NoteNextra-origin/pages/Math4111/Math4111_L20.md

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.

EOP

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.

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,

\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.

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

\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.

EOP