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 $n\geq N$ 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 $n\geq N$ then $\mid a_{n}-A\mid \geq \epsilon _{0}$. 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 $\displaystyle {\lim _{n\to \infty }a_{n}=\infty }$, that is $\forall M\in \mathbb {R}$ there exists an $N\in \mathbb {N}$ such that if $n\geq N$ then $a_{n}>M$. Similarly, $(a_{n})$ is said to be Properly Divergent to $-\infty$ if $\displaystyle {\lim _{n\to \infty }a_{n}=-\infty }$, that is $\forall M\in \mathbb {R}$ there exists an $N\in \mathbb {N}$ such that if $n\geq N$ 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 $n\geq M$ 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 $n\geq M$ 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 $a_{n}\leq b_{n}$ 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 $a_{n}\leq b_{n}$ 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 $n\geq N$ then $a_{n}\geq M$. But we have that $b_{n}\geq a_{n}$ for all $n\in \mathbb {N}$ and so for $n\geq N$ we have that $b_{n}\geq M$. 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 $a_{n}\leq b_{n}$ 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 $a_{n}\leq b_{n}$ 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 $n\geq N$ then $b_{n}\leq M$. But we have that $a_{n}\leq b_{n}$ for all $n\in \mathbb {N}$ and so for $n\geq N$ we have that $a_{n}\leq M$. 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 $n\geq N$ then $\mid {\frac {a_{n}}{b_{n}}}-L\mid <{\frac {L}{2}}$ or equivalently:

${\begin{aligned}{\frac {L}{2}}<{\frac {a_{n}}{b_{n}}}<{\frac {3L}{2}}\\{\frac {L}{2}}b_{n}<a_{n}<{\frac {3L}{2}}b_{n}\\\end{aligned}}$

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 $k\geq K$ then $\mid a_{n_{k}}-L\mid <\epsilon$, and so for $k\geq K$ 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 $n\geq N$ then $a_{n}>L+\epsilon$. So for $n_{k}\geq \mathrm {max} \{K,N\}$, 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 $\infty$.

We want to show that $\forall M\in \mathbb {R}$ there exists an $N\in \mathbb {N}$ such that if $n\geq N$ 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 $M\leq n$, and so $n^{2}>n\geq M$. Therefore the sequence $(n^{2})$ diverges properly to $\infty$.

Licensing

Content obtained and/or adapted from:

• Properly Divergent Sequences, mathonline.wikidot.com under a CC BY-SA license