorem 3.32
|
|
|
|
$e$ is irrational.
|
|
|
|
Q: How good is the approximation is $s_n$ to $e$?
|
|
|
|
A: Very good actually.
|
|
$$
|
|
\begin{aligned}
|
|
e-s_n &= \sum_{k=n+1}^{\infty} \frac{1}{k!} \\
|
|
&<\frac{1}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+\cdots\right) \\
|
|
&=\frac{1}{(n+1)!}\sum_{k=0}^{\infty}\left(\frac{1}{n+1}\right)^k \\
|
|
&=\frac{1}{(n+1)!}\frac{1}{1-\frac{1}{n+1}} \\
|
|
&=\frac{1}{n!}\cdot\frac{1}{n} \\
|
|
&<\frac{1}{n!n}
|
|
\end{aligned}
|
|
$$
|
|
|
|
Proof:
|
|
|
|
Suppose $e=\frac{p}{q}$ for some $p,q\in\mathbb{N}$.
|
|
|
|
Observe that:
|
|
|
|
$$
|
|
s_q=1+1+\frac{1}{2}+\cdots+\frac{1}{q!}
|
|
$$
|
|
|
|
So $q! s_q$ is an integer.
|
|
|
|
Since $e=\frac{p}{q}$, $q!e$ is an integer, $q!(e-s_q)$ is an integer.
|
|
|
|
However,
|
|
|
|
$$
|
|
0<q!(e-s_q)<\frac{q!}{q!q}<\frac{1}{q}
|
|
$$
|
|
|
|
Contradiction.
|
|
|
|
QED
|
|
|
|
### The root and ratio tests
|
|
|
|
This is a fancy way of using comparison test with geometric series.
|
|
|
|
#### Theorem 3.33 (Root test)
|
|
|
|
> $$ \sqrt[n]{|a_n|} \leq \alpha \implies |a_n|\leq \alpha^n$$
|
|
|
|
Given a series $\sum_{n=0}^{\infty} a_n$, put $\alpha = \limsup_{n\to\infty} \sqrt[n]{|a_n|}$.
|
|
|
|
Then
|
|
|
|
(a) If $\alpha < 1$, then $\sum_{n=0}^{\infty} a_n$ converges.
|
|
(b) If $\alpha > 1$, then $\sum_{n=0}^{\infty} a_n$ diverges.
|
|
(c) If $\alpha = 1$, the test gives no information
|
|
|
|
Proof:
|
|
|
|
(a) Suppose $\alpha < 1$. Then $\exists \beta$ such that $\alpha < \beta < 1$.
|
|
|
|
By **Theorem 3.17(b)**, $\forall n\geq N, \sqrt[n]{|a_n|} < \beta$.
|
|
|
|
So $\forall n\geq N, |a_n| < \beta^n$.
|
|
|
|
By comparison test, $\sum_{n=0}^{\infty} a_n$ converges.
|
|
|
|
(b) Suppose $\alpha > 1$. By **Theorem 3.17(a)**, $\{n\in \mathbb{N}: \sqrt[n]{|a_n|} > 1\}$ is infinite.
|
|
|
|
Thus $a_n\not\to 0$, $\sum_{n=0}^{\infty} a_n$ diverges.
|
|
|
|
(c) $\sum_{n=0}^{\infty} \frac{1}{n}$ and $\sum_{n=0}^{\infty} \frac{1}{n^2}$ both have $\alpha = 1$. but the first diverges and the second converges.
|
|
|
|
QED
|
|
|
|
#### Theorem 3.34 (Ratio test)
|
|
|
|
> $$ \left|\frac{a_{n+1}}{a_n}\right| \leq \alpha \implies |a_n|\leq \alpha^n$$
|
|
|
|
Given a series $\sum_{n=0}^{\infty} a_n$, $a_n\in\mathbb{C}\backslash\{0\}$.
|
|
|
|
Then
|
|
|
|
(a) If $\limsup_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| < 1$, then $\sum_{n=0}^{\infty} a_n$ converges.
|
|
(b) If $\left|\frac{a_{n+1}}{a_n}\right| \geq 1$ for all $n\geq n_0$ for some $n_0\in\mathbb{N}$, then $\sum_{n=0}^{\infty} a_n$ diverges.
|
|
|
|
Remark:
|
|
|
|
1. If $\limsup_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$, the test gives no information.
|
|
2. If $\limsup_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| > 1$, the test gives no information.
|
|
|
|
Proof:
|
|
|
|
(b) $\forall n\geq n_0, \left|\frac{a_{n+1}}{a_n}\right| \geq 1$.
|
|
|
|
So $a_{n_0}\not\to 0$, $\sum_{n=0}^{\infty} a_n$ diverges.
|
|
|
|
(a) $\beta \in(\limsup_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|, 1)$.
|
|
|
|
By **Theorem 3.17(b)**, $\exists N$ such that $\forall n\geq N, \left|\frac{a_{n+1}}{a_n}\right| < \beta < 1$.
|
|
|
|
So,
|
|
|
|
$$
|
|
\begin{aligned}
|
|
|a_N| &< \beta|a_N|\\
|
|
|a_{N+1}| &< \beta|a_{N+1}|\\
|
|
|a_{N+2}| &< \beta|a_{N+2}|\\
|
|
\end{aligned}
|
|
$$
|
|
|
|
i.e. $\forall n\geq N, |a_n| < \beta^{n-N}|a_N|=\beta^n(\beta^{-N}|a_N|)$.
|
|
|
|
Since $\sum_{n=N}^{\infty} \beta^n$ converges, by comparison test, $\sum_{n=0}^{\infty} a_n$ converges.
|
|
|
|
QED
|
|
|
|
We will skip **Theorem 3.37**. One implication is that if ratio test can be applied, then root test can be applied.
|
|
|
|
### Power series
|
|
|
|
#### Definition 3.38
|
|
|
|
Let $(c_n)$ be a sequence of complex numbers. A power series is a series of the form
|
|
|
|
$$
|
|
\sum_{n=0}^{\infty} c_n z^n
|
|
$$
|
|
|
|
#### Theorem 3.39
|
|
|
|
Given a power series $\sum_{n=0}^{\infty} c_n z^n$, let $R=\frac{1}{\limsup_{n\to\infty} \sqrt[n]{|c_n|}}$.
|
|
|
|
Then
|
|
|
|
(a) The series converges absolutely for all $z\in\mathbb{C}$ with $|z| < R$.
|
|
(b) The series diverges for all $z\in\mathbb{C}$ with $|z| > R$.
|
|
(c) If $0\leq r < R$, then the series converges uniformly on the closed disk $\{z\in\mathbb{C}: |z|\leq r\}$.
|
|
|
|
Proof:
|
|
|
|
$$
|
|
\begin{aligned}
|
|
\limsup_{n\to\infty} \sqrt[n]{|c_n z^n|} &= \limsup_{n\to\infty} \sqrt[n]{|c_n|} \cdot |z| \\
|
|
&= \frac{|z|}{R}
|
|
\end{aligned}
|
|
$$
|
|
|
|
By root test, the series converges absolutely for all $z\in\mathbb{C}$ with $|z| < R$.
|
|
|
|
QED
|