Properly Divergent Sequences

Properly Divergent Sequences

Recall that a sequence $(a_n)$ of real numbers is said to be convergent to the real number $A$ if $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - A \mid < \epsilon$.

If we negate this statement we have that a sequence $(a_n)$ of real numbers is divergent if $\forall A \in \mathbb{R}$ then $\exists \epsilon_0 > 0$ such that $\forall N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - A \mid ≥ \epsilon_0$. However, there are different types of divergent sequences. For example, a sequence can alternate between different points and be divergent such as the sequence $((-1)^n)$, or instead, the sequence can tend to infinity such as $(n)$ or negative infinity such as $(-n)$, or neither, such as $((-1)^n(n))$. We will now define properly divergent sequences.

Definition: A sequence of real numbers $(a_n)$ is said to be properly divergent to $\infty$ if $\lim_{n \to \infty} a_n = \infty$, that is $\forall M \in \mathbb{R}$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $a_n > M$. Similarly, $(a_n)$ is said to be properly divergent to $-\infty$ if $\lim_{n \to \infty} a_n = -\infty$, that is $\forall M \in \mathbb{R}$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $a_n < M$.

Now let's look at some theorems regarding properly divergent sequences.

Theorem 1: An increasing sequence of real numbers $(a_n)$ is properly divergent to $\infty$ if it is unbounded. A decreasing sequence of real numbers $(a_n)$ is properly divergent to $-\infty$ if it is unbounded.
  • Proof: Suppose that $(a_n)$ is a sequence of real numbers that is increasing. Since $(a_n)$ is unbounded, then for any $M \in \mathbb{R}$ there exists a term $a_M$ (dependent on $M$) such that $M < a_M$. Since $(a_n)$ is an increasing sequence, then for $n ≥ M$ we have that $M < a_n$ and since $M$ is arbitrary we have that $\lim_{n \to \infty} a_n = \infty$.
  • Similarly suppose that $(a_n)$ is a sequence of real numbers that is decreasing. Since $(a_n)$ is unbounded, then for any $M \in \mathbb{R}$ there exists a term $a_M$ (dependent on $M$ such that $M > a_M$. Since $(a_n)$ is a decreasing sequence, then for $n ≥ M$ we have that $M > a_n$ and since $M$ is arbitrary we have that $\lim_{n \to \infty} a_n = -\infty$. $\blacksquare$
Theorem 2: Let $(a_n)$ and $(b_n)$ be sequences of real numbers such that $a_n ≤ b_n$ for all $n \in \mathbb{N}$. Then if $\lim_{n \to \infty} a_n = \infty$ then $\lim_{n \to \infty} b_n = \infty$.
  • Proof: Let $(a_n)$ and $(b_n)$ be sequences of real numbers such that $a_n ≤ b_n$ for all $n \in \mathbb{N}$, and let $\lim_{n \to \infty} a_n = \infty$. Then it follows that for all $M \in \mathbb{R}$ that there exists an $N$ (dependent on $M$ such that if $n ≥ N$ then $a_n ≥ M$. But we have that $b_n ≥ a_n$ for all $n \in \mathbb{N}$ and so for $n ≥ N$ we have that $b_n ≥ M$. Since $M$ is arbitrary it follows that $\lim_{n \to \infty} b_n = \infty$. $\blacksquare$
Theorem 3: Let $(a_n)$ and $(b_n)$ be sequences of real numbers such that $a_n ≤ b_n$ for all $n \in \mathbb{N}$. Then if $\lim_{n \to \infty} b_n = -\infty$ then $\lim_{n \to \infty} a_n = \infty$.
  • Proof: Let $(a_n)$ and $(b_n)$ be sequences of real numbers such that $a_n ≤ b_n$ for all $n \in \mathbb{N}$, and let $\lim_{n \to \infty} b_n = -\infty$. Then it follows that for all $M \in \mathbb{R}$ that there exists an $N$ (dependent on $M$ such that if $n ≥ N$ then $b_n ≤ M$. But we have that $a_n ≤ b_n$ for all $n \in \mathbb{N}$ and so for $n ≥ N$ we have that $a_n ≤ M$. Since $M$ is arbitrary it follows that $\lim_{n \to \infty} a_n = -\infty$. $\blacksquare$
Theorem 4: If $(a_n)$ and $(b_n)$ are sequences of positive real numbers suppose that for some real number $L > 0$ that $\lim_{n \to \infty} \frac{a_n}{b_n} = L$. Then $\lim_{n \to \infty} a_n = \infty$ if and only if $\lim_{n \to \infty} b_n = \infty$.
  • Proof: Suppose that $(a_n)$ and $(b_n)$ are convergent sequences and that $\lim_{n \to \infty} \frac{a_n}{b_n} = L$ for $L \in \mathbb{R}$ and $L > 0$. Then for $\epsilon = \frac{L}{2} > 0$ we have that for some $N \in \mathbb{N}$ if $n ≥ N$ then $\mid \frac{a_n}{b_n} - L \mid < \frac{L}{2}$ or equivalently:
(1)
\begin{align} \frac{L}{2} < \frac{a_n}{b_n} < \frac{3L}{2} \\ \frac{L}{2}b_n < a_n < \frac{3L}{2}b_n \\ \end{align}
  • If $\lim_{n \to \infty} a_n = \infty$ then since $a_n < \frac{3L}{2} b_n$ it follows that $\lim_{n \to \infty} b_n = \infty$. Similarly if $\lim_{n \to \infty} b_n = \infty$ then since $\frac{L}{2} b_n < a_n$ it follows that $\lim_{n \to \infty} a_n = \infty$. $\blacksquare$
Theorem 5: If $(a_n)$ is a properly divergent subsequence then there exists no convergent subsequences $(a_{n_k})$ of $(a_n)$.
  • Proof: We will first deal with the case where $(a_n)$ is properly divergent to $\infty$. Suppose instead that there exists a subsequence $(a_{n_k})$ that converges to $L$. Then $\forall \epsilon > 0$ $\exists K \in \mathbb{N}$ such that if $k ≥ K$ then $\mid a_{n_k} - L \mid < \epsilon$, and so for $k ≥ K$ then $L - \epsilon < a_{n_k} < L + \epsilon$, and so $a_{n_k} < L + \epsilon$.
  • Now if $(a_n)$ diverges to $\infty$ then for $L + \epsilon \in \mathbb{R}$ $\exists N \in \mathbb{N}$ such that if $n ≥ N$ then $a_n > L + \epsilon$. So for $n_k ≥ \mathrm{max} \{ K, N \}$, we have that $L + \epsilon < a_{n_k} < L + \epsilon$ which is a contradiction. So our assumption that $(a_{n_k})$ converges was false, and so there exists no convergent subsequences $(a_{n_k})$. $\blacksquare$

Example 1

Show that the sequence $(n^2)$ is properly divergent to $\infty$.

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We want to show that $\forall M \in \mathbb{R}$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $n^2 > M$. Notice that $n^2 > n$ for all $n \in \mathbb{N}$. By the Archimedean property, since $M \in \mathbb{R}$ there exists an $n \in \mathbb{N}$ such that $M ≤ n$, and so $n^2 > n ≥ M$. Therefore the sequence $(n^2)$ diverges properly to $\infty$.

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