ANA1_L30

Quiz next week! likely on Tuesday 5:30. Uniform continuity, connectedness.
Test following week on Ch 5, 6.

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Series: Look ahead

Consider a series of functions $f_1,f_2,f_3,\cdots :X\to Y$. "$f_n\to f$" means for each $x\in X$, $f_n(x)\to f(x)$, i.e, point-wise convergence. In other words,

$$\forall x,\forall ϵ>0,\exists N\text{ such that }\forall n>N,d_Y(f_n(x),f(x))<ϵ.$$

Note that $N$ might depend on $x$ in this definition. If we can get $N$ which is independent of $x$, i.e, determined only by $ϵ$, we say that the sequence uniformly converges.

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Preliminaries

Definition 1(Definition).

Let $(b_n)$ be a sequence. An infinite series is an expression of the form

$$\sum _{n=1}^{\infty }{b}_n=b_1+b_2+b_3+\dots .$$

We define the corresponding sequence of partial sums $(s_m)$ by

$$s_m=b_1+b_2+\cdots +b_m,$$

and say that the series ${\sum }_{n=1}^{\infty }{b}_n$ converges to $B$ if the sequence $(s_m)$ converges to $B$. In this case, we write ${\sum }_{n=1}^{\infty }{b}_n=B$.

Note that only $a_n$ with $n>N_0$ for any finite $N_0$ matter for determining whether a series converges. In other words, we can drop a finite number of terms in the beginning of the series.

The Cauchy criterion can be restated for series:

Theorem 2(Theorem).

$\sum a_n$ converges $⟺$ $\forall ϵ>0$, $\exists N$ such that $\forall m\ge n>N$, we have

$$\left|\sum _{k=n}^m{a}_k\right|<ϵ.$$

The triangle inequality gives us

$$\sum |a_n|\text{ converges }⟹\sum a_n\text{ converges }.$$

Also, if we take $m=n$ in the above theorem, we get the following corollary.

Theorem 3(Corollary).

$\sum a_n$ converges $⟹$ ${\lim }_{n\to \infty }{a}_n=0$.

The converse is not true! $\sum \frac{1}{n}$ diverges.

Rudin, 3.24

Theorem 4(Theorem).

Suppose $a_n\ge 0$ for all $n$. Then, $\sum a_n$ converges $⟺$ $(s_n)$ is bounded.

Note that only the backward implication has any value; the forward implication is true for all convergent series, since convergent sequences are bounded. The former is true because $(s_n)$ is monotone.

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Convergence tests

Comparison test

Rudin, 3.25

Theorem 5(Theorem).

Suppose $|a_n|\le c_n$ for $n\ge$ some $N_0$.
a) $\sum c_n$ converges $⟹$ $\sum a_n$ converges.
b) $\sum a_n$ diverges $⟹$ $\sum c_n$ diverges.

Proof
Given $ϵ>0$, we know that $\exists N$ such that $\forall m\ge n>N$ we have

$$\sum _{k=n}^m{c}_k<ϵ$$

The same $N$ works for $\sum a_n$.

$$\left|\sum _{k=n}^m{a}_k\right|\le \sum _{k=n}^m|a_k|\le \sum _{k=n}^m{c}_k<ϵ.$$

(b) is just the contrapositive of (a). ❏

Rudin, 3.27

Rudin, 3.27

Theorem 6(Theorem).

Suppose $a_1\ge a_2\ge a_3\ge \cdots \ge 0$. Then the series $\sum a_n$ converges if and only if the series

$$\sum _{k=0}^{\infty }{2}^k{a}_{2^k}=a_1+2a_2+4a_4+8a_8+\dots$$

converges.

Proof of $⟸$
$s_n=a_1+a_2+\cdots +a_n$
$t_k=a_1+2a_2+\cdots +2^k{a}_{2^k}$
For $n<2^k$,

$$\begin{array}{rl}{s}_n& <a_1+(a_2+a_3)+(a_4+\cdots +a_7)+\cdots +(a_{2^k}+\cdots +a_{2^{k+1}-1})\\ & \le a_1+2a_2+4a_4+\cdots +2^k{a}_{2^k}\\ & =t_k\end{array}$$

Since $(t_n)$ converges, we know it is bounded, i.e, $t_k\le M$ for all $k$. We have shown that for every $n$, there exists a $k$ such that $s_n<t_k$. Thus, $(s_n)$ is bounded, and hence it must converge. ❏

Proof of $⟹$
Let $n,k\in {\mathbb{Z}}^{+}$. For $n\ge 2^k$,

$$\begin{array}{rl}{s}_n& \ge a_1+a_2+(a_3+a_4)+(a_5+\cdots +a_8)+\cdots +(a_{2^{k-1}+1}+\cdots +a_{2^k})\\ & \ge \frac{1}{2}{a}_1+a_2+2a_4+4a_8+\cdots +2^{k-1}{a}_{2^k}\\ & =\frac{1}{2}{t}_k\end{array}$$

So, for all $k$, there exists an $n$ such that $t_k\le 2s_n\le M$ for some $M$. ❏

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Standard series

To use the comparison test efficiently, we need to have plenty of series of nonnegative terms whose convergence or divergence is known.

The geometric series

Rudin, 3.26

Theorem 7(Theorem).

If $0\le x<1$, then

$$\sum _{n=0}^{\infty }{x}^n=\frac{1}{1-x}.$$

If $x\ge 1$, the series diverges.

Proof
If $x\ne 1$,

$$s_n=\sum _{k=0}^n{x}^k=\frac{1-x^{n+1}}{1-x}.$$

The result follows if we let $n\to \infty$.

The p-series

Rudin, 3.28

Theorem 8(Theorem).

$\sum \frac{1}{n^p}$ converges if $p>1$ and diverges if $p\le 1$.

Proof
If $p\le 0$, then the series diverges because ${\lim }_{n\to \infty }1/n^p\ne 0$. If $p>0$, we can use 3.27.

$$\begin{array}{r}{t}_k=\sum 2^k\left(\frac{1}{2^{kp}}\right)=\sum 2^{(1-p)k}.\end{array}$$

This is a geometric series. If $p\in (0,1]$, $2^{1-p}\ge 1$, and $t_k$ diverges, implying $\sum \frac{1}{n^p}$ diverges. Similarly, if $p>1$, $2^{1-p}<1$, and $t_k$ converges, implying $\sum \frac{1}{n^p}$ converges. ❏

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The number e

Rudin, 3,30

Definition 9(Definition).

$$\begin{array}{r}e=\sum _{n=0}^{\infty }\frac{1}{n!}.\end{array}$$

Since

$$\begin{array}{r}{s}_n<1+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots ,\end{array}$$

$s_n$ converges.

Rudin, 3.31

Theorem 10(Theorem).

$$\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n=e.$$

Proof
Let

$$s_n=\sum _{k=0}^n\frac{1}{k!},t_n={\left(1+\frac{1}{n}\right)}^n.$$

We have to show that ${\lim }_{n\to \infty }{t}_n$ exists, and that it is equal to ${\lim }_{n\to \infty }{s}_n$. Recall that a limit exists only when $\lim \sup =\lim \inf$. We know that for any sequence, $\lim \inf \le \lim \sup$. The goal is to show

$$e\le \underset{n\to \infty }{\mathrm{liminf}}{t}_n\le \underset{n\to \infty }{\mathrm{limsup}}{t}_n\le e.$$

Part 1: Showing $\lim \sup t_n\le e$

$$\begin{array}{rl}{t}_n& ={\left(1+\frac{1}{n}\right)}^n\\ & =\sum _{k=0}^n{\left(\frac{1}{n}\right)}^k\left(\begin{array}{c}n\\ k\end{array}\right)\\ & =\sum _{k=0}^n\frac{1}{n^k}\frac{n!}{k!(n-k)!}\\ & =1+1+\sum _{k=2}^n\frac{1}{k!}\frac{(n-k+1)\dots (n-1)}{n^{k-1}}\\ & =1+1+\sum _{k=2}^n\frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\dots \left(1-\frac{k-1}{n}\right)\\ & \le 1+1+\sum _{k=2}^n\frac{1}{k!}\\ & =s_n\end{array}$$

So, $t_n\le s_n$ for all $n$. It follows that $\lim \sup t_n\le \lim \sup s_n$. Since $(s_n)\to e$, $\lim \sup s_n=\lim \inf s_n=\lim s_n=e$. So we get $\lim \sup t_n\le e$.

Part 2: Showing $\lim \inf t_n\le e$
Fix $m$. For $n\ge m$,

$$\begin{array}{rl}{t}_n& \ge \underset{u_n}{\underbrace{1+1+\sum _{k=2}^m\frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\dots \left(1-\frac{k-1}{n}\right)}}.\end{array}$$

Note that $(u_n)$ is a convergent sequence.

$$\begin{array}{rl}\underset{n\to \infty }{\mathrm{liminf}}{t}_n& \ge \underset{n\to \infty }{\mathrm{liminf}}{u}_n\\ & =\sum _{k=0}^m\frac{1}{k!}\end{array}$$

This is true for every $m$. Thus, we have $\mathrm{liminf}{t}_n\ge e$.

❏

Rudin, 3.32

Theorem 11(Theorem).

$e$ is irrational.

Proof
If $s_n$ has the same definition as in the previous proof,

$$\begin{array}{rl}e-s_n& =\sum _{k=n+1}^{\infty }\frac{1}{k!}\\ & <\frac{1}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1{)}^2}+\dots \right)\\ & =\frac{1}{n!n}.\end{array}$$

Now, let $e=\frac{p}{q}$. Then,

$$0<q!(e-s_q)<\frac{1}{q}.$$

Note that $q!e$ and $q!s_q$ are integers. But there does not exist an integer between 0 and 1. ❏