Divergent Sequences

What It Means for a Sequence to Diverge

The previous tutorial focused on writing complete epsilon-N proofs of convergence. To show that a sequence diverges, the goal changes: we must show that there is no real number to which the sequence converges. Failing to prove convergence to one proposed value is not enough. A sequence might fail to converge to that value but converge to another.

For a proposed limit $L$, the Failure-Witness Criterion from earlier in this course describes exactly how convergence to $L$ can fail. There must be some fixed positive distance from $L$ that the sequence fails to enter, no matter how far along the sequence we go. Divergence requires this kind of failure for every proposed real limit, though the positive distance may depend on the proposed limit.

Definition: A real sequence $(a_n)$ is divergent if there is no $L\in\mathbb{R}$ such that $a_n\to L$. Equivalently, for every $L\in\mathbb{R}$, the sequence does not converge to $L$.

Characterization of Divergence: A real sequence $(a_n)$ is divergent if and only if, for every $L\in\mathbb{R}$, there is an $\varepsilon_L>0$ such that for every $N\in\mathbb{N}_0$, there is an $n\geq N$ satisfying $$ |a_n-L|\geq\varepsilon_L. $$

This characterization follows by applying the Failure-Witness Criterion to each proposed $L$. The order of the quantifiers is important: first fix any proposed limit $L$, then find a positive witness $\varepsilon_L$, and then show that terms at least that far from $L$ occur arbitrarily late. The witness may vary with $L$. It is not necessary to find one positive distance that works for all proposed limits.

The phrase “arbitrarily late” can also be stated as “infinitely often.” If there were only finitely many indices $n$ with $|a_n-L|\geq\varepsilon_L$, then all such indices would be below some threshold $N$. Beyond that threshold, every term would satisfy $|a_n-L|<\varepsilon_L$, contradicting the witness condition. Conversely, if the inequality holds at infinitely many indices, those indices cannot all lie below any fixed $N$, so one can find such an index beyond every threshold.

Use Subsequences to Rule Out a Limit

Often it is easier to study selected terms than to test the entire sequence against every possible limit. A subsequence retains terms at indices $n_0<n_1<n_2<\cdots$. If the original sequence converges to $L$, then every subsequence converges to that same $L$, by the earlier theorem Subsequences of a Convergent Sequence. Consequently, two subsequences with different limits rule out convergence of the original sequence. This is the theorem Distinct Subsequence Limits Obstruct Convergence, established earlier in the course.

Worked Example: An Alternating Sequence

Define $a_n=(-1)^n$ for $n\in\mathbb{N}_0$. The even-indexed terms form a subsequence, since the indices $0,2,4,\ldots$ are strictly increasing. For every $k\in\mathbb{N}_0$,

a_{2k}=(-1)^{2k}=1.

Thus the even-indexed subsequence is constant and converges to $1$. The odd-indexed terms also form a subsequence, with indices $1,3,5,\ldots$, and for every $k\in\mathbb{N}_0$,

a_{2k+1}=(-1)^{2k+1}=-1.

This subsequence is constant and converges to $-1$. Since $1\neq-1$, the two subsequences have distinct limits. The theorem Distinct Subsequence Limits Obstruct Convergence therefore shows that $(a_n)$ is divergent. The sequence is bounded, so this example also shows that divergence does not require terms to become arbitrarily large in absolute value.

This method is especially effective for sequences whose terms follow different patterns along different sets of indices. To use it, verify both parts: the selected index lists really do define subsequences, and the subsequences really do converge to distinct real numbers. Merely finding two groups of terms that look different is not enough; the limits must be established.

Worked Example: Two Subsequence Limits After a Perturbation

Let $b_n=(-1)^n+1/(n+1)$. We will show that this sequence is divergent by examining its even- and odd-indexed terms. For even indices,

b_{2k}=1+\frac{1}{2k+1}.

As $k$ increases, $2k+1\geq k+1$, and so $0<1/(2k+1)\leq1/(k+1)$. The sequence $1/(k+1)$ converges to zero, hence the Squeeze Theorem gives $1/(2k+1)\to0$. The even-indexed subsequence therefore converges to $1$.

For odd indices,

b_{2k+1}=-1+\frac{1}{2k+2}.

Since $2k+2\geq k+1$, we have $0<1/(2k+2)\leq1/(k+1)$, so $1/(2k+2)\to0$. By the limit laws for sums, the odd-indexed subsequence converges to $-1$. The two subsequential limits are distinct, and therefore the original sequence diverges.

The small term $1/(n+1)$ changes every term, but it does not erase the difference between the even and odd subsequential limits. The key is not that the terms alternate exactly between two fixed values; it is that two subsequences still approach different values.

Unboundedness Is One Route to Divergence

A sequence can diverge by oscillating among bounded values, as the examples above illustrate. Another useful route is to show that a sequence is unbounded. A convergent sequence is bounded, by the earlier theorem Convergent Sequences Are Bounded. Its contrapositive gives a direct test for nonconvergence.

Theorem (Unbounded Sequences Diverge): Every unbounded real sequence is divergent.

Proof. Suppose $(a_n)$ is unbounded. If it converged to some $L\in\mathbb{R}$, then the theorem Convergent Sequences Are Bounded would imply that $(a_n)$ is bounded. This contradicts the hypothesis. Thus the sequence converges to no real number, which is exactly to say that it is divergent. $\square$

Worked Example: An Unbounded Sequence with Alternating Signs

Define $c_n=(-1)^n(n+1)$. For every $n\in\mathbb{N}_0$,

$$ |c_n|=|(-1)^n|(n+1)=n+1, $$

because $|(-1)^n|=1$. Given any $M>0$, the Archimedean property gives an $n\in\mathbb{N}_0$ with $n+1>M$. For this index, $|c_n|>M$, so the sequence is unbounded. By the theorem Unbounded Sequences Diverge, $(c_n)$ is divergent.

This proof does not need to determine what happens to the even- and odd-indexed terms separately. Establishing unboundedness is enough to rule out every real limit. The conclusion is only that no finite real limit exists; identifying more precise behavior is a separate question.

The converse of this theorem is false: boundedness does not guarantee convergence. The sequence $(-1)^n$ is bounded but divergent. Thus boundedness and convergence should not be treated as equivalent properties. Unboundedness is a sufficient test for divergence, not a necessary one.

Finite Changes Do Not Remove Divergence

Convergence is unaffected by changing finitely many terms: the earlier theorem Finite Changes Preserve Convergence formalizes that fact. Divergence has the corresponding invariance. If a sequence is divergent, changing only finitely many of its terms cannot make it converge. The reason is that every divergence witness occurs arbitrarily late, so one may look beyond all the changed indices.

Theorem (Finite Changes Preserve Divergence): Suppose two real sequences $(a_n)$ and $(b_n)$ agree for every $n\geq M$, for some $M\in\mathbb{N}_0$. If $(a_n)$ is divergent, then $(b_n)$ is divergent.

Proof. Let $L\in\mathbb{R}$ be arbitrary. Since $(a_n)$ is divergent, the Characterization of Divergence gives an $\varepsilon_L>0$ such that, for every $N\in\mathbb{N}_0$, some $n\geq N$ satisfies $|a_n-L|\geq\varepsilon_L$. Now fix any $N\in\mathbb{N}_0$ and set $N'=\max\{N,M\}$. Choose $n\geq N'$ with $|a_n-L|\geq\varepsilon_L$. Since $n\geq M$, the sequences agree at $n$, so $b_n=a_n$ and hence $|b_n-L|\geq\varepsilon_L$. This works for every $N$, and $L$ was arbitrary. The Characterization of Divergence now shows that $(b_n)$ is divergent. $\square$

Worked Example: Changing the First Term

Let $a_n=(-1)^n$, and define $d_0=12$ while $d_n=a_n$ for every $n\geq1$. The sequences $a_n$ and $d_n$ agree for every $n\geq1$, so the theorem Finite Changes Preserve Divergence shows that $(d_n)$ is divergent, since $(a_n)$ is divergent.

The subsequences also display why this conclusion holds. For $k\geq1$, $d_{2k}=1$, so the even-indexed tail has limit $1$. For every $k\in\mathbb{N}_0$, $d_{2k+1}=-1$, so the odd-indexed subsequence has limit $-1$. The exceptional value $d_0=12$ belongs to neither of these selected tails and cannot make the two distinct subsequential limits agree.

Choosing a Useful Divergence Argument

There are several ways to prove that a sequence diverges, and the most efficient choice depends on its structure:

Look for two subsequential limits. This is often effective for oscillating sequences or sequences that split into different patterns on even and odd indices.
Check for unboundedness. If the absolute values can be made arbitrarily large, the theorem Unbounded Sequences Diverge applies immediately.
Use a witness for a proposed limit. If a problem asks whether the sequence converges to a particular $L$, show that some fixed positive distance from $L$ is maintained at arbitrarily late indices.
Ignore finitely many exceptional terms when appropriate. Agreement from some index onward is enough to transfer divergence.

A common logical pitfall is to prove only that a sequence does not converge to one chosen value and then call it divergent. Divergence rules out every real value. The subsequence method handles this by finding two incompatible limits; the unboundedness method rules out convergence altogether because every convergent sequence is bounded; and the quantified characterization makes the required witness explicit for each proposed $L$.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

State the quantified characterization of divergence, including the order of its quantifiers.
Why do two subsequences with distinct real limits show that the original sequence is divergent?
Does every divergent sequence have to be unbounded? Give a reason for your answer.
Why does unboundedness imply divergence, and why does the reverse implication fail?
If two sequences agree for every $n\geq M$, why can changing the terms before $M$ not remove divergence?

What You'll Learn