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Math 4111 Exam 3 review

Relations between series and topology (compactness, closure, etc.)

Limit points E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\backslash\{x\}\cap E\neq\phi\}

Closure \overline{E}=E\cup E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\cap E\neq\phi\}

p_n\to p\implies \forall \epsilon>0, \exists N such that \forall n\geq N, p_n\in B_\epsilon(p)

Some interesting results

Lemma

p\in \overline{E}\iff \exists (p_n)\subseteq E such that p_n\to p

p\in E'\iff \exists (p_n)\subseteq E\backslash\{p\} such that p_n\to p (you cannot choose p in the sequence)

Bolzano-Weierstrass Theorem

Let E be a compact set and (p_n) be a sequence in E. Then \exists (p_{n_k})\subseteq (p_n) such that p_{n_k}\to p\in E.

Rudin Proof:

Rudin's proof uses a fact from Chapter 2.

If E is compact, and S\subseteq E is infinite, then S has a limit point in E (S'\cap E\neq\phi).

Examples of Cauchy sequence that does not converge

Cauchy sequence in (X,d),\forall \epsilon>0, \exists N such that \forall m,n\geq N, d(p_m,p_n)<\epsilon

Let X=\mathbb{Q} and (p_q)=\{1,1.4,1.41,1.414,1.4142,1.41421,\dots\} The sequence is Cauchy but does not converge in \mathbb{Q}.

This does not hold in \mathbb{R} because compact metric spaces are complete.

Fact: Every Cauchy sequence is bounded.

Proof that e is irrational

e=\sum_{n=0}^\infty \frac{1}{n!}

Let s_n=\sum_{k=0}^n \frac{1}{k!}

So e-s_n=\left(\sum_{k=n+1}^\infty \frac{1}{k!}\right)<\frac{1}{n!n}

If e is rational, then \exists p,q\in\mathbb{Z} such that e=\frac{q}{p} and q!s_q\in\mathbb{Z}, q!e=q!\frac{p}{q}\in \mathbb{Z}, so q!(e-s_q)\in\mathbb{Z}

0<q!(e-s_q)<\frac{1}{n!n} leads to contradiction.

\limsup and \liminf

Let (a_n)=(-1)^n

\limsup a_n=1 and \liminf a_n=-1

Let (a_n)\to a

\limsup a_n=\liminf a_n=a

Facts about \limsup and \liminf

Convergence of subsequence

\limsup is the largest value that subsequence of a_n can approach to.

\liminf is the smallest value that subsequence of a_n can approach to.

Elements of sequence

\forall x>s^*,\{n:a_n>x\} is finite. \exists N such that \forall n\geq N, a_n\leq x

\forall x<s^*,\{n:a_n>x\} is infinite.

One example is (a_n)=(-1)^n\frac{n}{n+1}

\limsup a_n=1 and \liminf a_n=-1

So the size of set of elements of a_n that are greater than any x<1 is infinite. and the size of set of elements of a_n that are greater than any x>1 is finite.

\limsup(a_n+b_n)\leq \limsup a_n+\limsup b_n

One example for smaller than is (a_n)=(-1)^n and (b_n)=(-1)^{n+1}

\limsup(a_n+b_n)=0 and \limsup a_n+\limsup b_n=2

(\forall n,s_n\leq t_n) \implies \limsup s_n\leq \limsup t_n

One example of using this theorem is (s_n)=\left(\sum_{k=1}^n\frac{1}{k!}\right) and (t_n)=\left(\frac{1}{n}+1\right)^n

Rearrangement of series

Will not be tested.

infinite sum is not similar to finite sum. For infinite sum, the order of terms matters. But for finite sum, the order of terms does not matter, you can rearrange the terms as you want.

Ways to prove convergence of series

n-th term test (divergence test)

If \lim_{n\to\infty}a_n\neq 0, then \sum a_n diverges.

Definition of convergence of series (convergence and divergence test)

If \sum a_n converges, then \lim_{n\to\infty}\sum_{k=1}^n a_k=0.

Example: Telescoping series and geometric series.

Comparison test (convergence and divergence test (absolute convergence))

Let (a_n) be a sequence in \mathbb{C} and (c_n) be a non-negative sequence in \mathbb{R}. Suppose \forall n, |a_n|\leq c_n.

(a) If the series \sum_{n=1}^{\infty}c_n converges, then the series \sum_{n=1}^{\infty}a_n converges.
(b) If the series \sum_{n=1}^{\infty}a_n diverges, then the series \sum_{n=1}^{\infty}c_n diverges.

Ratio test (convergence and divergence test (absolute convergence))

\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.

Root test (convergence and divergence test (absolute convergence))

\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

Cauchy criterion

Geometric series

P-series

(a) \sum_{n=0}^{\infty}\frac{1}{n} diverges.
(b) \sum_{n=0}^{\infty}\frac{1}{n^2} converges.

Cauchy condensation test (convergence test)

Suppose (a_n) is a non-negative sequence. The series \sum_{n=1}^{\infty}a_n converges if and only if the series \sum_{k=0}^{\infty}2^ka_{2^k} converges.

Dirichlet test (convergence test)

Suppose

(a) the partial sum A_n of \sum a_n form a bounded sequence.
(b) b_0\geq b_1\geq b_2\geq \cdots (non-increasing)
(c) \lim_{n\to\infty}b_n=0.

Then \sum a_nb_n converges.

Example: \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} converges.

Abel's test (convergence test)

Let (b_n)^\infty_{n=0} be a sequence such that:

(a) b_0\geq b_1\geq b_2\geq \cdots (non-increasing) (b) \lim_{n\to\infty}b_n=0

Then if |z|=1 and z\neq 1, \sum_{n=0}^\infty b_nz^n converges.