Basic Analysis: Sequence Convergence (1)

Much of analysis deals with the study of R, the set of real numbers. It provides a rigourous foundation of concepts which we usually take for granted, e.g. continuity, differentiation, sequence convergence etc. One should have a mental picture of the set of rational numbers Q having “gaps” in its order structure, while R “fills up” these gaps. We shall now describe this in a more rigourous manner.

First some basic definitions. Suppose X = R or Q and S is a non-empty subset of X.

Definitions.

An upper bound of S is an element $x\in X$ such that x ≥ y for all $y\in S$ .If S has an upper bound, we say it is upper bounded.

If $x \in S$ is an upper bound of S, we call it the maximum of S, denoted max(S).

A lower bound of S is an element $x\in X$ such that x ≤ y for all $y\in S$ . If S has a lower bound, we say it is lower bounded.

If $x\in S$ is a lower bound of S, we call it the minimum of S, denoted min(S).

If S is both upper- and lower-bounded, we just say it is bounded.

Let S be upper-bounded, and T the set of upper bounds of S. The minimum of T is called the supremum of S, denoted sup(S).

Let S be lower-bounded, and T the set of lower bounds of S. The maximum of T is called the infimum of S, denoted inf(S).

Some basic properties include:

The maximum (resp. minimum) is unique if it exists.
Hence, the infimum (resp. supremum) is unique if it exists.
If max(S) exists, then so does sup(S), and max(S) = sup(S).
If min(S) exists, then so does inf(S), and min(S) = inf(S).
A set may be upper-bounded without possessing a maximum, e.g. the set of real (or rational) numbers 0 < x < 1. Likewise, it may be upper-bounded without possessing a minimum.

A useful picture to visualise all these definitions is:

Example. The open interval (0, 1) is bounded but has no maximum or minimum. On the other hand, its infimum is 0 and supremum is 1. The closed interval [0, 1] has maximum (and hence supremum) 1.

Now we shall state the property which distinguishes R from Q.

Completeness Property. Every upper-bounded subset S of R has a supremum. We say that R is complete.

Replacing S with –S, we see that every lower-bounded subset S of R has an infimum. We’re going to take this property for granted and proceed from here. By the way, to see that Q is not complete, consider the set of rational numbers r such that r² < 2. Clearly this is a bounded subset of Q, but it doesn’t have a supremum in Q (and since R is complete, this set has a supremum in R, namely √2).

We end this section with a quick observation. By definition L = sup(S) iff L is an upper bound of S and there is no smaller upper bound L-ε (for ε>0). Since L-ε is not an upper bound, there exists x in S, x > L-ε.

Thus, L = sup(S) iff (i) L is an upper bound, (ii) for each ε>0, there exists x in S, x > L-ε.

Definition of Convergence

A sequence in R is given by (a₁, a₂, a₃, …), where each a_i is in R. One can think of it as a function N → R, where N is the set of positive integers. Our focus here is to provide a rigourous foundation for the statement “sequence (a_n) → L as n → ∞”.

Definition (Convergence). Let (a_n) be a sequence. We say that (a_n) converges to a real number L (written a_n → L) if:

for every positive real ε, there exists N, such that whenever n>N, we have |a_n – L| < ε.

We say a sequence is convergent if it converges to some real L.

The definition takes some getting used to, so first we’ll use this to prove that the limit, if exists, must be unique.

Example 0. If (a_n) → K and (a_n) → L, then K=L.

Proof. If K ≠ L, take ε=|K–L|/2. By definition,

we can find N such that when n > N, we have |a_n–K| < ε;
we can find M such that when n > M, we have |a_n–L| < ε.

Then we can find n > max(M, N) such that

$|a_n - K| < \epsilon, \ |a_n-L|< \epsilon \implies |K-L| \le |K-a_n|+|a_n-L| < 2\epsilon = |K-L|,$

which is clearly absurd. Thus K=L. ♦

Example 1. Prove that if a_n= 1/n, then (a_n) → 0.

Proof. For each ε>0, we need to establish an N such that when n > N, we have |a_n – 0| < ε. But |a_n| = 1/n, so this is easy: just set N = 1/ε. Now:

$n > N\implies |a_n - 0| = |a_n| = 1/n < 1/N = \epsilon$

so by definition the sequence converges to 0. ♦

Example 2. Prove that if a_n= 3n/(n+2), then (a_n) → 3.

Proof. Let’s first write $|a_n - 3| = |\frac{3n}{n+2}-3| = \frac 6{n+2}$ . Now, for any ε>0, we set N = 6/ε. Then:

$n>N \implies |a_n - 3| = \frac 6{n+2} <\frac 6 n <\frac 6 N = \epsilon$ ,

which shows that the sequence converges to 3. ♦

Example 3. Prove that the sequence a_n= (-1)ⁿ is not convergent.

Proof. We need to take the negation of the definition, i.e. what does it mean to say that a_ndoes not converge to L? There’s a nice trick to take the negation of such statements:

“It is not true that for every x, P(x) holds.” → “There exists x, P(x) doesn’t hold.”
“It is not true that there exists x, P(x) holds.” → “For any x, P(x) doesn’t hold.”

Applying this rule-of-thumb recursively, we see that:

Tip. The sequence a_ndoes not converge to L if and only if:

there exists an ε, such that for any N, we can find n>N satisfying |a_n-L| ≥ ε. [ Or equivalently, there exists an ε and infinitely many n such that |a_n-L| ≥ ε. ]

For our problem, suppose (a_n) → L. Take ε = 1. If L ≥ 0, then for any odd n, we have |a_n–L| = |-1-L| ≥ 1. And we’re done since there’re infinitely many odd n. On the other hand, if L < 0, then pick the even n‘s: we have infinitely many n for which |a_n–L| = |1-L| ≥ 1. ♦

Arithmetic Properties of Limits

For more complicated expressions of a_n, it becomes a pain to consistently use the ε-N definition, so we need further properties to help us.

Theorem. Let $(a_n), (b_n)$ be sequences converging to K, L respectively. Then:

$(a_n + b_n)$ converges to K+L;

a convergent sequence is bounded;

$(a_n b_n)$ converges to KL;

if L ≠ 0, then $(1/b_n)$ converges to 1/L.

Proof. The proofs are conceptually easy but a bit tedious.

1. For addition: given ε>0, since ε/2>0, we can

find M such that when n > M, |a_n – K| < ε/2;
find M’ such that when n > M’, |b_n – L| < ε/2.

Hence if we let N = max{M, M’}, then whenever n > N, we have:

$|(a_n + b_n) -(K+L)| \le |a_n-K| + |b_n-L| < \epsilon/2 + \epsilon/2 = \epsilon.$

2. Let’s show $(a_n)$ is bounded. Since $(a_n)\to K$ , we pick ε=1, and by definition of convergence, there exists N such that when n > N, we have

$|a_n - K| < 1 \implies |a_n| \le |a_n-K| + |K| < |K|+1.$

Now there are only finitely many a_n for n≤N, so we can let B be the maximum of these |a_n|. Then the set of all |a_n| is bounded by max{B, |K|}.

3. To show $(a_n b_n) \to KL$ , just use:

$|a_n b_n - KL| = |a_n(b_n - L) + L(a_n - K)| \le |a_n|\cdot|b_n - L| + |L|\cdot|a_n - K|.$

Now use the fact that |a_n| is bounded, say by B. Then: $|a_n b_n - KL| \le B|b_n - L| + |L|\cdot|a_n - K|$ . Given any ε>0, pick M, M’ such that:

when n>M, we have |a_n–K| < ε/(2|L|);
when n>M’, we have |b_n–L| < ε/(2B).

Thus, letting N = max{M, M’}, we see that whenever n>N, $|a_n b_n - KL| \le \epsilon/2 + \epsilon/2 = \epsilon$ .

4. Finally, to show that $(1/b_n) \to 1/L$ , write:

$\left|\frac 1 {b_n}-\frac 1 L\right| = \frac{|L-b_n|}{|L|\cdot|b_n|}$ .

Now pick M’ such that whenever n>M’, we have $|b_n - L|<|L|/2$ and so $|b_n| \ge |L|-|L-b_n|> |L|/2$ .

Suppose we’re given ε>0. Since $(b_n)\to L$ , there exists M such that whenever n>M, we get $|b_n - L| < \epsilon|L|^2/2$ . Now letting N = max{M, M’}, whenever n>N, we get:

$\left|\frac 1 {b_n} - \frac 1 L\right| < \epsilon|L|^2/2\cdot(|L| \cdot |L|/2)^{-1}=\epsilon.$ ♦

Having proven these basics, some consequences are immediate.

Corollary. Let $(a_n), (b_n)$ be sequences converging to K, L respectively. Then:

for any real c, $(c\cdot a_n)$ converges to cK;

$(a_n - b_n)$ converges to K-L;

if L ≠ 0, then $a_n/b_n$ converges to K/L.

Proof.

For the first statement note that the constant sequence (c, c, … ) converges to c. Hence, the result follows by multiplying (a_n) and (c). For the second statement, write a_n–b_n = a_n + (-1)b_n and apply the first statement. For the third, write a_n/b_n = a_n·(1/b_n). ♦

Now we can prove Example 2 using Example 1. Indeed, 3n/(n+2) = 3/(1+2/n). Since 1/n → 0 by Example 1, we have 2/n → 0. Thus 1+(2/n) → 1 and 3/(1+2/n) → 3, which is much easier.

Example 4. Prove that a_n= (n² + 2n – 1)/(2n² – n – 2) converges. Find its limit.

Answer. Divide the numerator and denominator by n²to obtain $a_n = \frac{1 + 2/n - 1/n^2}{2 - 1/n - 2/n^2}$ . Use the fact that 1/n² → 0 and we get a_n→ 1/2. ♦

Example 5. Find the limit of (2012ⁿ + 2011ⁿ)/(2012ⁿ – 2011ⁿ) as n → ∞.

Answer. Divide the numerator and denominator by 2012ⁿ to obtain $a_n = \frac{1+\alpha^n}{1-\alpha^n}$ where α = 2011/2012. Since $\alpha^n \to 0$ , a_n→ 1. ♦

The final example would have been hellish to prove with the ε-N definition.