Lecture 20

Review

Using the binomial theorem, prove that

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:

Since $j\geq 1$, $\frac{n-j+1}{n} \leq1$.

New material

Series

Definition 3.30

Lemma 3.30

$\sum_{n=0}^{\infty} \frac{1}{n!}$ converges.

Proof:

If $n\geq 2$,

So $\sum_{n=0}^{\infty} \frac{1}{n!}$ converges.

Theorem 3.31

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 $\limsup_{n\to\infty} t_n \geq e$.

Idea: (special case of the argument)

If $n\geq 2$, then

Let $n\to\infty$, then

Fix $m\geq 2$, for any $n\geq m$,

Let $n\to\infty$, then

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 $\lim_{n\to\infty} t_n = e$.

EOP

Theorem 3.32

$e$ is irrational.

Q: How good is the approximation is $s_n$ to $e$?

A: Very good actually.

Proof:

Suppose $e=\frac{p}{q}$ for some $p,q\in\mathbb{N}$.

Observe that:

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,

Contradiction.

EOP

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.

EOP

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,

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.

EOP

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

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:

By root test, the series converges absolutely for all $z\in\mathbb{C}$ with $|z| < R$.

EOP