Strategies for Determining Convergence/Divergence of a Sequence

Table of Contents

We will now review some various methods for determining the convergence or divergence of a sequence $$(a_n)$$ of real numbers.

Evaluate the Limit. This is the most common technique for determining the convergence or divergence of a sequence. For example, if we wanted to determine whether the sequence $(a_n) = \left ( 1 - \frac{1}{n} \right )$ converges or diverges, we could simply take $\lim_{n \to \infty} 1 - \frac{1}{n}$ . If we apply the various theorems from the Properties of Convergent Sequences - Sum and Multiple Laws, Properties of Convergent Sequences - Product and Quotient Laws, and Properties of Convergent Sequences - Comparison Laws, we will see that $\lim_{n \to \infty} 1 - \frac{1}{n} = lim_{n \to \infty} 1 - \lim_{n \to \infty} \frac{1}{n} = 1 - 0 = 1$ , and so $$(a_n)$$ converges.

Use the Definition of a Convergent Sequence Directly. This method will allow us to actually prove that a sequence $$(a_n)$$ has a limit $$L$$ . Recall that $$(a_n)$$ converges to $$L$$ if $\forall \epsilon > 0$ $\exists N \in \mathbb{N}$ such that if $$n \geq N$$ then $\mid a_n - L \mid < \epsilon$ . We need to thus show that if we're given any $\epsilon > 0$ we can give an $N \in \mathbb{N}$ which satisfies the condition above. It's important to note that $$N$$ is usually dependent on $\epsilon$ , so the notation $N = N(\epsilon)$ is common. For example, suppose we wanted to actually prove that $(a_n) = \left ( 1 - \frac{1}{n} \right )$ converges to $$1$$ . Here's how we'd prove this. Let $\epsilon > 0$ . We need to find $N \in \mathbb{N}$ such that if $$n \geq N$$ then $\biggr \rvert 1 - \frac{1}{n} - 1 \biggr \rvert < \epsilon$ . Notice that this simplifies down to $\biggr \rvert -\frac{1}{n} \biggr \rvert = \frac{1}{n}$ and that $\frac{1}{n} < \epsilon$ if and only if $n > \frac{1}{\epsilon}$ , so choose $N > \frac{1}{\epsilon}$ . Then if $n ≥ N > \frac{1}{\epsilon}$ we have that $\biggr \rvert 1 - \frac{1}{n} - 1 \biggr \rvert = \frac{1}{n} ≤ \frac{1}{N} < \epsilon$ , and so $$(a_n)$$ converges to $$1$$ .

Utilize Sequence Comparison Theorems. Sometimes the Squeeze Theorem can be rather useful provided we can find two other sequences that converge to the same limit for which our unknown sequence is "squeezed" between. When it comes to properly divergent sequences, if $$a_n \leq b_n$$ ultimately and $$(a_n)$$ properly diverges to $\infty$ then $$(b_n)$$ properly diverges to $\infty$ . Similarly, if $$(b_n)$$ properly diverges to $-\infty$ then $$(a_n)$$ properly diverges to $-\infty$ .

Determine if the Sequence is Unbounded. From The Boundedness of Convergent Sequences Theorem, a sequence is bounded if it is convergent. The contrapositive of this statement says that a sequence is divergent if it is unbounded. Thus if we wanted to prove a sequence is divergent and know it is unbounded, then suppose it is bounded and show that is results in a contradiction. Then you can conclude $$(a_n)$$ is unbounded and hence divergent.

Determine if the Sequence is Monotone and Bounded. If a sequence $$(a_n)$$ is monotone and bounded, then by The Monotone Convergence Theorem, $$(a_n)$$ must be convergent. This method is particularly useful for recursive sequences.

Apply the Ratio Test. Recall from The Ratio Test for Sequence Convergence page that if we have a positive sequence of real numbers, $$(a_n)$$ and $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L < 1$ , then $$(a_n)$$ converges to $$0$$ . This method is particularly useful for sequences that cannot easily be evaluated with limits, for example functions with powers to the $$n$$ or terms including $$n!$$ .

Use Subsequences. By The Divergence Criteria for Sequences, if $$(a_n)$$ is a sequence, then if we can show $$(a_n)$$ has two subsequences $(a_{n_k})$ and $(a_{n_p})$ which converge to different limits, then $$(a_n)$$ diverges. Furthermore, if $$(a_n)$$ contains a divergent subsequence, then $$(a_n)$$ diverges.

Determine if a Sequence is Cauchy. Sometimes we don't know the limit of a sequence $$(a_n)$$ . In such a case, if we can show that $$(a_n)$$ is Cauchy then by The Cauchy Convergence Criterion, $$(a_n)$$ is convergent. Also note that if it's not readily seeable that $$(a_n)$$ is a Cauchy Sequence and difficult to prove so, then from the Additional Cauchy Sequence Proofs, if we can find Cauchy sequences that create $$(a_n)$$ then $$(a_n)$$ will also be Cauchy.

Mathonline