Convergence

Definition

A sequence S is a function whose domain is the set of natural numbers, \mathbb{N}.

A sequence S is usually notated (s_n) or as a list of elements

(s_1,s_2,s_3,\ldots)

Example

The function S \colon \mathbb{N} \to \mathbb{Q} such that n \rightarrowtail \frac{1}{n} is a sequence. We use the notation (1/n) as an abbreviation for

\left(1,\frac{1}{2},\frac{1}{3},\ldots\right)

Similarly, the sequence S \colon \mathbb{N} \to \mathbb{N} where n \rightarrowtail 2n is denoted (2n), abbreviating

(2,4,6,\ldots)

Normally we will consider sequences where the codomain is \mathbb{R}.

Definition and properties

Definition

A sequence (s_n) of real numbers converges to a real number s if

\forall \varepsilon > 0, \exists N \in \mathbb{N} \;s.t.\; \forall n \geq N, |s_n - s| < \varepsilon

Where |s_n - s| is the distance between the points s_n and s. Essentially, a sequence is convergent if the elements of (s_n) get arbitrarily close to a point s as the sequence progresses.

If a sequence does not converge, than we say it is divergent.

Definition

If (s_n) converges to s, then we say that s is the limit of (s_n) and we write

\lim_{n \to \infty} s_n = s

We use (s_n) to denote the sequence as a whole, where as s_n refers to an indivual element of the sequence. We also typically abbreviate \lim_{n\to\infty}s_n = s as \lim s_n = s

Suppose we have a sequence (s_n) which appears to converge to a number s. How would we go about proving the sequence fufills the definition of convergence?

Example

Claim: Given the sequence s_n = 1/n, we have

\lim_{n \to \infty} \frac{1}{n} = 0.

In order to show this sequence converges, we need to prove that for all \varepsilon > 0 there exists some N \in \mathbb{N} such that for all n \geq N, we have

|s_n - s| = \left|\frac{1}{n} - 0\right| = \left|\frac{1}{n}\right| = \frac{1}{n} < \varepsilon.

By the Archimedean property, we know there always exists some n \in \mathbb{N} such that xn > y for any positive x,y \in \mathbb{R}. It is then clear there exists some n such that 1/n < y/x.

Therefore for any \varepsilon > 0, we simply choose N such that 0 < \frac{1}{N} < \varepsilon. Then for any n \geq N, we have |1/n - 0| = 1/n \leq 1/N < \varepsilon. Thus our sequence converges, and \lim s_n = 0.

See Practice for more examples.

Theorem 1.1.1

Let (s_n) and (a_n) be sequences of real numbers and let s \in \mathbb{R}. If for some k > 0 and some m \in \mathbb{N} we have

|s_n - s| \leq k|a_n|, \forall n \geq m

and if \lim a_n = 0, then it follows that \lim s_n = s.

Proof

Given any \varepsilon > 0, since \lim a_n = 0 there exists N_1 \in \mathbb{N} such that n \geq N_1 implies |a_n| < \varepsilon/k. Let N = \max\{m,N_1\}. Then for n \geq \mathbb{N} we have n \geq m and n \geq N_1, so

|s_n - s| \leq k|a_n| < k\left(\frac{\varepsilon}{k}\right) = \varepsilon.

Thus \lim s_n = s.

This Theorem can be applied quite liberally to help with a great variety of limit problems.

Example

Claim: The \lim(4n^2 - 3)/(5^2 -2n) = 4/5.

We start with some algebraic manipulation

\left|\frac{4n^2 - 3}{5n^2 - 2n} - \frac{4}{5}\right| = \left| \frac{8n-15}{5(5n^2-2n)}\right|

We aim to find an upper bound to simplify the fraction. For the numerator, |8n - 15| < 8n for all n \in \mathbb{N} (Note 0 \not \in \mathbb{N}). For the denominator, we want some relation in the form 5n^2 - 2n \geq kn^2 for some k > 0. We use k = 4 to obtain

5n^2 - 2n \geq 4n^2 \implies n^2 \geq 2n \implies n \geq 2.

When n \geq 2, we now have the relation

\left| \frac{8n-15}{5(5n^2-2n)}\right| < \frac{8n}{5(4n^2)} = \frac{2}{5}\left(\frac{1}{n}\right).

We can turn this into a formal proof. If n \geq 2, then n^2 \geq 2n and 5n^2 -2n\geq 4n^2 such that

\left|\frac{4n^2 - 3}{5n^2 - 2n} - \frac{4}{5}\right| = \left| \frac{8n-15}{5(5n^2-2n)}\right| < \frac{8n}{5(4n^2)} = \frac{2}{5}\left(\frac{1}{n}\right)

Since \lim (1/n) = 0 as shown in our last example, Theorem 1.1.1 states that

\lim \frac{4n^2-3}{5n^2 - 2n} = \frac{4}{5}.

Definition

A sequence (s_n) is bounded if there exists M \geq 0 such that |s_n| \leq M.

Theorem 1.1.2

Every convergent sequence is bounded.

Proof

Let (s_n) be a convergent sequence and let \lim s_n = s. Choose \varepsilon = 1. Because (s_n) is convergent, there exists some N \in \mathbb{N} such that for all n \geq N, |s_n - s| < 1.

Through the triangle inequality, we have that, for all n \geq N

|s_n| - |s| \leq |s_n - s| < 1 \iff |s_n| < |s| + 1.

Let M = \max\{|s_1|,|s_2|,\ldots,|s_N|,|s| + 1\}. Thus for all n \in \mathbb{N}, |s_n| \leq M and (s_n) is bounded.

The contrapositive is of course also true, any unbounded sequence is divergent. However, not all divergent sequences are unbounded.

Example

Let (s_n) = (-1)^n. This sequence is clearly bounded by M = 1, but not convergent.

Theorem 1.1.3

If a sequence converges, its limit is unique.

Proof

Let (s_n) be a convergent sequence and suppose \lim s_n = s and \lim s_n = t. We want to show that s and t are arbitrarily close and therefore s = t.

By the definition of convergence we have

\exists N_1 \in \mathbb{N} \text{ such that }\forall n \geq N_1, |s_n - s| < \frac{\varepsilon}{2}.

and

\exists N_2 \in \mathbb{N} \text{ such that }\forall n \geq N_2, |s_n - t| < \frac{\varepsilon}{2}.

Let N = \max\{N_1,N_2\}. From the triangle inequality we have

\begin{align*} |s-t| &= |s - s_n + s_n - t|\\ &\leq |s-s_n|+|s_n-t|\\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \end{align*}

Since this holds for all \varepsilon > 0, we have s = t.

Practice

Problem \star

Prove if |x| < 1, then \lim_{n\to\infty} x^n = 0 using Bernoulii’s inequality,

(1 + y)^n \geq 1 + ny.

Solution

If x = 0, the proof is trivial, so we need to prove the case where 0 < |x| < 1. Bernoulii’s inequality states

(1 + y)^n \geq 1 + ny.

If we let x = 1/(1+y) for some y > 0, we get

(1 + \frac{1}{x} - 1)^n \geq 1 + n\left(\frac{1}{x} - 1\right) \implies \frac{1}{x^n} \geq 1 + \frac{n}{x} - n

Flipping both sides of the inequality gets us

x^n \leq \frac{1}{1 + \frac{n}{x} - n}.

Taking the right hand side of the equation, and using our substitution for y, we obtain

\frac{1}{1 + \frac{n}{x} + n} = \frac{1}{1 + yn + 2n}.

Let 1 + yn + 2n = k For any \varepsilon > 0, there exists N \in \mathbb{N} such that for all n \geq N,

\left|\frac{1}{1 + yn + 2n} - 0\right| = \left|\frac{1}{k} - 0 \right| \leq \frac{1}{k} \leq \varepsilon

We know by the Archimedian property there exists some n giving us a suffienct value of k. Thus \lim 1/(1 + n/x - n) = 0. We have

|x^n - 0| = x^n \leq \frac{1}{1 + \frac{n}{x} - n}

Which by Theorem 1.1.1, gives us \lim x^n = 0.

Recap

In this section we proved the following theorems and results.

Theorem 1.1.1

Let (s_n) and (a_n) be sequences of real numbers and let s \in \mathbb{R}. If for some k > 0 and some m \in \mathbb{N} we have

|s_n - s| \leq k|a_n|, \forall n \geq m

and if a_n = 0, then it follows that \lim s_n = s.

Theorem 1.1.2

Every convergent sequence is bounded.

Theorem 1.1.3

If a sequence converges, its limit is unique.

Result

If |x| < 1, then \lim_{n\to\infty} x^n = 0.