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10.31 Subsequences

Sequences

Introduction

\(e\) is by definition \(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}\)

In order to be able to give this definition, we need to prove that \(a_n=(1+\frac1n)^n\) is convergent.

We need to prove 1. \(a_n\) is monotonic increasing 2. \(a_n\) is bounded above \(\leq 3\)

  1. We want to show \(a_{n}>a_{n-1},\forall n\in\mathbb{N}\Leftrightarrow\frac{a_{n}}{a_{n-1}}>1,\forall n\in\mathbb{N}\)​

    Since \(a_{n}=\left(\frac{n+1}{n}\right)^{n}\), \(a_{n-1}=\left(\frac{n}{n-1}\right)^{n-1}\)

    Analysis: First, i think that i can use binomial theorem or transform into a n square

    I prefer the second because it is not too complex. Then i will have the form like image, it remind me to use Bernoulli's Inequality

    \(\frac{a_{n}}{a_{n-1}}=\frac{\frac{n}{n-1}\left(\frac{n+1}{n}\right)^{n}}{\left(\frac{n}{n-1}\right)^{n}}=\frac{n}{n-1}\left(\frac{\frac{n+1}{n}}{\frac{n}{n-1}}\right)^{n}=\frac{n}{n-1}\left(\frac{\left(n+1\right)\left(n-1\right)}{n^2}\right)^{n}=\frac{n}{n-1}(\frac{n^2-1}{n^2})^{n}=\frac{n}{n-1}\left(1-\frac{1}{n^2}\right)^{n}\)​

    Since Bernoulli's inequality, then \(\frac{n}{n-1}\left(1-\frac{1}{n^2}\right)^{n}>\frac{n}{n-1}\left(1-\frac{1}{n}\right)=\frac{n}{n-1}\left(\frac{n-1}{n}\right)=1\) when \(n>2\)

  2. \(( 1 + \frac{1}{n} )^n, \, n \to \infty\)

    \(= C_n^0 \left( \frac{1}{n} \right)^0 + C_n^1 \left( \frac{1}{n} \right)^1 + C_n^2 \left( \frac{1}{n} \right)^2 + \cdots + C_n^n \left( \frac{1}{n} \right)^n + \cdots\)

    \(= 1 + \frac{n!}{(n-1)!1!} \cdot \frac{1}{n} + \frac{n!}{2!(n-2)!} \cdot \frac{1}{n^2} + \cdots + \frac{n!}{n!} \left( \frac{1}{n} \right)^n + \cdots\)

    \(= 1 + 1 + \frac{n(n-1)}{2!} \cdot \frac{1}{n^2} + \frac{n(n-1)(n-2)}{3!} \cdot \frac{1}{n^3} + \cdots + \frac{1}{n^n} + \cdots\)

    \(< 1 + 1 + \frac{n^2}{2!} \cdot \frac{1}{n^2} + \frac{n^3}{3!} \cdot \frac{1}{n^3} + \cdots + \frac{n^n}{n!} \cdot \frac{1}{n^n} + \cdots\)

    \(= 2 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!} + \cdots\)

    \(= 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!} + \cdots\)

    \(< 1 + 1 + \frac{1}{2^1} + \frac{1}{2^2} + \cdots + \frac{1}{2^{n-1}} + \cdots\)

    \(= 2 + \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 3\)

    Prove binomial theorem

    \(C_m^n = \frac{n!}{m!(n-m)!}\)

    Use induction

    Suppose it is true for \((x+1)^{n}=\sum_{i=0}^{n}C_{n}^{i}x^{i}\)

    \((x + 1)^n \cdot (x + 1) = \sum_{i=0}^{n} C_n^i x^i \cdot (x + 1)\)

    \(= \sum_{i=0}^{n} C_n^i x^i + \sum_{i=0}^{n} C_n^i x^{i+1}\)

    \(= C_n^0 x^0 + (C_n^0 + C_n^1) x^1 + (C_n^1 + C_n^2) x^2 + \cdots\)

    \(=C_{n+1}^0x^0+C_{n+1}^1x^1+C_{n+1}^2x^2+\cdots\)

    \(=\sum_{i=0}^{n+1}C_{n+1}^{i}x^{i}\)​

Summary

\(a_{n}=\left(1+\frac{1}{n}\right)^{n}\) is bounded above and monotonic increasing \(\Rightarrow\) is convergent

Definition

Let's call \(e\) the \(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}\), we can see that \(e\geq2.7182....\)​

Subsequences

If \(a_{n}=\left(a_1,a_2,a_3,\ldots\right)\), then a subsequence of \(a_n\) is something like \(b_{n}=(a_2,a_3,a_7,a_{19},\ldots)\)

Definition

A subsequence of \((a_n)\) is a sequence \(b_{k}=a_{g\left(k\right)}\) with \(g:\mathbb{N}\rightarrow\mathbb{N}\) strictly increasing

In the example, \(g(1)=2,g(2)=3,g(3)=7,g(4)=19,...\)

We usually write: \(b_{k}=a_{n_k},n_k=g(k)\)​

Note \(n_1<n_2<n_3<\ldots\) (\(g\) is strictly increasing)

Example

\(b_{k}=a_{2k+1}\)

\(b_{k}=\left(a_3,a_5,a_7,\ldots\right)\)

In particular: If \(a_n=(-1)^n\) and \(b_k=-1,\forall k\)

Theorem

If \((a_n)\) has \(\lim_{n\to\infty}a_{n}=l\), then \(\lim_{n\to\infty}b_{k}=l,\forall\)subsequence \(b_{k}=a_{n_k}\)

Proof

We know \(\lim_{n\to\infty}a_{n}=l\Rightarrow\forall\varepsilon>0,\exists n_0\) such that \(\left|a_{n}-l\right|<\varepsilon,\forall n>n_0\)

We want to prove that \(\lim_{k\to\infty}b_{k}=l\)

Let \(\varepsilon>0\) we want \(k_0\) such that \(\left|b_{k}-l\right|<\varepsilon,\forall k>k_0\)

image​

Analysis here:

((Then here is the awkward thing, we have \(|a_{n}-l|<\varepsilon,\forall n>n_0\), then need \(\left|b_{k}-l\right|<\varepsilon,\forall k>k_0\)

Recall the definition: \(b_k=a_{n_k}\), thus \(\left|b_{k}-l\right|<\varepsilon,\forall k>k_0\Rightarrow\left|a_{n_{k}}-l\right|<\varepsilon,\forall n_{k}>n_{k_0}\)

Then we need to prove that \(n_{k}>n_{k_0}>n_0\), we can let \(n_{k_{0}}>n_0\) since we can take any \(k_0\equiv n_{k_0}\)

Then by increasing, \(n_{k}>n_{k_0}\)))

Continuing proof

Since \(g\) is increasing (\(n_1<n_2<n_3<....\)), there is \(k_0\) such that \(n_{k_0}>n_0\Rightarrow k>k_0\) then \(n_{k}>n_{k_0}>n_0\)

And since \(\forall\varepsilon>0,\exists n_0\) such that \(\left|a_{n}-l\right|<\varepsilon,\forall n>n_0\), then \(|a_{n_{k}}-l|<\varepsilon\Rightarrow|b_{k}-l|<\varepsilon\)

Remark

The converse is false as the previous example shows: \((-1)^n\) not convergent, \(b_k=-1\) is convergent

Lemma

Any sequence \((a_n)\) has a subsequence \((a_{n_k})\) that is either increasing or decreasing

IMG_20241031_160156​

We say that \(a_n\) is a top if \(a_n\geq a_m,\forall m>n\)

Let \(S=\text{the tops }=\{a_{n_1}\geq a_{n_2}\geq a_{n_3}\geq...\}\) could be \(\begin{cases}\emptyset\\ finite\\ infinite\end{cases}\)​

If \(S\) is infinite, then \(\{a_{n_k}\}\) is monotonic decreasing

Suppose \(S=\begin{cases}\emptyset\\ finite\end{cases}\Rightarrow\exists n_0\) there are no tops in \((a_{n_0},a_{n_1},a_{n_2},...)\Rightarrow \exists n_1>n_0\) such that \(a_{n_1}>a_{n_0}\)

Also \(\exists n_2>n_1\) such that \(a_{n_2}>a_{n_1}\) and \(\exists n_3>n_2\) such that \(a_{n_3}>a_{n_2}\)......

Thus we find \(a_{n_0}<a_{n_1}<a_{n_2}<...\) which is monotonic increasing

Theorem

Every bounded sequence \((a_n)\) has a convergent subsequence

Proof

From the lemma we know that there is a subsequence \((a_{n_k})\) that is either monotonic increasing or monotonic decreasing.

Since \((a_n)\) is bounded, then \((a_{n_k})\) is bounded. Then by theorem, \((a_{n_k})\) is convergent.

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