16 - Convergence of Sequences

Convergence of a Sequence

Let $(a_n)$ be a real sequence
Let $L\in \mathbb{R}$ then

$(a_n)$ converges to $L$ as $n\to \infty$ if

$$\forall ϵ>0:\exists n\in \mathbb{N}\text{ s.t. }\forall n\ge N,|a_n-L|<ϵ$$

It can also be written as

$$(a_n)\to L\text{ as }n\to \infty \text{ or }\underset{n\to \infty }{\lim }{a}_n=L$$

Alternative definition for convergence $(a_n)$ converges to $L$ as $n\to \infty$ if

$$\forall ϵ>0:\exists n\in \mathbb{N}\text{ s.t. }\forall n>N,|a_n-L|\le ϵ$$

Tip - for when proving a sequence convergence $|a_n-L|<cϵ$ where $c$ is some constant that is good enough! This is because you can just scale everything back to get $ϵ$

If you can show that

Convergence vs Divergence of a Sequence

Let $(a_n)$ be a real sequence then

1. $(a_n)$ converges, or is convergent, if there is

$$L\in \mathbb{R}\text{ s.t. }{a}_n\to L\text{ as }n\to \infty$$

2. $(a_n)$ diverges, or is divergent if $(a_n)$ doesn’t converge

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Tails Lemma lemma

Let $(a_n)$ be a sequence

1. If $(a_n)$ converges to a limit $L$ then every tail of $(a_n)$ converges and to $L$

2. If a tail $(b_n)=(a_{n+k})$ of $(a_n)$ converges, then $(a_n)$ converges

Proof

1. Take a tail of $(a_n)$ so for $k\ge 1$ and let $b_n=a_{n+k}$ for $n\ge 1$
   Assume that $(a_n)$ converges to a limit L
   Take $ϵ>0$
   Then there is $N\in \mathbb{N}$ such that

$$\text{if }n\le N\text{ then }|a_n-L|<ϵ$$

But if $n\ge N$ then $n+k\ge N$ hence

$$|a_{n+k}-L|<ϵ⟹|b_n-L|<ϵ$$

Hence $(b_n)$ converges and $b_n\to L$ as $n\to \infty$
2) Assume that $(b_n)=(a_{n+k})$ converges
Then there is $L\in \mathbb{R}$ such that

$$b_n\to L\text{ as }n\to \infty$$

Take $ϵ>0$
Then there is $N$ such that if $m\ge N$ then

$$|b_m-L|<ϵ⟹|a_{m+k}-L|<ϵ$$

Now if $n\ge N+k$ then $n=m+k$ where $m\ge N$ so

$$|a_n-L|<ϵ$$

So $(a_n)$ converges and

$$a_n\to L\text{ as }n\to \infty$$

When showing convergence you don't have to find the smallest $n$

You can also show it works for $0<ϵ<1$ which naturally extends for $ϵ>0$

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Useful Convergence lemma

1. Take $c\in \mathbb{R}$ with $|c|<1$ then

$$c^n\to 0\text{ as }n\to \infty$$

2. Let $a_n=\frac{n}{2^n}$ for $n\ge 1$ then

$$a_n\to 0\text{ as }n\to \infty$$

Proof

1. Write $|c|=\frac{1}{1+y}$ where $y>0$
   Take $ϵ>0$
   Let $N=\lceil \frac{1}{y\cdot ϵ}\rceil +1$
   Take $n\ge N$
   By Bernoulli’s Inequality (since $y>0$ and $n\ge 1$ we have

$$(1+y{)}^n\ge 1+ny$$

Then

$$|c^n|=\frac{1}{(1+y{)}^n}\le \frac{1}{1+ny}\le \frac{1}{Ny}<ϵ$$

So $c^n\to 0$ as $n\to \infty$
2) If $n\ge 2$ then $2^n=(1+1{)}^n\ge (\genfrac{}{}{0ex}{}{n}{2})$ (by the binomial theorem)
Take $ϵ>0$
Let $N=\lceil 2+\frac{2}{ϵ}\rceil$
For $n\ge N$ then

$$|a_n-0|=\frac{n}{2^n}\le \frac{n}{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}=\frac{2}{n-1}\le \frac{2}{N-1}<ϵ$$

Hence $a_n\to 0$ as $n\to \infty$

Modulus on a sequence

Let $(a_n)$ be a convergent sequence then $(|a_n|)$ also converges
Moreover if $a_n\to L$ as $n\to \infty$ then

$$|a_n|\to L\text{ as }n\to \infty$$

Proof

Say $a_n\to L$ as $n\to \infty$
Take $ϵ>0$ then
There is $N$ such that if $n\ge N$ then $|a_n-L|<ϵ$
If $n\ge N$ then by the Reverse Triangle Inequality then

$$||a_n|-|L||\le |a_n-L|<ϵ$$

So $(|a_n|)$ converges and $|a_n|\to |L|$ as $n\to \infty$

Can also be proved using Sandwiching Lemma

Limits preserve weak inequalities

Let $(a_n)$ and $(b_n)$ be real sequences, and assume that

$$a_n\to L\text{ and }{b}_n\to M\text{ as }n\to \infty \text{ and }{a}_n\le b_n\text{ for all }n$$

Then

$$L\le M$$

Proof

Suppose, for a contradiction, that it is not the case that $L\le M$ so

$$L>M\text{ (by trichotomy)}$$

Let $ϵ=\frac{1}{2}(L-M)>0$
Since $a_n\to L$ as $n\to \infty$, there is $N_1$ such that

$$\text{if }n\ge N_1\text{ then }|a_n-L|<ϵ$$

Since $b_n\to M$ as $n\to \infty$, there is $N_2$ such that

$$\text{if }n\ge N_1\text{ then }|a_n-M|<ϵ$$

For $n\ge \max \{N_1,N_2\}$, we have

$$a_n>L-ϵ\text{ and }{b}_n<M+ϵ$$

Hence

$$\begin{array}{rl}L-ϵ& <{\alpha }_n\le b_n<M+ϵ\\ L-M& <2ϵ=L-M\end{array}$$

This is a contradirction hence $L=M$

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Limits of $n^{\alpha }$ lemma

1. If $\alpha <0$, then $n^{\alpha }\to 0$ as $n\to \infty$

2. If $\alpha >0$, then $n^{\alpha }\to \infty$ as $n\to \infty$

Proof

1. Take $ϵ\in (0,1)$ we have

$$\begin{array}{rl}{n}^{\alpha }& <ϵ\\ ⟺e^{\alpha \log n}& <ϵ\\ ⟺\alpha \log n& <\log ϵ\\ ⟺\log n& >\frac{1}{\alpha }\log ϵ(\text{note }\alpha <0)\\ ⟺n& >e^{\frac{1}{\alpha }\log ϵ}\end{array}$$

So let $N=1+\lceil e^{\frac{1}{\alpha }\log ϵ}\rceil$
2) Take $M>0$ we have

$$\begin{array}{rl}{n}^{\alpha }& >M\\ ⟺e^{\alpha \log n}& >M\\ ⟺\alpha \log n& >\log M\\ ⟺\log n& >\frac{1}{\alpha }\log M(\text{note }\alpha >0)\\ ⟺n& >e^{\frac{1}{a}\log ϵ}\end{array}$$

So let $N=1+\lceil e^{\frac{1}{a}\log ϵ}\rceil$

Limits of $c^n$ ($c\in {\mathbb{R}}^{>0}$)

1. If $c<1$, then $c^n\to 0$ as $n\to \infty$

2. If $c=1$, then $c^n\to 1$ as $n\to \infty$

3. If $c>1$, then $c^n\to \infty$ as $n\to \infty$

Proof

1. Shown in Useful Convergences (1)
2. Clear from definition as $c^n=1$ for all $n\in \mathbb{N}$

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Reciprocals and infinite/zero limits

Let $(a_n)$ be a sequence of positive real numbers then

$$a_n\to \infty \text{ as }n\to \infty ⟺\frac{1}{a_n}\to 0\text{ as n→∞}$$

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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

$$\frac{L}{2}=L⟹L=0$$