Properties of Convergent Sequences - Comparison Laws

Properties of Convergent Sequences - Comparison Laws

We will now look at some more theorems regarding comparisons, this time with emphasis on the comparison between various sequences. If you haven't already, be sure to check out the following theorems and proofs regarding sequences:

Theorem 1: Let $(a_n)$ be a convergent sequence, that is $\lim_{n \to \infty} a_n = A$, and suppose that $a_n ≥ 0$ for all $n \in \mathbb{N}$. Then $A = \lim_{n \to \infty} ≥ 0$.
  • Proof: We will show this by proof by contradiction. Let $(a_n)$ be a convergent sequence such that $\lim_{n \to \infty} a_n = A$ and suppose that $A < 0$. The $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - A \mid < \epsilon$. This implies that:
(1)
\begin{align} A - \epsilon < a_n < A + \epsilon \end{align}
  • Now we note that since $A < 0$, then $-A > 0$, and so suppose we consider $-A = \epsilon_0 > 0$. Then for some $K \in \mathbb{N}$, it follows that if $n ≥ K$ then $\mid a_n - A \mid < \epsilon_0$ or equivalently:
(2)
\begin{align} \quad A - \epsilon_0 < a_K < A + \epsilon_0 = A + (-A) = 0 \\ a_K < 0 \end{align}
  • But this contradicts the fact that $a_n ≥ 0$ for all $n \in \mathbb{N}$. Therefore our assumption that $A < 0$ was false, and so $A ≥ 0$. $\blacksquare$

A similar proof can be constructed to show that if $(a_n)$ is a convergent sequence such that $a_n ≤ 0$ for all $n \in \mathbb{N}$ then $\lim_{n \to \infty} a_n = A ≤ 0$.

Theorem 2: Let $(a_n)$ and $(b_n)$ be convergent sequences such that $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$. If $a_n ≤ b_n$ for all $n \in \mathbb{N}$ then $A ≤ B$.
  • Proof: Let $(a_n)$ and $(b_n)$ be convergent sequences such that $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$, and suppose that $a_n ≤ b_n$ for all $n \in \mathbb{N}$.
  • Now consider the sequence $c_n = (b_n - a_n)$ whose limit $\lim_{n \to \infty} c_n = B - A$. Since $a_n ≤ b_n$ for all $n \in \mathbb{N}$ it follows that $c_n ≥ 0$ for all $n \in \mathbb{N}$, and so by theorem 1 above, $0 ≤ B - A$ or equivalently, $A ≤ B$. $\blacksquare$
Theorem 3 (Squeeze Theorem): Let $(a_n)$, $(b_n)$, and $(c_n)$ be convergent sequences such that $\lim_{n \to \infty} a_n = L$, $\lim_{n \to \infty} b_n = B$, and $\lim_{n \to \infty} c_n = L$. If $a_n ≤ b_n ≤ c_n$ for all $n \in \mathbb{N}$ then $B = L$.
  • Proof: Let $(a_n)$, $(b_n)$ and $(c_n)$ be convergent sequences such that $\lim_{n \to \infty} a_n = L$ and $\lim_{n \to \infty} b_n = B$, and $\lim_{n \to \infty} c_n = L$, and let $a_n ≤ b_n ≤ c_n$ for all $n \in \mathbb{N}$.
  • Now since $(a_n)$ and $(c_n)$ are convergent, it follows that for some $N \in \mathbb{N}$ that if $n ≥ N$ then both $\lim_{n \to \infty} a_n = L$ and $\lim_{n \to \infty} c_n = L$, or in other words, $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - L \mid < \epsilon$ and $\mid c_n - L \mid < \epsilon$, which is equivalent to $L -\epsilon < a_n < L + \epsilon$ and $L - \epsilon < c_n < L + \epsilon$. Now since $a_n ≤ b_n ≤ c_n$ for all $n \in \mathbb{N}$ it follows that:
(3)
\begin{align} L - \epsilon < a_n ≤ b_n ≤ c_n < L + \epsilon \\ -\epsilon < a_n - L ≤ b_n - L ≤ c_n - L < \epsilon \end{align}
  • Therefore $- \epsilon < b_n - L < \epsilon$ and so $\mid b_n - L \mid < \epsilon$, or in other words, $\lim_{n \to \infty} b_n = B = L$. $\blacksquare$
Theorem 4: Let $(a_n)$ be a convergent sequence, that is $\lim_{n \to \infty} a_n = A$. The sequence $(\mid a_n \mid)$ then also converges, and $\lim_{n \to \infty} \mid a_n \mid = \mid A \mid$.
  • Proof: Let $(a_n)$ be a convergent sequence so that $\lim_{n \to \infty} a_n = A$. We want to show that $\lim_{n \to \infty} \mid a_n \mid = \mid A \mid$, that is $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid \mid a_n \mid - \mid A \mid \mid < \epsilon$.
(4)
\begin{align} \quad \mid \mid a_n \mid - \mid A \mid \mid ≤ \mid a_n - A \mid < \epsilon \end{align}
  • Therefore as an immediate consequence, $\lim_{n \to \infty} \mid a_n \mid = \mid A \mid$. $\blacksquare$
Theorem 5: Let $(a_n)$ be a convergent sequence such that $a_n ≥ 0$ for all $n \in \mathbb{N}$. Then the sequence $(\sqrt{a_n})$ also converges, and $\lim_{n \to \infty} \sqrt{a_n} = \sqrt{A}$.
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