Lecture 6

Convergence of sequences

Definition: a sequence is a function, $f: \mathbb{N} \rightarrow \mathbb{R}, n \rightarrow f(n)$

We denote a sequence with one of the following:

• $\big\{{a_{n}}\big\}_{n \in \mathbb{N}}$
• $/{a_{n}/}$
• ${(a_{n})}_{n}$

We say that $a_{n} is the n^{th} term. We can start at index = 0, 1, ... any number really.

Examples:

• ${(a_{n})} = \frac{1}{n}$
• ${(a_{n})} = (-1)^n$
• ${(a_{n})} =$ the largest integer less than $\sqrt{n}$

Sequences can be defined by induction: For example, take $u_{1} = 1$. Given a defined $u_{n}$, let $u_{n+1} = 5u_{n} + 7$. This gives $u_{n+1}$ in terms of $u_{n}$

Example: Consider the Fibonacci sequence. $u_{0} = 1, u_{1} = 1$ and $u_{n+2} = u_{n+1} + u_{n}$. Values: 1, 1, 2, 3, 5, 8, 13, 21, ...

Example: If ${(a_{n})}$ is a sequence, let for $n \in \mathbb{N}: b_{n} = \sum_{k=1}^{n} a_k$

$b_n$ is called the n^{th} partial sum. The "$\sum_{k=1}^{n} a_k$" part is an infinite sequence (or series)

Example: If $r \in \mathbb{R}$, let $b_n = \sum_{k=1}^{n} r_k$ is a geometric series. A geometric series can converge or diverge.

Example: $b_n = \sum_{k=1}^{n} \frac{1}{k}$ is a harmonic series. A harmonic series diverges.

Example: If $p>0$, $b_n = \sum_{k=1}^{n} \frac{1}{k^p}$ is a "p" series. Converges or diverges.

Question: Let ${(a_{n})}$ a sequence. What happens as $n$ gets larger? How do you express that the terms $a_{n}$ approach some value, lets call it "$a$", as the index $n$ increases? Where do the terms of this sequence go?

Definition: The sequence ${(a_{n})}$ converges if $\exists a \in \mathbb{R}$ such that $\forall \epsilon > 0, \exists N \in \mathbb{N}, n \geq N \implies |a_{n} - a| < \epsilon$. We also write: $\lim_{x\to\infty} a_{n} = a$

Viz to go here

If we now take $\epsilon / 2$, the proprty gives another $N$ that yields another $|a_{n}-a| < \epsilon / 2$

SURPRISE: Not every sequence converges. $\lim_{x\to\infty} a_{n}$ may not exist.

Claim: If $\lim_{x\to\infty} a_{n}$ exists, then it is unique.

Proof:

Suppose by contradiction, $\lim_{x\to\infty}$ converges to both "$a$" and "$a'$" and that $a \neq a'$
(1) $\forall \epsilon > 0, \exists N \in \mathbb{N}, n \geq N \implies |a_{n} - a| < \epsilon$
(2) $\forall \epsilon > 0, \exists N' \in \mathbb{N}, n \geq N' \implies |a_{n} - a'| < \epsilon$
Let $\epsilon = \frac{|a' - a|}{4} > 0$
By (1), $\exists N_{1} \in \mathbb{N}, n \geq N_{1} \implies |a_{n} - a| < \epsilon$
By (2), $\exists N_{2} \in \mathbb{N}, n \geq N_{2} \implies |a_{n} - a'| < \epsilon$
Let n = max($N_{1}, N_{2}$), then $|a_{n} - a| < \epsilon = \frac{|a' - a|}{4}$ and $|a_{n} - a'| < \epsilon = \frac{|a' - a|}{4}$
Using the triangle inequality, $|a' - a| = |(a - a_{n}) -(a' - a_{n})| \leq |a - a_{n}| + |a' - a_{n}| < \frac{|a' - a|}{4} + \frac{|a' - a|}{4} = \frac{|a' - a|}{2}$
$\implies |a' - a| < \frac{|a' - a|}{2} $ which is a contradiction! So $a \neq a'$ is impossible and $\lim_{n\to\infty} a_{n}$ is unique. $\qquad \qquad \qquad \square$

Example: prove that $\lim_{n\to\infty} \frac{1}{n} = 0$

Want to show: $\forall \epsilon > 0, \exists N \in \mathbb{N}, n \geq N \implies |\frac{1}{n} - 0| < \epsilon$

Notice that: $|\frac{1}{n} - 0| = |\frac{1}{n}| = \frac{1}{n}$, so really we want to show: $\forall \epsilon > 0, \exists N \in \mathbb{N}, n \geq N \implies \frac{1}{n} < \epsilon$. Will use the Archimedian Principle.

Let $\epsilon > 0$. By the Archimedian Principle, $\exists N$ such that $\frac{1}{N} < \epsilon$, then, if $n \geq N, \frac{1}{n} \leq \frac{1}{N} < \epsilon$. $\qquad \qquad \qquad \square$

Example: The sequence $a_{n} = (-1)^n$ does not converge. $a_{1} = -1, a_{2} = 1, a_{3} = -1, ...$ We don't expect it to approach any limit.

Claim: If $\lim_{n\to\infty} a_{n}$ exists, then $\lim_{n\to\infty} |a_{n+1} - a_{n}| = 0$.

If $\lim_{n\to\infty} |a_{n+1} - a_{n}| \neq 0$ then $\lim_{n\to\infty} a_{n}$ does not exist.

If ${/{b_{n}/}_{n \in \mathbb{N}}$ is such that $b_{n} = b \in \mathbb{R}, \forall n \in \mathbb{N} \implies \lim_{n\to\infty} b_{n} = b$,

Example: The sequence $a_{n} = (-1)^n$ does not converge. In this case, $|a_{n+1} - a_{n}| = |(-1)^{n+1} - (-1)^{n}| = 2$. So $\lim_{n\to\infty} |a_{n+1} - a_{n}| = 2 \neq 0$ and the sequence does not converge.

Example: Show $\lim_{n\to\infty} (\frac{2}{n^2} + \frac{4}{n} + 3 ) = 3.

Want to show: $\forall \epsilon > 0, \exists N \in \mathbb{N}, n \geq N \implies |\frac{2}{n^2} + \frac{4}{n} + 3 - 3| < \epsilon$

Notice that: |\frac{2}{n^2} + \frac{4}{n} + 3 - 3| = |\frac{2}{n^2} + \frac{4}{n}| = \frac{2}{n^2} + \frac{4}{n}

Notice that: n^2 \geq n \implies \frac{1}{n^2} \leq \frac{1}{n}. So \frac{2}{n^2} + \frac{4}{n} \leq \frac{6}{n}

Proof: Let $\epsilon > 0$. Since $\lim_{n\to\infty} \frac{1}{n} = 0, \exists N, \frac{1}{N} < \frac{\epsilon}{6} \iff \frac{6}{N} < \epsilon.

Then, $\forall n \geq N, \frac{2}{n^2} + \frac{4}{n} \leq \frac{6}{n} \leq \frac{6}{N} < \epsilon $ $\qquad \qquad \qquad \square$

The Comparison Lemma

The Comparison Lemma: let two sequences, $a_{n}, b_{n}, a,b \in \mathbb{R}$. Suppose $\lim_{n\to\infty} a_{n} = a$ and $\exists c > 0, N_{0} \in \mathbb{N}, \forall n \geq N_{0}, |b_{n}-b| \leq c |a_{n} - a|$. Then $\lim_{n\to\infty} b_{n} = b$

Proof:

Want to show: $\forall \epsilon > 0, \exists N \in \mathbb{N}, n \geq N \implies |b_{n} - b| < \epsilon$
Let $\epsilon > 0$. Since $\lim_{n\to\infty} a_{n} = a, \forall \epsilon > 0, \exists n \geq N, |a_{n} - a| < \epsilon$/
Consider $\frac{\epsilon}{c} > 0$. $\exists N_{1}, \forall n \geq N_{1}, |a_{n} - a| < \frac{\epsilon}{c}$
Set N = max(N_{0}, N_{1}). Then, $\forall n \geq N, |b_{n}| \leq c |a_{n} - a| < c \cdot \frac{\epsilon}{c} = c$ $\qquad \qquad \qquad \square$