25. Monotonic Sequences

Monotonic Increasing / Decreasing

Let $(a_n)$ be a real sequence

1. Monotonic Increasing / Monotone Increasing / Increasing:

$$a_n\le a_{n+1}\text{ for all }n$$

2. Monotonic Decreasing / Monotone Decreasing / Decreasing

$$a_n\ge a_{n+1}\text{ for all }n$$

3. Monotonic / Monotone

$$\text{Either increasing or decreasing}$$

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Strictly Increasing / Decreasing

Let $(a_n)$ be a real sequence
4) Strictly Increasing:

$$a_n<a_{n+1}\text{ for all }n$$

5. Strictly Decreasing

$$a_n>a_{n+1}\text{ for all }n$$

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13 - Monotone Sequences Theorem

Monotone Sequences Theorem

Let $(a_n)$ be a real sequence

1. If $(a_n)$ is increasing and bounded above, then $(a_n)$ converges
2. If $(a_n)$ is decreasing and bounded below, then $(a_n)$ converges

TLDR: Bounded monotone seqence converges

Proof

1. Assume that $(a_n)$ is increasing and bounded above
   Define set $S:=\{a_n:n\ge 1\}$ then it is non-empty and bounded above
   By Completeness Axiom then $\sup S$ exists

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Take $ϵ>0$
By the Approximation Property, there is $N$ such that

$$\sup S-ϵ<a_N\le \sup S$$

As $(a_n)$ is increasing then for $n\ge N$ then

$$\sup S-ϵ<a_N\le a_n\le \sup S$$

Hence

$$|a_n-\sup S|<ϵ$$

So $(a_n)$ converges, and $a_n\to \sup S$ as $n\to \infty$

2. If $(a_n)$ is decreasing and bounded below, then $(-a_n)$ is increasing and bounded above so $(1)$

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Convergence of $\frac{\log n}{n}$ lemma

$$\frac{\log n}{n}\to 0\text{ as }n\to \infty$$

Proof

Let $a_n=\frac{\log n}{n}$
Then $a_n\ge 0$ for all $n$ so $a_n$ is bounded below

By properties of $\log$ then $(a_n{)}_{n\ge 100}$ is decreasing
Hence by the Monotone Sequence Theorem, $(a_n)$ converges

Suppose $\frac{\log n}{n}\to L$ as $n\to \infty$
Since limits preserve weak inequalities, we have $L\ge 0$

Now

$$a_{2n}=\frac{\log (2n)}{2n}=\frac{\log 2+\log n}{2n}\to 0+\frac{L}{2}\text{ by AOL}$$

But as $a_{2n}$ is a subsequence of $(a_n)$ then $a_{2n}\to L$ as $n\to \infty$
Hence by uniqueness of limits

$$\frac{L}{2}=L⟹L=0$$

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