The comparison theorem for sequences is a fundamental tool in the study of limits. It allows us to determine the behaviour of a sequence by comparing it with one or more sequences whose limits are already known.
The underlying idea is simple: if, from a certain index onwards, the terms of one sequence are ordered with respect to the terms of another, then their limits—when they exist—must respect that same ordering. In a particularly important form, known as the squeeze theorem, a sequence is trapped between two sequences that converge to the same limit; in this case the sequence in the middle is forced to converge to that limit as well.
On this page we shall study the comparison theorem in its principal forms: the comparison of convergent sequences, the comparison with sequences diverging to \(+\infty\) or to \(-\infty\), and the squeeze theorem. We shall pay particular attention to the meaning of the word eventually, that is, to the fact that the required inequalities need not hold for every index, but only from a certain index onwards.
Contents
- Comparison of sequences and the meaning of eventually
- The comparison theorem for convergent sequences
- Proof of the comparison theorem
- The squeeze theorem for sequences
- Proof of the squeeze theorem
- Comparison with divergent sequences at infinity
- Worked examples on the comparison theorem
- Common mistakes in applying the theorem
Comparison of sequences and the meaning of eventually
The comparison theorem concerns real sequences whose terms can be ordered, at least from a certain index onwards. For this reason, before stating the theorem, it is important to clarify precisely what it means to compare two sequences.
Let \((a_n)\) and \((b_n)\) be two real sequences. To say that
\[ a_n \le b_n \]
eventually means that there exists an index \(N\in\mathbb{N}\) such that, for every \(n\ge N\),
\[ a_n \le b_n. \]
In symbols:
\[ \exists N\in\mathbb{N}\ \text{such that}\ \forall n\ge N,\quad a_n\le b_n. \]
Thus the inequality need not hold for every index \(n\), but only from a certain point onwards. The first few terms of the sequences may well fail to satisfy the comparison, since a finite number of initial terms does not affect the limit.
For instance, if \(a_n\le b_n\) for every \(n\ge 5\), then we may say that \(a_n\le b_n\) eventually. It does not matter that the inequality fails for \(n=0,1,2,3,4\), because the limiting behaviour depends on the terms of the sequence as \(n\) becomes arbitrarily large.
This observation is essential: theorems on the limits of sequences do not describe the behaviour of the first few terms, but the behaviour of the sequence at infinity. Therefore, in applications of the comparison theorem, what matters is establishing an ordering between the sequences for all sufficiently large indices.
Similarly, to write
\[ a_n \le b_n \le c_n \]
eventually means that there exists an index \(N\in\mathbb{N}\) such that, for every \(n\ge N\), both inequalities
\[ a_n \le b_n \qquad\text{and}\qquad b_n \le c_n \]
hold simultaneously. In this case the sequence \((b_n)\) is eventually squeezed between \((a_n)\) and \((c_n)\). This is the typical situation of the squeeze theorem: if the two outer sequences converge to the same limit, then the sequence in the middle is forced to converge to that limit as well.
Comparison of sequences, then, is not a mere term-by-term comparison taken in isolation. It is a comparison that is stable from a certain index onwards, and it is precisely this stability that makes it possible to transfer information about the limit from one sequence to another.
The comparison theorem for convergent sequences
The first form of the comparison theorem concerns two convergent real sequences. It states that an ordering between the terms of the sequences that holds eventually is preserved in passing to the limit.
Let \((a_n)\) and \((b_n)\) be two real sequences such that
\[ a_n \le b_n \]
eventually. If
\[ \lim_{n\to+\infty} a_n=\ell \qquad\text{and}\qquad \lim_{n\to+\infty} b_n=m, \]
then
\[ \ell \le m. \]
In other words, if from a certain index onwards each term of \((a_n)\) is less than or equal to the corresponding term of \((b_n)\), then the limit of \((a_n)\) cannot exceed the limit of \((b_n)\).
This result is very natural, but it must be interpreted with care: the theorem does not say that, knowing only that \(a_n\le b_n\), we can compute the two limits. Rather, it says that, if the two limits exist, then they must respect the same ordering.
It is also important to observe that a strict inequality between the terms does not necessarily produce a strict inequality between the limits. If
\[ a_n < b_n \]
eventually and the two sequences converge to \(\ell\) and \(m\) respectively, we can only conclude that
\[ \ell \le m, \]
not necessarily that \(\ell<m\).
For instance, for every \(n\ge 1\) we have
\[ 0 < \frac{1}{n}. \]
However
\[ \lim_{n\to+\infty}0=0 \qquad\text{and}\qquad \lim_{n\to+\infty}\frac{1}{n}=0. \]
Thus the strict inequality between the terms may turn into an equality between the limits.
The comparison theorem for convergent sequences therefore expresses a property of compatibility between the ordering of the real numbers and the passage to the limit: the eventual ordering between convergent sequences cannot be reversed in the limit.
Proof of the comparison theorem
We prove the theorem in the form stated in the previous section. Let \((a_n)\) and \((b_n)\) be two real sequences such that
\[ a_n \le b_n \]
eventually, and suppose that
\[ \lim_{n\to+\infty}a_n=\ell \qquad\text{and}\qquad \lim_{n\to+\infty}b_n=m. \]
We wish to show that
\[ \ell \le m. \]
We argue by contradiction. Suppose, then, that the conclusion is false, that is, suppose that
\[ \ell>m. \]
Consider the midpoint of \(\ell\) and \(m\):
\[ \alpha=\frac{\ell+m}{2}. \]
Since \(\ell>m\), we have
\[ m<\alpha<\ell. \]
From the convergence of \((a_n)\) to \(\ell\), together with the fact that \(\alpha<\ell\), there exists an index \(N_1\in\mathbb{N}\) such that, for every \(n\ge N_1\),
\[ a_n>\alpha. \]
Indeed, since
\[ \varepsilon=\ell-\alpha>0, \]
it follows from the definition of limit that eventually
\[ |a_n-\ell|<\varepsilon, \]
and hence
\[ a_n>\ell-\varepsilon=\alpha. \]
Similarly, from the convergence of \((b_n)\) to \(m\), together with the fact that \(m<\alpha\), there exists an index \(N_2\in\mathbb{N}\) such that, for every \(n\ge N_2\),
\[ b_n<\alpha. \]
Indeed, since
\[ \varepsilon=\alpha-m>0, \]
it follows from the definition of limit that eventually
\[ |b_n-m|<\varepsilon, \]
and hence
\[ b_n<m+\varepsilon=\alpha. \]
Moreover, by hypothesis, \(a_n\le b_n\) eventually. Hence there exists an index \(N_0\in\mathbb{N}\) such that, for every \(n\ge N_0\),
\[ a_n\le b_n. \]
Now take an index \(n\) greater than or equal to all three indices \(N_0\), \(N_1\), \(N_2\). For such an \(n\) the following inequalities hold simultaneously:
\[ \alpha<a_n\le b_n<\alpha. \]
This is impossible, since a quantity cannot at the same time be greater than \(\alpha\) and less than \(\alpha\). The contradiction arises from having assumed \(\ell>m\).
Therefore we must have
\[ \ell\le m. \]
This completes the proof.
The squeeze theorem for sequences
One of the most important forms of the comparison theorem is the squeeze theorem. It allows us to compute the limit of a sequence when it is eventually trapped between two sequences having the same limit.
Let \((a_n)\), \((b_n)\) and \((c_n)\) be three real sequences such that
\[ a_n \le b_n \le c_n \]
eventually. If
\[ \lim_{n\to+\infty}a_n=\ell \qquad\text{and}\qquad \lim_{n\to+\infty}c_n=\ell, \]
then the sequence in the middle \((b_n)\) also converges to \(\ell\), that is,
\[ \lim_{n\to+\infty}b_n=\ell. \]
The meaning of the theorem is the following: if a sequence is squeezed, from a certain index onwards, between two sequences that approach the same real number, then it has no possibility of tending to a different limit. The two outer sequences compel the sequence in the middle to approach the same value.
The fundamental hypothesis is that the two outer sequences have the same limit. It is not enough to know that \((a_n)\) and \((c_n)\) are convergent: if their limits differ, the sequence in the middle may behave in different ways.
For instance, from the comparison
\[ 0\le b_n\le 1 \]
alone we cannot conclude that \((b_n)\) is convergent. The sequence might oscillate, as happens for
\[ b_n=\frac{1+(-1)^n}{2}, \]
which takes the values \(1\) and \(0\) alternately. In this case the two outer sequences are constant, but they have different limits:
\[ \lim_{n\to+\infty}0=0 \qquad\text{and}\qquad \lim_{n\to+\infty}1=1. \]
A very commonly used form of the squeeze theorem is the following. If \((r_n)\) is a real sequence such that
\[ r_n\ge 0 \]
eventually,
\[ \lim_{n\to+\infty}r_n=0 \]
and
\[ |b_n-\ell|\le r_n \]
eventually, then
\[ \lim_{n\to+\infty}b_n=\ell. \]
Indeed, the inequality
\[ |b_n-\ell|\le r_n \]
is equivalent to saying that
\[ \ell-r_n\le b_n\le \ell+r_n. \]
Since both outer sequences \((\ell-r_n)\) and \((\ell+r_n)\) tend to \(\ell\), the squeeze theorem forces \((b_n)\) to tend to \(\ell\).
This formulation is particularly useful when it is not easy to study \(b_n\) directly, but one can estimate the distance between \(b_n\) and the candidate limit \(\ell\) by means of a positive null sequence.
Proof of the squeeze theorem
We prove the squeeze theorem. Let \((a_n)\), \((b_n)\) and \((c_n)\) be three real sequences such that
\[ a_n \le b_n \le c_n \]
eventually. Suppose, moreover, that
\[ \lim_{n\to+\infty}a_n=\ell \qquad\text{and}\qquad \lim_{n\to+\infty}c_n=\ell. \]
We wish to show that
\[ \lim_{n\to+\infty}b_n=\ell. \]
By the definition of limit, we must prove that, for every \(\varepsilon>0\), there exists an index \(N\in\mathbb{N}\) such that, for every \(n\ge N\),
\[ |b_n-\ell|<\varepsilon. \]
So let \(\varepsilon>0\). Since \(a_n\to \ell\), there exists an index \(N_1\in\mathbb{N}\) such that, for every \(n\ge N_1\),
\[ |a_n-\ell|<\varepsilon. \]
In particular, for every \(n\ge N_1\),
\[ \ell-\varepsilon<a_n. \]
Since \(c_n\to \ell\), there exists an index \(N_2\in\mathbb{N}\) such that, for every \(n\ge N_2\),
\[ |c_n-\ell|<\varepsilon. \]
In particular, for every \(n\ge N_2\),
\[ c_n<\ell+\varepsilon. \]
Moreover, by hypothesis, \(a_n\le b_n\le c_n\) eventually. Hence there exists an index \(N_0\in\mathbb{N}\) such that, for every \(n\ge N_0\),
\[ a_n\le b_n\le c_n. \]
Now consider
\[ N=\max\{N_0,N_1,N_2\}. \]
Then, for every \(n\ge N\), the following inequalities hold simultaneously:
\[ \ell-\varepsilon<a_n\le b_n\le c_n<\ell+\varepsilon. \]
In particular,
\[ \ell-\varepsilon<b_n<\ell+\varepsilon. \]
This double inequality is equivalent to
\[ |b_n-\ell|<\varepsilon. \]
We have thus shown that, for every \(\varepsilon>0\), there exists an index \(N\in\mathbb{N}\) such that, for every \(n\ge N\),
\[ |b_n-\ell|<\varepsilon. \]
By the definition of limit, it follows that
\[ \lim_{n\to+\infty}b_n=\ell. \]
This completes the proof.
Comparison with divergent sequences at infinity
The comparison theorem also admits very useful forms for sequences that diverge to \(+\infty\) or to \(-\infty\). In these cases the comparison does not serve to establish the order between two finite limits, but to deduce that a sequence diverges when it is dominated, in the appropriate direction, by another divergent sequence.
The first form concerns divergence to \(+\infty\). Let \((a_n)\) and \((b_n)\) be two real sequences such that
\[ a_n \le b_n \]
eventually. If
\[ \lim_{n\to+\infty}a_n=+\infty, \]
then
\[ \lim_{n\to+\infty}b_n=+\infty. \]
Indeed, if \((a_n)\) tends to \(+\infty\), its terms eventually become greater than any fixed real number. Since, from a certain index onwards, \(b_n\) is greater than or equal to \(a_n\), \(b_n\) too must eventually become greater than any fixed real number.
More explicitly, let \(M\in\mathbb{R}\). Since \(a_n\to+\infty\), there exists an index \(N_1\in\mathbb{N}\) such that, for every \(n\ge N_1\),
\[ a_n>M. \]
Moreover, since \(a_n\le b_n\) eventually, there exists an index \(N_0\in\mathbb{N}\) such that, for every \(n\ge N_0\),
\[ a_n\le b_n. \]
Hence, for every \(n\ge \max\{N_0,N_1\}\), we have
\[ b_n\ge a_n>M. \]
By the definition of divergence to \(+\infty\), it follows that
\[ b_n\to+\infty. \]
The second form concerns divergence to \(-\infty\). Let \((a_n)\) and \((b_n)\) be two real sequences such that
\[ a_n \le b_n \]
eventually. If
\[ \lim_{n\to+\infty}b_n=-\infty, \]
then
\[ \lim_{n\to+\infty}a_n=-\infty. \]
In this case the argument is symmetric: if \((b_n)\) tends to \(-\infty\), its terms eventually become less than any fixed real number. Since, from a certain index onwards, \(a_n\) is less than or equal to \(b_n\), \(a_n\) too must tend to \(-\infty\).
More explicitly, let \(M\in\mathbb{R}\). Since \(b_n\to-\infty\), there exists an index \(N_1\in\mathbb{N}\) such that, for every \(n\ge N_1\),
\[ b_n<M. \]
Moreover, since \(a_n\le b_n\) eventually, there exists an index \(N_0\in\mathbb{N}\) such that, for every \(n\ge N_0\),
\[ a_n\le b_n. \]
Hence, for every \(n\ge \max\{N_0,N_1\}\), we have
\[ a_n\le b_n<M. \]
By the definition of divergence to \(-\infty\), it follows that
\[ a_n\to-\infty. \]
It is important to note the direction of the inequalities. To prove that a sequence tends to \(+\infty\), it suffices to find a smaller (or equal) sequence that tends to \(+\infty\). Likewise, to prove that a sequence tends to \(-\infty\), it suffices to find a larger (or equal) sequence that tends to \(-\infty\).
In symbols:
\[ a_n\le b_n,\quad a_n\to+\infty \quad\Longrightarrow\quad b_n\to+\infty, \]
and
\[ a_n\le b_n,\quad b_n\to-\infty \quad\Longrightarrow\quad a_n\to-\infty. \]
In other words, for \(+\infty\) one needs a lower bound, whereas for \(-\infty\) one needs an upper bound.
Worked examples on the comparison theorem
Let us now look at some typical examples. The aim is not merely to compute the limits, but to understand which form of the comparison theorem is being used and why the hypotheses are satisfied.
Example 1. We compute the limit
\[ \lim_{n\to+\infty}\frac{\sin n}{n}. \]
For every \(n\ge 1\) we know that
\[ -1\le \sin n\le 1. \]
Dividing every term by \(n\), which is positive for every \(n\ge 1\), we obtain
\[ -\frac{1}{n}\le \frac{\sin n}{n}\le \frac{1}{n}. \]
Now
\[ \lim_{n\to+\infty}\left(-\frac{1}{n}\right)=0 \qquad\text{and}\qquad \lim_{n\to+\infty}\frac{1}{n}=0. \]
The sequence \(\displaystyle \frac{\sin n}{n}\) is therefore squeezed between two sequences that both tend to \(0\). By the squeeze theorem,
\[ \lim_{n\to+\infty}\frac{\sin n}{n}=0. \]
Example 2. We compute the limit
\[ \lim_{n\to+\infty}\left(2+\frac{(-1)^n}{n}\right). \]
The oscillating part is \((-1)^n\), but it is multiplied by \(\displaystyle \frac{1}{n}\), which tends to \(0\). To make this observation rigorous, we consider the distance of the sequence from the candidate limit \(2\):
\[ \left|2+\frac{(-1)^n}{n}-2\right| = \left|\frac{(-1)^n}{n}\right| = \frac{1}{n}. \]
Since
\[ \lim_{n\to+\infty}\frac{1}{n}=0, \]
it follows that the distance between \(\displaystyle 2+\frac{(-1)^n}{n}\) and \(2\) tends to \(0\). By the absolute-value form of the squeeze theorem,
\[ \lim_{n\to+\infty}\left(2+\frac{(-1)^n}{n}\right)=2. \]
Example 3. We study the limit of the sequence
\[ b_n=n^2+\sin n. \]
Since
\[ \sin n\ge -1, \]
for every \(n\in\mathbb{N}\) we have
\[ n^2+\sin n\ge n^2-1. \]
Moreover
\[ \lim_{n\to+\infty}(n^2-1)=+\infty. \]
Thus \((b_n)\) is bounded below by a sequence that tends to \(+\infty\). By comparison with sequences diverging to \(+\infty\), we obtain
\[ \lim_{n\to+\infty}(n^2+\sin n)=+\infty. \]
Example 4. We study the limit of the sequence
\[ c_n=-n+\cos n. \]
Since
\[ \cos n\le 1, \]
for every \(n\in\mathbb{N}\) we have
\[ -n+\cos n\le -n+1. \]
Moreover
\[ \lim_{n\to+\infty}(-n+1)=-\infty. \]
Thus \((c_n)\) is bounded above by a sequence that tends to \(-\infty\). By comparison with sequences diverging to \(-\infty\), we obtain
\[ \lim_{n\to+\infty}(-n+\cos n)=-\infty. \]
These examples show that the comparison theorem is useful not only when a sequence is explicitly trapped between two sequences. Often the decisive point is to construct a suitable estimate: a two-sided estimate in order to apply the squeeze theorem, a lower estimate to prove divergence to \(+\infty\), or an upper estimate to prove divergence to \(-\infty\).
Common mistakes in applying the theorem
The comparison theorem is very powerful, but it must be applied with precise attention to its hypotheses. Many mistakes arise from neglecting the direction of the inequalities, the meaning of eventually, or the role played by the limits of the sequences being compared.
Confusing a strict inequality with a strict inequality between the limits
If \(a_n<b_n\) eventually and the two sequences converge to \(\ell\) and \(m\) respectively, one cannot necessarily conclude that \(\ell<m\). One can only conclude that
\[ \ell\le m. \]
Indeed, a strict inequality between the terms may turn into an equality between the limits. For instance, for every \(n\ge 1\),
\[ 0<\frac{1}{n}, \]
but both sequences tend to \(0\).
Using the wrong direction in comparison at infinity
To prove that a sequence tends to \(+\infty\), it is not enough to find an upper bound for it that tends to \(+\infty\). What is needed instead is a lower estimate by means of a sequence that tends to \(+\infty\).
Likewise, to prove that a sequence tends to \(-\infty\), it is not enough to find a lower bound for it that tends to \(-\infty\). What is needed instead is an upper estimate by means of a sequence that tends to \(-\infty\).
In symbols:
\[ a_n\le b_n,\quad a_n\to+\infty \quad\Longrightarrow\quad b_n\to+\infty, \]
whereas
\[ a_n\le b_n,\quad b_n\to-\infty \quad\Longrightarrow\quad a_n\to-\infty. \]
Applying the squeeze theorem without the same limit at the ends
In the squeeze theorem it is not enough to have
\[ a_n\le b_n\le c_n \]
eventually. It is necessary that the two outer sequences tend to the same limit:
\[ a_n\to \ell \qquad\text{and}\qquad c_n\to \ell. \]
If, on the other hand, the limits of the outer sequences differ, the comparison can only provide a region in which the terms of \((b_n)\) lie, but it does not necessarily determine the limit of \((b_n)\).
Forgetting that the comparison must hold eventually
The inequalities required by the theorem need not hold for every index, but they must hold from a certain index onwards. Nevertheless, it is not enough to verify them for many values of \(n\), or for a few numerical examples: one must prove that there exists an index \(N\in\mathbb{N}\) such that the inequality is true for every \(n\ge N\).
This point is essential because the limit describes the behaviour of the sequence for arbitrarily large indices. The first few terms may be irrelevant, but the comparison must stabilise eventually.
Using too weak an estimate
An estimate is useful only if it contains enough information to apply the theorem. For instance, knowing that a sequence is bounded is not enough to conclude that it is convergent. In the same way, knowing that a sequence lies between two convergent sequences is not enough, if these do not have the same limit.
The comparison theorem does not replace the study of the limit: it provides a rigorous criterion when the comparison is constructed in the correct direction and with sequences whose behaviour is known.
In conclusion, to apply the theorem correctly means identifying three elements: an eventual inequality, the right direction of comparison, and the limit of the sequences used as a reference.