NoteNextra-origin/content/Math4111/Math4111_L18.md

Lecture 18

Review

Let (s_n) be a sequence in \mathbb{R}, and suppose \limsup_{n\to\infty} s_n=1. Consider the following four sets:

1. \{n\in\mathbb{N}:s_n>2\}
2. \{n\in\mathbb{N}:s_n<2\}
3. \{n\in\mathbb{N}:s_n>0\}
4. \{n\in\mathbb{N}:s_n<0\}

For each set, determine if the set (1) must be infinite, or (2) must be finite, or (3) could be either finite or infinite, depending on the sequence (s_n).

If \liminf_{n\to\infty} s_n=1, then \lim_{n\to\infty} \sup\{s_n,s_{n+1},s_{n+2},\dots\}=1.

So 1 must be finite, since if it is infinite, then \limsup_{n\to\infty} s_n\geq 2, which contradicts the given \limsup_{n\to\infty} s_n=1.

2 and 3 are infinite.

since \liminf_{n\to\infty} s_n=1, there exists infinitely many n such that 2>s_n>0.

4 could be either finite or infinite.

• s_n=(-1)^n is example for 4 being infinite.
• s_n=1 is example for 4 being finite.

Continue on Limit Superior and Limit Inferior

Limit Superior

Definition 3.16

Let (s_n) be a sequence of real numbers.

S^* is the largest possible value that a subsequence of (s_n) can converge to.

(Normally, we need to be careful about the definition of "largest possible value", but in this case it does exist by Theorem 3.7.)

Abbott's definition:

S^*=\limsup_{n\to\infty}\{s_k:k\geq n\}.

Theorem 3.17

Let (s_n) be a sequence of real numbers.

S^* is the unique number satisfying the following:

1. \forall x<S^*, \{n\in\mathbb{N}:s_n\geq x\} is infinite. (same as saying \forall N\in\mathbb{N},\exists n\geq N such that s_n\geq x)
2. \forall x>S^*, \{n\in\mathbb{N}:s_n\geq x\} is finite. (same as saying \exists N\in\mathbb{N} such that n\geq N\implies s_n<x)

In other words, S^* is the boundary between when \{n\in\mathbb{N}:s_n\geq x\} is infinite and when it is finite.

Proof:

(a) Case 1: S^*=\infty.

Then \sup E=\infty, so (s_n) is not bounded above.

So \forall x\in\mathbb{R}, \{n\in\mathbb{N}:s_n\geq x\} is infinite.

Case 2: S^*\in\mathbb{R}.

Then S^*=\sup E\in \overline{E}=E, by Theorem 3.7.

So \exists (s_{n_k}) such that s_{n_k}\to S^*.

So \forall x<S^*, \{n\in\mathbb{N}:s_n\geq x\} is infinite.

Case 3: S^*=-\infty: The statement is vacuously true. (\nexists x<-\infty)

(b) We'll prove the contrapositive: If \{n\in\mathbb{N}:s_n\geq x\} is infinite, then S^*\leq x.

Case 1: (s_n) is not bounded above.

Then \exists subsequence (s_{n_k}) such that s_{n_k}\to\infty. Thus S^*=\infty\leq x.

Case 2: (s_n) is bounded above.

Let M be an upper bound for (s_n). Then \{n\in\mathbb{N}:s_n\in[M,x]\} is infinite, by Theorem 3.6 (b) (\exists subsequence (s_{n_k}) in [x,M) and \exists t\in[x,M] such that s_{n_k}\to t. This implies t\in E, so x\leq t\leq S^*).

QED

Theorem 3.19 ("one-sided squeeze theorem")

Let (s_n) and (t_n) be two sequences such that s_n\leq t_n for all n\in\mathbb{N}, then

\limsup_{n\to\infty} s_n\leq \limsup_{n\to\infty} t_n

\liminf_{n\to\infty} s_n\leq \liminf_{n\to\infty} t_n

Proof:

By transitivity of \leq, for all x\in\mathbb{R},

|\{n\in\mathbb{N}:s_n\geq x\}|\leq |\{n\in\mathbb{N}:t_n\geq x\}|

By Theorem 3.17, \{n\in\mathbb{N}:t_n\geq x\} is finite \implies \{n\in\mathbb{N}:s_n\geq x\} is finite.

Thus \limsup_{n\to\infty} s_n\leq \limsup_{n\to\infty} t_n.

QED

Normal squeeze theorem: If s_n\leq t_n\leq u_n for all n\in\mathbb{N}, and \lim_{n\to\infty} s_n=\lim_{n\to\infty} u_n=L, then \lim_{n\to\infty} t_n=L.

Proof: Exercise, hint: u_n\to L\implies \limsup_{n\to\infty} u_n=\liminf_{n\to\infty} u_n=L.

Theorem 3.20

Binomial theorem: (1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k.

Special sequences:

(a) If p>0, then \lim_{n\to\infty}\frac{1}{n^p}=0.

We want to find \frac{1}{n^p}<\epsilon\iff n\geq\frac{1}{\epsilon^{1/p}}.

(b) If p>0, then \lim_{n\to\infty}\sqrt[n]{p}=1.

We want to find \sqrt[n]{p}-1<\epsilon\iff p<(1+\epsilon)^n.

Bernoulli's inequality: for \epsilon>0,n\in\mathbb{N}, (1+\epsilon)^n\geq 1+n\epsilon.

So it's enough to have p<1+n\epsilon

So we can choose N>\frac{p-1}{\epsilon}.

Another way of writing this: Let x_n=\sqrt[n]{p}-1.

Then p=(1+x_n)^n\geq 1+nx_n.

So 0\leq x_n\leq\frac{p-1}{n}.

By the squeeze theorem, x_n\to 0.s

(c) \lim_{n\to\infty}\sqrt[n]{n}=1.

We want to find \sqrt[n]{n}-1<\epsilon\iff n<(1+\epsilon)^n. (this will not work for bernoulli's inequality)

So it's enough to have n<\frac{n(n-1)}{2}\epsilon^2\iff n>1+\frac{2}{\epsilon^2}. So choose N>1+\frac{2}{\epsilon^2}.

(d) If p>0 and \alpha is real, then \lim_{n\to\infty}\frac{n^\alpha}{(1+p)^n}=0.

With binomial theorem, (1+p)^n\geq \binom{n}{k}p^k(k\leq n).

\binom{n}{k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}.

If n\geq 2k, then n-k+1\geq n-\frac{n}{2}+1\geq\frac{n}{2}.

So \binom{n}{k}\geq\frac{(n/2)^k}{k!}.

Continue on next class.