Introduction
A sequence can approach a limit while its terms vary along the way. Convergence places an important restriction on that variation: a convergent sequence cannot have terms that grow without bound. The reverse, however, is not true. A sequence can stay within a fixed interval and still fail to approach any single number.
This distinction matters whenever you need to decide whether a sequence could converge. Boundedness is a necessary condition for convergence, so an unbounded sequence cannot converge. But boundedness by itself does not settle the question.
Definitions: Boundedness and Convergence
A real sequence is a list of real numbers \(a_1,a_2,a_3,\ldots\), indexed here by \(n\in\mathbb{N}=\{1,2,3,\ldots\}\).
The bound \(M\) does not have to be the smallest possible bound. It only needs to work for every term, including the first terms.
Convergence concerns all terms from some point onward: they must lie as close as desired to \(L\). Boundedness concerns every term in the sequence. The proof that convergence implies boundedness must therefore account for both the terms sufficiently far along and the finitely many terms before that point.
Why Convergence Forces Boundedness
Suppose a sequence converges to \(L\). Apply the definition of convergence with the particular choice \(\varepsilon=1\). Then there is an index \(N\) such that every term from \(a_N\) onward is less than \(1\) unit from \(L\). By the triangle inequality, those terms are within \(|L|+1\) of zero. The terms before \(a_N\) are finite in number, so they also have a finite bound.
Proof of the Theorem
Assume \(a_n\to L\). Taking \(\varepsilon=1\) in the definition gives an \(N\in\mathbb{N}\) such that \(n\geq N\) implies \(|a_n-L|<1\).
For each \(n\geq N\), the triangle inequality gives \(|a_n|\leq |a_n-L|+|L|<1+|L|\). Thus all terms from \(a_N\) onward share the finite bound \(|L|+1\).
If \(N>1\), the terms \(a_1,\ldots,a_{N-1}\) form a finite set, so their absolute values have a finite maximum \(B=\max\{|a_1|,\ldots,|a_{N-1}|\}\). Set \(M=\max\{B,|L|+1\}\). If \(N=1\), set \(M=|L|+1\); there are no initial terms to include.
For \(n<N\), \(|a_n|\leq B\leq M\) when there are initial terms. For \(n\geq N\), \(|a_n|<|L|+1\leq M\). Hence \(|a_n|\leq M\) for every \(n\in\mathbb{N}\), so the sequence is bounded.
The key idea is the split between a finite beginning and an eventually controlled tail. Convergence directly controls the tail, not necessarily the earlier terms. The finite beginning cannot make the sequence unbounded, because a finite set of real numbers always has a finite maximum absolute value. This completes the proof.
Worked Example: A Convergent Sequence with an Early Outlier
Worked Example: A Convergent Sequence with an Early Outlier
Define \(a_1=100\), and for \(n\geq 2\), define \(a_n=\frac{1}{n}\). Show that the sequence converges and is bounded.
To verify convergence to \(0\), let \(\varepsilon>0\). Choose \(N=\max\{2,\lfloor 1/\varepsilon\rfloor+1\}\), where \(\lfloor x\rfloor\) is the greatest integer less than or equal to \(x\). Then \(N>1/\varepsilon\), and \(n\geq N\) implies \(n>1/\varepsilon\). Therefore, for such \(n\),
So \(a_n\to 0\). To bound the sequence, the first term has \(|a_1|=100\). For \(n\geq 2\), \(0<1/n\leq 1/2\), so \(|a_n|\leq 1/2\leq 100\). Thus \(M=100\) is a bound for every term.
Check: The proposed limit satisfies the definition because every \(n\geq N\) gives \(|a_n-0|<\varepsilon\). The bound includes the exceptional first term as well as the tail: \(|a_1|=100\), and every later absolute value is at most \(1/2\). Ignoring the first term would give an invalid bound for the full sequence.
The Converse Is False
The theorem gives one implication: convergence guarantees boundedness. It does not say that boundedness guarantees convergence. A bounded sequence may keep moving between different values rather than settling near one number.
Worked Example: Bounded but Not Convergent
Consider \(a_n=(-1)^n\). For every positive integer \(n\), \(a_n\) is either \(1\) or \(-1\). Therefore \(|a_n|=1\), and \(M=1\) bounds the sequence.
To show that it does not converge, suppose instead that \(a_n\to L\). Use the definition with \(\varepsilon=1/2\). There would be an \(N\) such that every \(n\geq N\) satisfies \(|a_n-L|<1/2\). There is an even integer \(p\geq N\) and an odd integer \(q\geq N\). Since \(a_p=1\) and \(a_q=-1\), the supposed condition gives \(|1-L|<1/2\) and \(|-1-L|<1/2\). But the triangle inequality then implies
This says \(2<1\), a contradiction. Therefore the sequence is bounded but does not converge.
Check: The terms really do satisfy the bound \(|a_n|\leq 1\) for every index, while the assumed limit would force both \(1\) and \(-1\) to lie within \(1/2\) of the same \(L\). The contradiction confirms nonconvergence.
Using Unboundedness to Rule Out Convergence
The theorem can also be used in reverse by its contrapositive: if a sequence is unbounded, then it cannot converge. This conclusion is valid because a convergent sequence would have to be bounded.
Worked Example: An Unbounded Sequence Cannot Converge
Consider \(a_n=n\). To show that it is unbounded, let \(M\geq 0\) be any proposed bound. Choose \(n=\lfloor M\rfloor+1\). Then \(n\in\mathbb{N}\) and \(n>M\), so \(|a_n|=n>M\). Since this works for every proposed \(M\), no finite number bounds the sequence.
If \(a_n\) converged, the theorem would imply that it was bounded. That contradicts the calculation above. Hence \(a_n=n\) does not converge.
Check: For every proposed bound \(M\), the chosen term has \(|a_n|>M\), so the sequence is indeed unbounded. The theorem then rules out convergence; no guess about a possible finite limit is needed.
A Common Mistake
A frequent error is to treat boundedness as if it were a test that proves convergence. For example, one might reason:
The first statement is true, but the conclusion does not follow from it. The sequence \((-1)^n\) alternates between \(1\) and \(-1\), and the proof above shows that it has no limit. The correct conclusion from the displayed inequalities is only that the sequence is bounded.
The implication that does hold goes in the other direction: if a sequence converges, then it must be bounded. Thus, finding an unbounded sequence proves it cannot converge, while finding a bounded sequence alone does not establish convergence.
Why This Result Matters
This theorem gives a quick necessary condition for convergence. Before trying to identify a limit, you can ask whether the terms have any finite bound at all. If the sequence is unbounded, the theorem rules out convergence immediately.
When a sequence is bounded, more analysis is needed. The alternating example shows why: boundedness controls how far the terms can go, but not whether they gather around one value. In later work on sequences, boundedness will be considered alongside additional properties; it is not a substitute for checking the definition of convergence.
Check Your Understanding
Answer these questions using the definitions and theorem from this tutorial.
- In the proof that convergence implies boundedness, why is it not enough to bound only the terms \(a_n\) for \(n\geq N\)?
- Why can the terms \(a_1,\ldots,a_{N-1}\) always be included in a finite bound?
- Give a bound for the sequence \(b_n=3+(-1)^n\).
- Explain why boundedness alone cannot prove that a sequence converges.
- What can you conclude if you prove that a sequence is unbounded?
