# 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 $x\leq t\leq S^*$).

EOP

#### 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 $\limsup_{n\to\infty} s_n\leq \limsup_{n\to\infty} t_n$.

EOP

> 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.
