# 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
