Why the Supremum Gives the Limit
A bounded increasing sequence has a natural candidate for its limit: the supremum of all its terms. The supremum is the smallest number that is at least as large as every term. It may be a term of the sequence, but it need not be. The key fact is that terms of an increasing sequence eventually come arbitrarily close to this smallest upper bound.
This tutorial proves that claim from the least-upper-bound property of the real numbers. It also proves the matching result for a decreasing sequence and its infimum. The argument explains not only that convergence occurs, but exactly which number the sequence approaches.
Definitions and the Completeness Property
A sequence \((a_n)\) of real numbers is increasing if \(a_n\le a_{n+1}\) for every \(n\in\mathbb{N}\). It is decreasing if \(a_{n+1}\le a_n\) for every \(n\in\mathbb{N}\). These definitions allow consecutive terms to be equal.
A real number \(u\) is an upper bound for a set \(S\subseteq\mathbb{R}\) if \(x\le u\) for every \(x\in S\). The supremum of \(S\), written \(\sup S\), is its least upper bound: it is an upper bound, and it is no greater than any other upper bound. A lower bound is defined by reversing the inequality; the greatest lower bound is the infimum, written \(\inf S\).
The convergence statement used below is the epsilon-\(N\) definition: \((a_n)\) converges to \(L\) if, for every \(\varepsilon>0\), there is an \(N\in\mathbb{N}\) such that \(n\ge N\) implies \(|a_n-L|<\varepsilon\). Thus the proof must show that every sufficiently late term lies within any prescribed positive distance of the proposed limit.
The Increasing Case: A Complete Proof
Let \((a_n)\) be increasing and bounded above, and let \(S=\{a_n:n\in\mathbb{N}\}\) be its set of terms. The set \(S\) is nonempty and bounded above, so it has a supremum. Write \(s=\sup S\). We will prove directly that \(a_n\to s\).
Let \(\varepsilon>0\) be arbitrary. The target is to find one index after which every term differs from \(s\) by less than \(\varepsilon\).
The number \(s-\varepsilon\) is not an upper bound for \(S\). If it were, it would be an upper bound strictly smaller than \(s\), contradicting that \(s\) is the least upper bound. Therefore some term \(a_N\) satisfies \(a_N>s-\varepsilon\).
For every \(n\ge N\), increasingness gives \(a_n\ge a_N>s-\varepsilon\). Also \(a_n\le s\), because \(s\) is an upper bound for every term.
Combining them gives \(0\le s-a_n<\varepsilon\), so \(|a_n-s|<\varepsilon\) for every \(n\ge N\). This is exactly the definition of \(a_n\to s\).
The proof depends on two ingredients that play different roles. The supremum property guarantees that there is a term above \(s-\varepsilon\); increasingness guarantees that all later terms remain above that term. The upper bound \(s\) then traps the entire tail in an interval of width \(\varepsilon\).
The Decreasing Case: The Infimum Proof
Now let \((a_n)\) be decreasing and bounded below. Its set of terms \(S=\{a_n:n\in\mathbb{N}\}\) has an infimum; write \(m=\inf S\). The proof follows the same structure, with the inequalities reversed. For any \(\varepsilon>0\), the number \(m+\varepsilon\) cannot be a lower bound: if it were, it would be a lower bound greater than the greatest lower bound \(m\). Therefore there is some \(N\) for which \(a_N<m+\varepsilon\).
For every \(n\ge N\), decreasingness gives \(a_n\le a_N<m+\varepsilon\). Since \(m\) is a lower bound, \(a_n\ge m\). Hence \(0\le a_n-m<\varepsilon\), which proves \(|a_n-m|<\varepsilon\) for every \(n\ge N\).
For an increasing sequence, the terms approach the least upper bound from at or below. For a decreasing sequence, they approach the greatest lower bound from at or above. The endpoint need not itself occur among the terms.
Worked Examples
Worked Example: An Increasing Sequence Whose Supremum Is Not a Term
Consider \(a_n=1-\frac{1}{2^n}\) for \(n\in\mathbb{N}\). The first terms are \(a_1=\frac12\), \(a_2=\frac34\), and \(a_3=\frac78\). To check increasingness, calculate the consecutive difference:
Thus the sequence is increasing. Every term is less than \(1\), so \(1\) is an upper bound. To show that it is the least upper bound, take any \(u<1\) and set \(\delta=1-u>0\). Choose \(n\) so large that \(n+1>1/\delta\). Since \(2^n\ge n+1\), we have \(2^n>1/\delta\), hence \(1/2^n<\delta\). It follows that \(a_n=1-1/2^n>1-\delta=u\). Therefore no number less than \(1\) is an upper bound, and \(\sup\{a_n:n\in\mathbb{N}\}=1\).
The theorem now gives \(a_n\to1\). The epsilon condition can also be checked directly: \(|a_n-1|=1/2^n\). Given \(\varepsilon>0\), choose \(N\) with \(N+1>1/\varepsilon\). For \(n\ge N\), \(2^n\ge n+1\ge N+1>1/\varepsilon\), so \(1/2^n<\varepsilon\). This verifies the limit, and no term equals \(1\), since \(1/2^n>0\) for every \(n\).
Worked Example: A Decreasing Sequence and Its Infimum
Let \(a_n=3+\frac1n\) for \(n\in\mathbb{N}\). Its first terms are \(4\), \(7/2\), and \(10/3\). For each \(n\),
so the sequence is decreasing. Since \(1/n>0\), every term is greater than \(3\), making \(3\) a lower bound. If \(\ell>3\), then \(\ell-3>0\). Choose \(n>1/(\ell-3)\). This gives \(1/n<\ell-3\), and therefore \(a_n=3+1/n<\ell\). Thus \(\ell\) is not a lower bound; no number greater than \(3\) is a lower bound. Consequently, \(\inf\{a_n:n\in\mathbb{N}\}=3\).
The decreasing-case theorem gives \(a_n\to3\). Directly, \(|a_n-3|=1/n\). For any \(\varepsilon>0\), choose \(N>1/\varepsilon\). If \(n\ge N\), then \(1/n\le1/N<\varepsilon\), so \(|a_n-3|<\varepsilon\). The limit check agrees with the infimum, and no term equals \(3\).
Worked Example: A Supremum That Is Attained
Define \(a_1=1\), \(a_2=3/2\), and \(a_n=5/3\) for every \(n\ge3\). The first terms are \(1,3/2,5/3,5/3,\ldots\). They are nondecreasing because \(1\le3/2\), \(3/2\le5/3\), and every subsequent pair is equal. Every term is at most \(5/3\), and \(a_3=5/3\); therefore the supremum is \(5/3\), and it is a term of the sequence.
For every \(n\ge3\), \(|a_n-5/3|=0\). Given any \(\varepsilon>0\), take \(N=3\). Then \(n\ge N\) implies \(|a_n-5/3|=0<\varepsilon\), proving convergence to the supremum. This example contrasts with the first: a supremum may or may not be attained, but either way it is the limit in the increasing bounded case.
A Common Pitfall: Confusing a Supremum with a Maximum
A common incorrect step is to claim that because \(s\) is the supremum of the terms, there must be some index \(N\) with \(a_N=s\). That conclusion is false. In the first example, every term \(1-1/2^n\) is strictly less than \(1\), even though the supremum is \(1\).
The correct statement is weaker and exactly what the proof needs: for every \(\varepsilon>0\), some term is greater than \(s-\varepsilon\). It follows from the definition of least upper bound, not from the existence of a largest term. Once that term has been found, monotonicity keeps all later terms at least as large.
Why the Result Matters
This proof reveals the role of completeness in real analysis. Monotonicity alone does not identify a limit, and boundedness alone does not make a sequence converge. Together they place the terms on one side of a real least upper bound and force them eventually within any positive distance of it. The least-upper-bound property ensures that the candidate exists as a real number.
The theorem is also a practical way to identify a limit when a sequence is already known to be monotone and bounded: determine the supremum or infimum of its terms, then the theorem identifies the limit. Establishing monotonicity and boundedness may require separate arguments; the supremum proof begins once those properties are available.
Check Your Understanding
Use the definitions and the proof structure above to answer the following questions.
- Why can \(s-\varepsilon\) not be an upper bound when \(s\) is the supremum?
- In the increasing case, which step ensures that every term after \(a_N\) remains above \(s-\varepsilon\)?
- State the corresponding epsilon argument for a decreasing sequence with infimum \(m\).
- Does a sequence have to contain its supremum in order to converge to it? Explain using an example.
- What role does the least-upper-bound property of \(\mathbb{R}\) play in the increasing-case proof?
