Sequences

f

a_1 = f(1), a_2 = f(2), a_3 = f(3), \dots

We say that a sequence has a limit $L$ if we can make terms arbitrarily close to $L$ by taking $n$ to be sufficiently large. More precisely, the limit of a sequence $a_n$ as $n$ approaches infinity is $L$ ( $\lim_{n \to \infty} a_n = L$ ) if for every $\epsilon > 0$ there exists an $N \in \naturals$ such that $\abs{a_n -L} < \epsilon$ whenever $n \geq N$ .

If a sequence's limit exists and is finite, we say the sequence converges to the limit and is a convergent sequence.

Example: The limit of the sequence $a_n = \frac{3n}{n+1}$ is $3$ . Proof: We need to show that given $\epsilon > 0$ , there is some $N \in \naturals$ where if $n > N$ , $\abs{\frac{3n}{n+1} - 3} < \epsilon$ . First, note that

$\abs{\frac{3n}{n+1} - 3} = \abs{\frac{3n - 3(n+1)}{n+1}} = \abs{\frac{-3}{n+1}} = \frac{3}{n+1}.$

Now,

$\frac{3}{n+1} < \epsilon \iff 3 < \epsilon (n+1) \iff 3 < \epsilon n + \epsilon \iff -\epsilon n > \epsilon - 3 \iff n > \frac{3 - \epsilon}{\epsilon}.$

Therefore, if we pick a natural number $N \geq \frac{3 - \epsilon}{\epsilon}$ , then $\abs{a_n - 3} < \epsilon$ whenever $n \geq N$ , and $\lim_{n \to \infty} \frac{3n}{n+1} = 3.$

We can simplify this proof a bit by recognizing that because $\frac{3}{n+1} < \frac{3}{n}$ , if $\frac{3}{n} < \epsilon$ , then $\frac{3}{n+1}$ is also less than $\epsilon$ . $\frac{3}{n} < \epsilon$ when $n < \frac{3}{\epsilon},$ so we can choose $N$ as a natural number greater than $\frac{3}{\epsilon}.$

Uniqueness of the Limit of a Sequence

Note that the limit of a sequence is unique, that is, if $\lim_{n \to \infty} = L$ and $\lim_{n \to \infty} = M,$ then $L = M$ . Proof:. Suppose $L > M$ . Then $\frac{L - M}{2} > 0,$ and for some $N_1, N_2 \in \naturals$ we have

$n > N_1 \implies \abs{a_n - L} < \frac{L - M}{2}, \quad n > N_2 \implies \abs{a_n - M} < \frac{L - M}{2}.$

Now, let $N = \max{(N_1, N_2)}.$ We have that whenever $n > N$ ,

$\abs{a_n - L} < \frac{L - M}{2}, \quad \abs{a_n - M} < \frac{L - M}{2},$

which can be rewritten as

$- \frac{L - M}{2} < a_n - L < \frac{L - M}{2}, \text{ and } - \frac{L - M}{2} < a_n - M < \frac{L - M}{2}.$

Therefore, whenever $n \geq N$ ,

$- \frac{L - M}{2} + L < a_n, \text{ and } a_n < \frac{L - M}{2} + M.$

Hence we have that $\frac{L+M}{2} < a_n < \frac{L+M}{2}$ , but this is impossible, so our supposition that $L > M$ must be false.

We can make a similar argument assuming that $M > L$ , and this leads to the conclusion that $L = M$ , and therefore $L$ is the unique limit of the sequence.

Limit Laws for Sequences

Constant Sequence: If $a_n = L$ for all $n \in N$ , then $\lim_{n \to \infty} a_n = L.$

Limit of a Sum is the Sum of the Limits: If $a_n$ and $b_n$ are convergent sequences, then $\lim_{n \to \infty}{(a_n + b_n)} = \lim_{n \to \infty}{a_n} + \lim_{n \to \infty}{b_n}$ .

Limit of a Difference is the Difference of the Limits: If $a_n$ and $b_n$ are convergent sequences, then $\lim_{n \to \infty}{(a_n - b_n)} = \lim_{n \to \infty}{a_n} - \lim_{n \to \infty}{b_n}$ .

Factoring a Constant Through the Limit: If $a_n$ is a convergent sequence and $c$ is a real number, then $\lim_{n \to \infty} ca_n = c \lim_{n \to \infty} a_n.$

Limit of a Product is the Product of the Limits: If $a_n$ and $b_n$ are convergent sequences, then $\lim_{n \to \infty}{(a_n \cdot b_n)} = \lim_{n \to \infty}{a_n} \cdot \lim_{n \to \infty}{b_n}$ .

Limit of a Quotient is the Quotient of the Limits: If $a_n$ and $b_n$ are convergent sequences, and $\lim_{n \to \infty}{(b_n)} \neq 0$ , then $\lim_{n \to \infty}{(\frac{a_n}{b_n})} = \frac{\lim_{n \to \infty}{a_n}}{\lim_{n \to \infty}{b_n}}$ .

Squeeze Theorem: If $a_n$ and $c_n$ are convergent sequences with the same liimt $L$ , and $b_n$ is a sequence for which $a_n \leq b_n \leq c_n$ for all $n$ , then $\lim_{n \to \infty}{b_n} = L$ .

Bounded Sequences

A sequence is said to be a bounded sequence if there exists a real number $M$ such that $\abs{a_n} \leq M$ for all natural numbers $n$ . In other words, $a_n$ is never further from $0$ than $\pm M$ . If $a_n$ is less than or equal to some real number for all $n$ , then $a_n$ is said to be bounded above, and if $a_n$ is greater than or equal to some real number for all $n$ , it is said to be bounded below.

Divergent Sequences

In general, any sequence that does not converge is said to diverge and is a divergent sequence.

The real sequence $a_n$ is said to diverge to infinity if for every real number $M$ , there exists a natural number $N$ such that $a_n > M$ whenever $n \geq N.$ In this case, we write $\lim_{n->\infty}{a_n} = \infty$ . This means that for any given real number, not matter how large, we can find an $n$ that makes $a_n > M$ .

The real sequence $a_n$ is said to diverge to minus infinity if for every real number $M$ , there exists a natural number $N$ such that $a_n < M$ whenever $n \geq N.$ In this case, we write $\lim_{n->\infty}{a_n} = -\infty$ . This means that for any given real number, not matter how small, we can find an $n$ that makes $a_n < M$ .

A sequence can also diverge but not to infinity or minus infinity. For example, $a_n = (-1)^n$ does not converge, so it is divergent, but it does not diverge to $\pm \infty$ .

Monotone Sequences

The real sequence $a_n$ is said to be increasing if $a_n \leq a_{n+1}$ for all natural $n$ , and is said to be decreasing if $a_n \geq a_{n+1}$ for all natural $n$ .

A real sequence is $a_n$ is said to be monotone if it is either increasing or decreasining.

The monotone convergence theorem says that if a sequence is monotone and bounded, then it converges. Proof: Suppose that the sequence $a_n$ is bounded and increasing. Then, by the Completeness axiom, the set $\set{a_n | n \in \naturals}$ has a least upper bound; let's call it $L .$ Then, for any positive $\epsilon$ , $L - \epsilon$ can't be an upper bound of the set, since $L - \epsilon < L$ . Therefore, for some $N \in \naturals$ , $L - \epsilon < a_N$ , and since $a_n$ increases monotonically, for all $n \geq N$ , we have $L - \epsilon < a_n < L + \epsilon$ , which implies $\abs{a_n - L} < \epsilon,$ that is, that $a_n$ converges to $L$ . $\square$

Subsequences

The sequence $a_{f(n)}$ is called a subsequence of the sequence $a_n$ if $f : \naturals \to \naturals$ is a strictly increasing function. Another way to put this is that if $a_n$ is a sequence, if we construct $a'_n$ by removing $0$ or more elements from $a_n$ and leaving the remaining elements in the same order, then $a'_n$ is a subsequence of $a_n.$

The Limit of a Subsequence Theorem says that if $a_n$ is a convergent sequence with limit $L$ , then any subsequence $a_{f(n)}$ of $a_n$ will also converge to $L .$ Proof: Suppose $a_n$ converges to $L .$ Let $\epsilon > 0.$ Then for some $N \in \naturals$ , for $n \geq N,$ $\abs{a_n - L} < \epsilon$ . Since $f(n)$ is strictly increasing, $f(n) \geq n$ for all $n,$ so if $n \geq N,$ then $f(n) \geq n \geq N,$ and $\abs{a_{f(n)} - L} < \epsilon.$ $\square$

This theorem also implies that a sequence is not convergent if it has a subsequence that does not converge, or if two subsequences converge to different limits, for example, $a_n = (-1)^n,$ where $a_{2n}$ converges to $1$ but $a_{2n+1}$ converges to $-1$ .

We can also use this to find some limits. For example,

$\lim_{n \to \infty} \parens{1 + \frac{1}{n+3}}^n$

can be factored into

$\lim_{n \to \infty} \parens{1 + \frac{1}{n+3}}^{n+3} \cdot \parens{1 + \frac{1}{n+3}}^{-3}$

The left factor is a subsequence of $\parens{1 + \frac{1}{n}}^n$ which goes to $e,$ and the right factor goes to $1,$ so the limit of the whole expression is $e .$

Bolzano-Weirstrass Theorem

The Bolzano-Weirstrass Theorem states that if a sequence of real numbers is bounded, then it has a convergent subsequence.

Cauchy Sequences

A sequence is said to be Cauchy if for every $\epsilon > 0$ , there exists a natural $N$ for which $\abs{a_m - a_n} < \epsilon$ whenever $m, n > N.$ In other words, as we go further in the sequence, the terms get arbitrarily closer together.

A sequence is Cauchy if and only if it is convergent.

Lim Sup and Lim Inf

Suppose the real sequence $a_n$ is bounded. Then

$\limsup_{n \to \infty} a_n = \lim_{n \to \infty} \sup\set{a_n, a_{n+1}, \dots}$

$\liminf_{n \to \infty} a_n = \lim_{n \to \infty} \inf\set{a_n, a_{n+1}, \dots}$

We can think of $\limsup_{n \to \infty}{a_n}$ as giving the smallest upper bound that the terms of $a_n$ get arbitrarily close to, but do not exceed, as $n$ increases indefinitely, and $\liminf_{n \to \infty}{a_n}$ as giving the greatest lower bound that the terms of $a_n$ get close to, but do not exceed, as $n$ increases indefinitely.

If we define

$b_n = \sup{\set{a_n, a_{n+1}, \dots}},$

then we know $b_n$ is bounded because $a_n$ is bounded. Now, $b_1 = \sup{\set{a_1, a_2, \dots}}$ is the least upper bound of all the terms in $a_n$ and $b_2 = \sup{\set{a_2, a_3, \dots}}$ is the least upper bound of all the terms in $a_n$ except $a_1$ and $a_2$ ; therefore $b_2$ must be less than or equal to $b_1$ since removing terms can only lower the least upper bound. The patterns holds for all $b_n$ , so $b_n$ is a decreasing sequence. Since it is decreasing and bounded, the Monotone Convergence Theorem tells us that $b_n$ converges. We can make a similar argument about

$c_n = \inf{\set{a_n, a_{n+1}, \dots}},$

which is bounded and increasing and also converges.

If a sequence $a_n$ is convergent, then $\limsup_{n \to \infty}{a_n} = \liminf_{n \to \infty}{a_n}.$

If a sequence $a_n$ is bounded, then $\liminf_{n \to \infty}{a_n} \leq \limsup_{n \to \infty}{a_n}.$

If a sequence $a_n$ is bounded above, then $\limsup_{n \to \infty}{a_n} = \lim_{n \to \infty}\sup{\set{a_n, a_{n+1}, \dots}}.$

If a sequence $a_n$ is not bounded above, then $\limsup_{n \to \infty}{a_n} = \infty.$

If a sequence $a_n$ is bounded below then $\liminf_{n \to \infty}{a_n} = \lim_{n \to \infty}\inf{\set{a_n, a_{n+1}, \dots}}.$

If a sequence $a_n$ is not bounded below, then $\liminf_{n \to \infty}{a_n} = -\infty.$

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