The Divergence Criteria for Sequences

# The Divergence Criteria for Sequences

Thus far we have looked at criteria for sequences to be convergent. We will now begin to look at some criteria which will tell us if a sequence is divergent.

 Theorem 1 (Divergence Criteria for Sequences): A sequence $(a_n)$ of real numbers is divergent if either one of the following hold: 1. $(a_n)$ has two subsequences $(a_{n_k})$ and $(a_{n_p})$ that converge to two different limits. 2. $(a_n)$ has a subsequence that is divergent. 3. $(a_n)$ is unbounded.

Notice that if either (1) or (2) hold then this immediately contradicts the fact that if a sequence $(a_n)$ is convergent then all of its subsequences $(a_{n_k})$ converge to the same limit.

Recall by The Boundedness of Convergent Sequences Theorem that if a sequence is convergent that it is bounded. The contrapositive of this statement is that is a sequence is not bounded then it is divergent, and so then (3) is justified as well.

We will now look at some examples of apply the Divergence Criteria for Sequences.

## Example 1

Show that the sequence $((-1)^n)$ is divergent.

To show this sequence is divergent, consider the subsequence of even terms which is $(1, 1, 1, ... )$ which converges to the real number $1$. Now consider the subsequence of odd terms which is $(-1, -1, -1, ... )$ which converges to the real number $-1$.

Since these two subsequences converge to different limits, we conclude that $((-1)^n)$ is divergent.

## Example 2

Show that the sequence $(a_n)$ defined by $a_n = \left\{\begin{matrix} 1/n & \mathrm{if \: n \: is \: even} \\ n & \mathrm{if \: n \: is \: odd}\end{matrix}\right.$.

Let's first look at a few terms of this sequence. We have that $(a_n) = \left(1, \frac{1}{2}, 3, \frac{1}{4}, 5, \frac{1}{6}, ... \right)$. We can see this sequence is not bounded above and hence not bounded, which we will prove.

Suppose that there exists an $M \in \mathbb{N}$ such that $\mid a_n \mid < M$ for all $n \in \mathbb{N}$. By the Archimedean property since $M \in \mathbb{R}$ there exists a natural number $n_M \in \mathbb{N}$ such that $M ≤ n_M$. We also note that $M ≤ n_M < n_M + 1$. So either $n_M$ or $n_M + 1$ is a term in the sequence $(a_n)$ which contradicts the sequence $(a_n)$ from being bounded.

## Example 3

Show that the sequence $(n)$ is divergent.

Once again, this sequence is unbounded. Suppose that instead $(n)$ is bounded, that is $\mid n \mid = n < M$ for some $M \in \mathbb{R}$. But this contradicts the Archimedean property which says that for any $M \in \mathbb{R}$ there exists an $n_M \in \mathbb{N}$ such that $M ≤ n_M$, and so in fact $(n)$ is not bounded and by the divergence criteria, is divergent as well.