We are primarily interested in infinite sequences, and their behaviour as their terms approach infinity. The sequence
(1,2,3,4,\dots)
appears to grow without bound, but the sequence
(1,1.4,1.41,1.414,\dots)
appears to be approaching some actual number. We formalize this through convergence.
A sequence (s_n) of real numbers is said to converge to a real number s if, for every \varepsilon > 0, there exists some natural number N such that n > N implies |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 (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 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?
Claim: Given the sequence s_n = 1/n, we claim
\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.
Properties
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.
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 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.
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}.
A sequence (s_n) is bounded if there exists M \geq 0 such that |s_n| \leq M.
Every convergent sequence is bounded.
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.
Let (s_n) = (-1)^n. This sequence is clearly bounded by M = 1, but not convergent.
If a sequence converges, its limit is unique.
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
Prove if |x| < 1, then \lim_{n\to\infty} x^n = 0 using Bernoulii’s inequality,
(1 + y)^n \geq 1 + ny.
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.
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.
Every convergent sequence is bounded.
If a sequence converges, its limit is unique.
If |x| < 1, then \lim_{n\to\infty} x^n = 0.