Theorem – Sequence Limit Theorems

Contents
1. Sum of limits
2. Proof

This page is longer than most theorem pages as the basic limit theorems for sequences are grouped here to aid speed of reference. The definition of a limit will not be stated here, but please find the definition given on this page.

Theorem 1 – $\lim_{n \rightarrow \infty} (s_n + t_n) = \lim_{n \rightarrow \infty} s_n + \lim_{n \rightarrow \infty} t_n$

Let $xleq epsilon + y$ be true for all $x,y,epsilon in mathbb{R}$ and $epsilon > 0$ . Then, $x leq y$ .

Proof

We will prove the theorem by contradiction. The statement is that

$forall epsilon >0, ~x leq epsilon + y implies x leq y$ .

Thus, the negation of this statement is that

$forall epsilon >0, ~x leq epsilon + y implies x > y$ .

However, the consequent of the implication can be re-written as $x-y>0$ . But the hypothesis of the implication gives $forall epsilon>0, x-y leq epsilon$ . Restating these equations clearly, we have

$0<x-y$ , and therefore $0<kx-ky$ where $k>0$ .
$x-y leq epsilon$

Adding these two equations together gives

$x-y leq epsilon +kx -ky Leftrightarrow (k+1)x-(k+1)y = (k+1)(x-y)leq epsilon$ . Therefore, since $epsilon > 0$ , and $(k+1)>0$ , it follows that $x-yleq 0 implies x leq y$ . Thus we have the desired contradiction, and so it must be the case that the theorem holds. $square$

Theorems

Theorem 2 – If $s_n < t_n$ for all $n$ , then, if both limits exist,

$lim_{n\rightarrow\infty}~s_n \leq lim_{n\rightarrow\infty}~t_n$

Proof – First, let $lim_{n\rightarrow\infty}~s_n = s$ and $lim_{n\rightarrow\infty}~t_n = t$ . Since both limits exist by hypothesis, we have from the definition of the limit of a sequence, for all $n>max(N_s, N_t)$ ,

$|s_n - s| < \epsilon_s$ and $|t_n - s| < \epsilon_t$

which is equivalent to

$(-\epsilon_s +s < s_n < \epsilon_s + s)$ and $(-\epsilon_t +t < t_n < \epsilon_t + t)$ .

By assumption however, we know that $s_n < t_n$ , so it follows that

$-\epsilon_s + s < \epsilon_t + t \implies s < (\epsilon_s+\epsilon_t )+ t \implies s < t$ .

We are not done, however. We first assumed that $n>max(N_s, N_t)$ and then used this to prove that the quantity $\epsilon_s+\epsilon_t$ may become infinitely small. This subtlety is easily confirmed to be of no concern, as for a given pair of $\epsilon_s+\epsilon_t$ we would first have to work out both $N_s$ and $N_t$ , but this is possible by from the definition of a limit, and so we have taken no circular steps in the proof. This concludes the proof of the theorem.