Why Monotonicity and Boundedness Matter
A sequence can have infinitely many terms, so checking a long list of values cannot establish convergence. The Monotone Convergence Theorem gives a powerful way to prove that a sequence converges without finding its limit: show that its terms move consistently in one direction and cannot move beyond a fixed bound.
The theorem is especially useful when a sequence is given by a formula or recurrence for which a limit is difficult to calculate. It guarantees that a limit exists, but it does not generally tell us the limit’s value. Establishing existence and calculating a value are distinct tasks.
Monotonicity and Bounds
A number \(M\) is an upper bound for \((a_n)\) if \(a_n\le M\) for every \(n\). A number \(m\) is a lower bound if \(m\le a_n\) for every \(n\). A sequence is bounded if it has both an upper and a lower bound. For an increasing sequence, an upper bound is the relevant bound; its first term already gives a lower bound. For a decreasing sequence, a lower bound is the relevant bound.
The one-sided versions are often the most convenient to apply. For example, an increasing sequence need not be checked for a separate lower bound: \(a_n\ge a_1\) for every \(n\). Similarly, for a decreasing sequence, \(a_n\le a_1\) for every \(n\).
Why the Theorem Works
The reason depends on the completeness of the real numbers. One formulation of completeness is the Cauchy criterion: every Cauchy sequence of real numbers converges. A sequence is Cauchy if, for every \(\varepsilon>0\), all terms sufficiently far along are within \(\varepsilon\) of one another.
For an increasing sequence, later terms can only be larger than earlier terms. If the sequence were not Cauchy, then arbitrarily far along there would still be pairs of terms separated by a fixed positive distance. Monotonicity would force the sequence to make that much upward progress repeatedly. That is impossible if it is bounded above.
Then there is some \(\varepsilon_0>0\) such that, no matter how far along we go, two later terms differ by at least \(\varepsilon_0\).
For each such pair, choose indices \(m_j>n_j\). Since the sequence is increasing, \(a_{m_j}-a_{n_j}\ge\varepsilon_0\).
After choosing \(n_j,m_j\), search from index \(m_j\) onward for the next pair. This gives \(n_{j+1}\ge m_j\), so the intervals of increase do not overlap.
For every positive integer \(k\), the total increase across the first \(k\) chosen intervals is at least \(k\varepsilon_0\), but it cannot exceed the sequence’s total available range.
To make the last step precise, suppose \(M\) is an upper bound. The nonoverlapping intervals and monotonicity give
The final quantity is fixed, while \(k\varepsilon_0\) can be made larger than it by choosing \(k\) sufficiently large. This contradiction shows that the increasing sequence is Cauchy. Completeness of \(\mathbb{R}\) then implies that it converges.
For a decreasing sequence bounded below, consider the sequence \((-a_n)\). It is increasing and bounded above, so it converges by the increasing case. Therefore \((a_n)\) converges as well, since \(a_n=-(-a_n)\) and the negative of a convergent sequence converges. This proves both forms of the theorem.
Applying the Theorem Without Finding a Limit
To use the theorem, establish two facts: the sequence is monotone in one direction, and it has a bound in the direction required by the theorem. Once those conditions are verified, the conclusion is convergence. The conclusion does not require you to state the limit.
Worked Example: An Increasing Sequence Bounded Above
Consider \(a_n=\frac{n}{n+1}\), for \(n\in\mathbb{N}\). We will decide whether the sequence converges without calculating its limit.
First, compare consecutive terms:
For every \(n\in\mathbb{N}\), the denominator is positive, so \(a_{n+1}-a_n>0\). Thus the sequence is increasing. Also, \(n>0\) and \(n<n+1\), so
It is bounded above by \(1\). The theorem therefore guarantees convergence. As a check, \(a_1=\frac12\), \(a_2=\frac23\), and their difference is \(\frac16\), agreeing with the general consecutive-difference formula at \(n=1\). We have established convergence without identifying the limit.
Worked Example: A Decreasing Sequence Bounded Below
Consider \(b_n=5+\frac{2}{n+1}\), for \(n\in\mathbb{N}\). Compare consecutive terms:
This difference is negative for every \(n\in\mathbb{N}\), so the sequence is decreasing. Since \(n+1\ge2\), we have \(0<\frac{2}{n+1}\le1\), and hence
In particular, \(5\) is a lower bound. The decreasing form of the theorem guarantees that \((b_n)\) converges. As a check, \(b_1=6\), \(b_2=\frac{17}{3}\), and \(b_2-b_1=-\frac13\), which agrees with the formula when \(n=1\). No limit value was needed.
Worked Example: A Second Increasing Sequence
Let \(x_n=\frac{3n+1}{n+2}\), for \(n\in\mathbb{N}\). We again test the change between consecutive terms:
The denominator and numerator are positive, so \(x_{n+1}>x_n\) for every \(n\). The terms are also positive, and
Thus the sequence is increasing and bounded above by \(3\); it converges by the theorem. For a numerical check of the difference calculation, \(x_1=\frac43\) and \(x_2=\frac74\), and \(\frac74-\frac43=\frac5{12}\), matching \(\frac{5}{(1+2)(1+3)}=\frac5{12}\). Again, the argument proves convergence without calculating the limit.
A Common Misuse: Bounded Does Not Mean Convergent
Worked Example: Why Boundedness Alone Is Not Enough
A tempting but incorrect argument is: “The sequence \(c_n=(-1)^n\) is bounded, so the Monotone Convergence Theorem says it converges.” The sequence is indeed bounded, since every term is either \(-1\) or \(1\). But it is not monotone: \(c_1=-1\), \(c_2=1\), and \(c_3=-1\), so it first increases and then decreases.
The theorem therefore does not apply. In fact, the sequence does not converge: its even-indexed terms are all \(1\), while its odd-indexed terms are all \(-1\). These subsequences have different limits, which is incompatible with convergence of the original sequence.
Correction: boundedness is only one condition in the theorem. A valid application must also establish monotonicity. This example is bounded but not monotone, so it provides no conclusion through the Monotone Convergence Theorem—and its distinct subsequential values show that it actually diverges.
What the Theorem Does—and Does Not—Tell You
The theorem is an existence result. It can settle the question “Does this sequence converge?” once monotonicity and the appropriate bound have been proved. It does not, by itself, give a formula for the limit or an estimate of how close a particular term is to that limit.
For an increasing sequence, each term is at most every later term, and the terms remain below a fixed upper bound. For a decreasing sequence, each term is at least every later term, and the terms remain above a fixed lower bound. Those restrictions prevent the sequence from continuing to make substantial progress indefinitely. That is the practical reason the theorem is useful: it turns a pattern of one-directional movement plus a bound into a rigorous convergence conclusion.
When using the theorem, keep the direction of the bound matched to the direction of monotonicity:
| Monotonicity | Bound needed | Conclusion |
|---|---|---|
| Increasing | Bounded above | Converges |
| Decreasing | Bounded below | Converges |
If a sequence is not monotone, this theorem gives no conclusion, even if the sequence is bounded. If the needed bound cannot be established, the theorem also cannot be applied. These limitations do not show that the sequence diverges; they only mean that this particular result does not settle its convergence.
Check Your Understanding
For each question, identify which conditions are needed before the Monotone Convergence Theorem can be used.
- State the theorem for an increasing sequence, including the required bound.
- What bound is needed for a decreasing sequence, and why is an upper bound not the relevant condition?
- Suppose \((u_n)\) is increasing and \(u_n\le 8\) for every \(n\). What can the theorem conclude, and what does it not necessarily provide?
- Why is it incorrect to conclude that every bounded sequence converges?
- If a sequence is decreasing and bounded below by \(2\), which form of the theorem applies?
