Skip to main content

Chapter 23 Sequences

Section 23.1 Motivation

Notice that each Taylor polynomial is equal to the previous plus one more term:
\begin{equation*} T_{n,t^\ast}(t) = T_{n-1,t^\ast}(t) + \frac{f^{(n)}(t^\ast)}{n!} \; (t - t^\ast)^n \text{.} \end{equation*}
Suppose we want to approximate \(\ln (1.001) \) using Taylor polynomials. We can begin with the linear approximation based at \(1 \text{,}\) and then obtain better and better approximations by increasing the degree. Using the pattern from Example 22.3.9, we have
\begin{align*} \ln(1.001) \amp \approx T_{1,1}(1.001) = 0.001 \\ \ln(1.001) \amp \approx T_{2,1}(1.001) = 0.001 - \frac{{0.001}^2}{2} \\ \ln(1.001) \amp \approx T_{3,1}(1.001) = 0.001 - \frac{{0.001}^2}{2} + \frac{{0.001}^3}{3} \\ \amp \quad \vdots \text{.} \end{align*}
What if we could continue this process indefinitely? Since these approximations should be getting better, we would expect this sequence of approximate values to “trend toward” the value of \(\ln (1.001) \text{.}\) Is there some sense in which we could say that \(T_{\infty,1}(1.001) \) (whatever that means) would be exactly equal to \(\ln (1.001) \text{?}\)

Section 23.2 Definition, properties, and examples

Subsection 23.2.1 Defining sequences

Definition 23.2.1. Infinite sequence.

An infinite sequence is a function whose domain is the non-negative integers. The values in the range of a sequence are called the terms of the sequence.
When we use function notation for a sequence, we will typically use \(n \) for the independent variable instead of \(x \) or \(t \) to emphasize the integer domain. But we will more often use index notation instead of function notation to refer to the terms of a sequence. That is, if \(s(n) \) is a sequence, we will usually write
\begin{gather*} s_0 \text{ instead of } s(0) \text{,} \\ s_1 \text{ instead of } s(1) \text{,} \\ s_2 \text{ instead of } s(2) \text{,} \end{gather*}
and so on. Another common notation is to express the pattern of the sequence values as an infinite ordered list \(\nseq{s} \text{.}\) Sometimes we wish to omit the index \(n = 0 \text{,}\) usually to avoid division by \(0 \) in the formula for the terms of our sequence. In this case, we will write \(\nseqtail[0]{s} \) to clarify the domain of our sequence.

Example 23.2.2.

The terms of the sequence \(s(n) = 2^n \) are
\begin{align*} s_0 \amp = 1 \text{,} \amp s_1 = \amp 2 \text{,} \amp s_2 = \amp 4 \text{,} \amp s_3 = \amp 8 \text{,} \amp \dotsc\text{.} \end{align*}
Alternatively we may describe this sequence as \(\seq{2^n} \text{.}\)
Not all sequences can be described using an input-output formula — sequences can also be defined recursively.

Example 23.2.3. The Fibonacci sequence.

We start by specifying “initial values” \(s_0 = 1 \) and \(s_1 = 1 \text{,}\) and then require that each subsequent term satisfy
\begin{equation*} s_n = s_{n - 1} + s_{n - 2}, \qquad n \ge 2 \text{.} \end{equation*}
This creates a sequence with terms
\begin{equation*} 1, 1, 2, 3, 5, 8, 13, 21, \dotsc \text{.} \end{equation*}
Sometimes we only want to consider a “part” of a sequence.
We specify a tail by restricting the domain: \(\nseqtail{s} \text{.}\) Note that every tail of a tail is a tail of the original sequence.

Example 23.2.5. Positive and negative subsequences of an alternating sequence.

The sequence \(\lrseq{ \frac{(-1)^n}{n + 1} } \) has terms
\begin{equation*} 1, - \frac{1}{2}, \frac{1}{3}, - \frac{1}{4}, \dotsc \text{.} \end{equation*}
We can split this into two subsequences by considering only the even-indexed terms and skipping the odd-indexed terms, or vice versa:
\begin{align*} a_n \amp \colon \quad 1, \frac{1}{3}, \frac{1}{5}, \dotsc \\ b_n \amp \colon \quad -\frac{1}{2}, -\frac{1}{4}, -\frac{1}{6} \dotsc \end{align*}
We can describe \(\nseq{a} \) by the pattern \(\lrseq{ \frac{1}{2 n + 1} } \text{,}\) and can describe \(\nseq{b} \) by the pattern \(\lrseq{ - \frac{1}{2 (n + 1)} } \text{.}\)

Example 23.2.6. An “eventually” increasing sequence.

The sequence \(\lrseqtail[n \gt 0]{ \frac{2^n}{n^2} } \) has terms
\begin{equation*} 2, 1, \frac{8}{9}, 1, \frac{32}{25}, \frac{16}{9}, \dotsc \text{.} \end{equation*}
We may wish to omit the anomalous first two terms, as the sequence begins with a downward trend but then turns upward for \(n \ge 3 \text{.}\)
Plot of sequence values as points in the Cartesian plane.
So we may want to consider the sequence tail \(\lrseqtail[n \ge 3]{ \frac{2^n}{n^2} } \) instead.

Subsection 23.2.2 Bounded and monotonic sequences

Example 23.2.8.

  1. The sequence \(\seq{ \sqrt{n} } \) is bounded below by \(0 \text{.}\)
  2. The sequence \(\lrseq{ \frac{(-1)^n}{n + 1} } \) is bounded as it is bounded above/below by \(\pm 1 \text{,}\) for example.
  3. The sequence \(\lrseq{ \frac{(-3)^n}{n + 1} } \) has terms
    \begin{equation*} 1, - \frac{3}{2}, \frac{9}{3}, - \frac{27}{4}, \dotsc \text{,} \end{equation*}
    and is unbounded.

Definition 23.2.9. Increasing/decreasing sequences.

In all of these cases, the sequence is called monotonic.

Example 23.2.10.

  1. The sequence \(\lrseq{ \frac{3}{n + 5} } \) is decreasing because the denominator always grows from term to term.
  2. The sequence \(\lrseq{ 1 - \frac{1}{n + 1} } \) is increasing because from one term to the next we are subtracting a smaller amount from \(1 \text{.}\)
  3. The sequence \(\bbseq{ \cos(n\pi) } \) is not monotonic as it alternates between terms \(1 \) and \(-1 \text{.}\)
  4. The sequence \(\lrseq{ \floor{\frac{n}{2}} } \) is non-decreasing because no term is less than the previous, but it is not increasing because it repeats terms:
    \begin{equation*} 0, 0, 1, 1, 2, 2, 3, 3, \dotsc \text{.} \end{equation*}

Example 23.2.12.

We claimed in Example 23.2.6 that the tail \(\lrseqtail[n \ge 3]{ \frac{2^n}{n^2} } \) is an increasing sequence. We have
\begin{align*} s_{n+1} - s_n \amp = \frac{2^{n+1}}{{(n+1)}^2} - \frac{2^n}{n^2} \\ \amp = 2^n \left( \frac{2}{{(n+1)}^2} - \frac{1}{n^2} \right) \\ \amp = 2^n \left( \frac{2 n^2 - {(n+1)}^2}{n^2 {(n+1)}^2} \right) \\ \amp = 2^n \left( \frac{n^2 - 2n - 1}{n^2 {(n+1)}^2} \right) \text{.} \end{align*}
The denominator is obviously positive, as is the factor \(2^n \text{.}\) If we consider the numerator as a graph \(y = x^2 - 2 x - 1 \text{,}\) it is a concave up parabola with roots at \(x = 1 \pm \sqrt{2} \text{.}\) Since \(1 + \sqrt{2} \lt 3 \text{,}\) we can conclude \(s_{n+1} - s_n \) is positive for \(n \ge 3 \text{.}\)

Checkpoint 23.2.13.

Verify algebraically that the sequence \(\nseq{s} \) with
\begin{align*} s_0 \amp = 1 \text{,} \amp s_n \amp = \frac{n}{n^2 + 1} \text{ for } n \ge 1 \text{,} \end{align*}
is decreasing by verifying that
\begin{equation*} s_{n+1} - s_n \lt 0 \end{equation*}
always.
We can use the ideas of Chapter 19 to determine if a sequence is monotonic.

Example 23.2.14.

The sequence \(\lrseq{ \frac{n+1}{n+2} } \) is associated to the function \(s(n) = \frac{n+1}{n+2} \text{.}\) We can turn this into a continuous function:
\begin{equation*} f(x) = \frac{x + 1}{x + 2} \text{,} \qquad x \ge 0 \text{.} \end{equation*}
From
\begin{equation*} f'(x) = \frac{1}{{(x + 2)}^2} \gt 0 \end{equation*}
we see that the continuous version is always increasing, and so the sequence is increasing as well. Investigating a little further, the long-term behaviour of \(f \) as \(x \to \infty \) is that
\begin{equation*} f(x) \approx \frac{x}{x} = 1 \text{,} \end{equation*}
so the graph increases up to a horizontal asymptote at \(y = 1 \text{.}\) Double-checking, we have
\begin{equation*} f''(x) = \frac{-2}{{(x + 2)}^3} \end{equation*}
which is always negative on our stated domain for \(f \text{,}\) and so the graph of \(f \) is concave down on that domain. Putting all this together we have the following sketch of the graph.
Graph of function associated to a sequence, with sequence values plotted along the graph.
From the graph we may conclude that \(\nseq{s} \) is bounded below by \(1/2 \) and above by \(1 \text{.}\)
In the previous example, we were able to say that the sequence was bounded below simply because it was increasing. We can make similar conclusions about every monotonic sequence.

Note 23.2.16.

If we apply Fact 23.2.15 to a subsequence, then we should replace \(s_0 \) in the statement of the fact by the first term in the subsequence.

Example 23.2.17.

We verified in Example 23.2.12 that the sequence \(\lrseqtail[n \ge 3]{\frac{2^n}{n^2}} \) is increasing, and so it must be bounded below by \(s_3 = 8/9 \text{.}\)

Section 23.3 Sequence limits

Subsection 23.3.1 Definitions and examples

In Example 23.2.14 we determined that the sequence \(\lrseq{ \frac{n+1}{n+2} } \) satisfied \(s_n \approx 1 \) for large values of \(n \) by applying the concept of long-term behaviour from Chapter 4. If this is true, we should eventually see the terms of the sequence “clustered” near \(1 \text{.}\)

Definition 23.3.1. Limit of a sequence.

We say that \(\nseq{s} \) approaches \(L \) and write
\begin{equation*} \lim_{n \to \infty} s_n = L \end{equation*}
or simply \(\nseq{s} \to L \) if every open range around \(L \) contains all terms of some tail of the sequence \(\nseq{s} \text{.}\) In this case, \(L \) is called the limit of the sequence.

Warning 23.3.2.

The definition does not require some magic tail that is contained in every open range around \(L \text{.}\) Rather, for each specific open range one is required to demonstrate a specific tail that is contained therein. But the example tail demonstrated may vary as different open ranges are considered.

Example 23.3.3. Verifying the definition of limit.

Let’s verify that \(\lrseq{ \frac{n+1}{n+2} } \) converges to \(1 \text{.}\) For this, we consider an arbitrary open interval \((a, b) \) containing \(1 \text{,}\) so that \(a \lt 1 \lt b \) is true. We then must demonstrate that some tail of the sequence is contained entirely within that interval.
In Example 23.2.14 we observed that the associated continuous function
\begin{equation*} f(x) = \frac{x + 1}{x + 2} \end{equation*}
is an increasing function on domain \(x \ge 0 \) by considering its derivative. With some algebra we can determine that
\begin{equation*} f(x) = a \qquad \text{when} \qquad x = \frac{2a - 1}{1 - a} \text{.} \end{equation*}
Since \(f \) is increasing, we have
\begin{equation*} f(x) \gt a \qquad \text{when} \qquad x \gt \frac{2a - 1}{1 - a} \text{.} \end{equation*}
In particular, \(s_n \gt a \) for all positive integers satisfying
\begin{equation*} n \gt \frac{2a - 1}{1 - a} \text{.} \end{equation*}
Take \(N \) to represent the “next” integer that is greater than (or equal to) \((2a - 1)/(1 - a) \text{:}\)
\begin{equation*} N = \ceil{\frac{2a - 1}{1 - a}} \text{.} \end{equation*}
Then all terms of the tail \(\nseqtail{s} \) satisfy \(s_n \gt a \text{.}\)
On the other hand, \(s_n \lt 1 \) is true for all terms in the sequence, as the denominator is always one larger than the numerator. So we may say that
\begin{equation*} s_n \lt 1 \lt b\text{.} \end{equation*}
Combining, we have \(a \lt s_n \lt b \) for all terms in the tail \(\nseqtail{s} \text{.}\)

Checkpoint 23.3.4.

Verify that \(\nseq{s} \to 0 \) for \(s_n = \sqrt{n+1} - \sqrt{n} \text{.}\)
Hint.
Notice that all terms of this sequence are positive, so for an open interval \((a,b) \) containing zero, you only need to verify that eventually \(s_n \lt b \) becomes true for all terms past a certain point. (The endpoint \(a \) must be negative in this case, so \(s_n \gt a \) is always true.) With this in mind, rationalize the expression \(s_n / 1 \text{,}\) and then try to determine when the result will fall below \(b \text{.}\)
It is not possible for a sequence to cluster around two different values.

Justification.

Assume for the moment that \(L \lt K \text{.}\) Take \(b \) to be the average of \(L \) and \(K \text{.}\) Note that \(b \) is strictly greater than \(L \) and strictly less than \(K \text{.}\) If \(\nseq{s} \to L \text{,}\) then there is some tail \(\nseqtail{s} \) completely contained in the open range \((L - 1, b) \) around \(L \text{.}\) But then no tail can be completely contained in the open range \((b, K + 1) \) around \(K \text{,}\) because this range is disjoint from the range around \(L \text{.}\)
In the case \(L \gt K \text{,}\) make the same argument using range \((b, L + 1) \) around \(L \) and range \((K - 1, b) \) around \(K \text{.}\)

Remark 23.3.6.

If \(\nseq{s} \to L \text{,}\) then Fact 23.3.5 makes it appropriate to refer to \(L \) as the limit value of \(\nseq{s} \text{,}\) as there is only one such value possible.

Definition 23.3.7. Convergent/divergent sequences.

A sequence is called convergent if it has a limit, and divergent otherwise.

Example 23.3.8. Constant sequences are convergent.

A sequence \(\nseq{s} \) where \(s_n = c \) for all \(n \) converges to that constant term value \(c \text{,}\) because every open range containing \(c \) will contain the “tail” \(\nseqtail[0]{s} \text{.}\)

Example 23.3.9. An “oscillating” sequence.

The sequence \(\bbseq{\cos(\pi n / 2)} \) has terms
\begin{equation*} 1, 0, -1, 0, 1, 0, -1, 0, \dotsc \text{.} \end{equation*}
This sequence is divergent as its tails do not cluster around any one value.

Example 23.3.10. Increasing without bound.

In Example 23.2.12 we verified that the sequence \(\lrseq{ \frac{2^n}{n^2} } \) increases for \(n \ge 3 \) by looking at the difference between subsequent terms. Now let’s look at the ratio between subsequent terms:
\begin{equation*} \frac{s_{n+1}}{s_n} = \frac{2^{n+1}/{(n+1)}^2}{2^n/n^2} = 2 \; {\left(\frac{n}{n+1}\right)}^2 \text{.} \end{equation*}
Since \(n/(n+1) \approx 1 \) for large \(n \text{,}\) we can say that terms eventually almost double from one to the next. This sequence must be divergent, since the gap between terms is growing, and so they cannot cluster around any one value.
The previous example suggests “increasing without bound” is a situation we might sometimes wish to highlight.

Definition 23.3.11. Infinite limits.

  1. We write
    \begin{equation*} \lim_{n \to \infty} s_n = \infty \end{equation*}
    or simply \(\nseq{s} \to \infty \) if every open range \((M,\infty) \) contains all terms of some tail of the sequence \(\nseq{s} \text{.}\) In this case, we say that \(\nseq{s} \) diverges to infinity.
  2. We write
    \begin{equation*} \lim_{n \to \infty} s_n = -\infty \end{equation*}
    or simply \(\nseq{s} \to -\infty \) if every open range \((-\infty,M) \) contains all terms of some tail of the sequence \(\nseq{s} \text{.}\) In this case, we say that \(\nseq{s} \) diverges to negative infinity.

Example 23.3.12. Verifying the definition of divergent to infinity.

Suppose we want to verify \(\lrseq{ \frac{n^2 + 1}{n + 1} } \to \infty \text{,}\) which seems likely since the numerator grows larger than the denominator. The associated function
\begin{equation*} f(x) = \frac{x^2 + 1}{x + 1} \end{equation*}
is increasing for \(x \gt 1 \text{,}\) which can be verified by checking that \(f'(x) \gt 0 \) on this domain. We conclude that the tail \(\nseqtail[0]{s} \) is increasing, and so is bounded below by \(s_1 = 1 \text{.}\) (In fact, \(s_0 = 1 \) as well, so the full sequence is bounded below by \(1 \text{.}\))
Now we consider an arbitrary open range \((M,\infty) \text{.}\) If \(M \lt 1 \text{,}\) then all terms are within this range because of our established lower bound for the sequence. For \(M \ge 1 \text{,}\) we may algebraically solve
\begin{align*} f(x) \amp = M \amp \amp \implies \amp x \amp = \frac{M \pm \sqrt{M^2 + 4 (M - 1)}}{2}\text{.} \end{align*}
Note that the expression under the root sign is non-negative since we have assumed \(M \ge 1 \text{.}\) In fact, the expression under the root sign is at least \(M^2 \text{,}\) so one of these solutions is positive but the other is either \(0 \) (when \(M = 1 \)) or negative. Let \(x_+ \) represent the positive solution value, and let \(N \) represent the next largest integer. That is, \(N = \ceil{x_+} \text{.}\) Since \(f \) is increasing, we have
\begin{equation*} s_N = f(N) \gt f(x_+) = M \text{.} \end{equation*}
And since the sequence is increasing, for all \(n \ge N \) we will have
\begin{equation*} s_n \gt s_N \gt M \text{.} \end{equation*}
In other words, the tail \(\nseqtail{s} \) is entirely contained within the range \((M,\infty) \text{,}\) as required.

Subsection 23.3.2 Limits of subsequences

It should not be surprising that subsequences of convergent sequences follow their “parent” sequence to the limit value.

Justification.

Every tail of \(\nseq{a} \) will be a subsequence of a tail of \(\nseq{s} \text{.}\) For a given open range around \(L \text{,}\) there will be a tail of \(\nseq{s} \) that is entirely contained within that range, and therefore so will the tail \(\nseqtail{a} \) where \(a_N \) is the first term of \(\nseq{a} \) that appears in that tail of \(\nseq{s} \text{.}\)
Be careful that convergence of subsequences can be misleading: convergence of subsequences does not necessarily imply anything about convergence of the original sequence.

Example 23.3.14.

The sequence \(\bbseq{\cos(\pi n)} \) is divergent as it bounces between terms of \(1 \) and \(-1 \text{.}\) However, it has two obvious convergent subsequences:
\begin{gather*} 1, 1, 1, 1, \dotsc \to 1 \\ -1, -1, -1, -1, \dotsc \to -1 \text{.} \end{gather*}
However, there are some situations where convergence of subsequences is informative. For example, convergence of tails is equivalent to convergence of the original sequence.
The following two facts follow almost immediately from Fact 23.3.15.

Example 23.3.17.

The sequence \(\seq{1 / (n + 10)} \) contains the same terms as the tail of the sequence \(\seqtail[n \ge 1]{1 / n} \) that begins at index \(n = 10 \text{.}\) Since this second sequence converges to \(0 \text{,}\) so does the first.
And if we investigate enough subsequences, that is sufficient to determine convergence/divergence of the original sequence. Though what “enough” means depends on whether we are trying to establish convergence or divergence. For divergence, “enough” might mean only one or two. One in the case that you can identify a divergent subsequence, in which case the original sequence must diverge as well. Or it might mean two.

Justification.

Consider the contrapositive: If \(\nseq{s} \) converged, then subsequences \(\nseq{a} \) and \(\nseq{b} \) would necessarily converge to the same limit value as \(\nseq{s} \) (Pattern 23.3.13).
In the case of convergence, “enough” means that the subsequences we consider must collectively cover all terms of the original sequence (or at least all terms in some tail of the original sequence, by Fact 23.3.15). We will only state the fact below for the case of two such subsequences, but the statement could be extended to any number of subsequences, as long as they collectively include all terms of the original.

Justification.

First, if there is any overlap between \(\nseq{a} \) and \(\nseq{b} \text{,}\) replace \(\nseq{b} \) by a subsequence of itself with that overlap removed. If only a finite number of terms remain, that means that \(\nseq{a} \) contains a full tail of \(\nseq{s} \) and our conclusion follows from Fact 23.3.15. If an infinite number of terms remain in the new \(\nseq{b} \text{,}\) then that sequence still converges to the same limit as before (Pattern 23.3.13) and we still have that \(\nseq{a} \) and \(\nseq{b} \) together account for all of the terms of \(\nseq{s} \text{,}\) but now with no overlap.
Now let \(L \) represent the common limit value of \(\nseq{a} \) and \(\nseq{b} \text{,}\) and let \((c,d) \) be some open range containing \(L \text{.}\) Then there are tails \(\nseqtail[N_1]{a} \) and \(\nseqtail[N_2]{b} \) that are completely contained in \((c,d) \text{.}\) As \(\nseq{a} \) and \(\nseq{b} \) are subsequences, both \(a_{N_1} \) and \(b_{N_2} \) appear in \(\nseq{s} \text{,}\) but they represent different terms in \(\nseq{s} \) because we removed any overlap between \(\nseq{a} \) and \(\nseq{b} \text{.}\) Without loss of generality, assume that \(a_{N_1} \) appears in \(\nseq{s} \) before \(b_{N_2} \text{,}\) and let \(N \) be the index so that \(b_{N_2} = s_N \text{.}\) (Note that this does not imply that \(N_1 \lt N_2 \text{,}\) or that \(N = N_2 \text{.}\)) Every term in the tail \(\nseqtail{s} \) that appears in \(\nseq{a} \) must do so with index greater than \(N_1 \text{,}\) since such a term will appear after \(b_{N_2} \text{,}\) which appears after \(a_{N_1} \text{.}\) And every term in the tail \(\nseqtail{s} \) that appears in \(\nseq{b} \) must obviously do so with index at least \(N_2 \text{.}\) In other words, every term in the tail \(\nseqtail{s} \) is either in the tail \(\nseqtail[N_1]{a} \) or in the tail \(\nseqtail[N_2]{b} \text{,}\) each of which are completely contained in \((c,d) \text{.}\) Thus the tail \(\nseqtail{s} \) is completely contained in \((c,d) \text{.}\)

Remark 23.3.21.

Be careful about what “appears in” means, especially in the case of a sequence where certain values are repeated as terms. Effectively, you should think of that condition in the statement of Fact 23.3.20 to mean that the two subsequences can be “interleaved” together, discarding any overlap, to recover the original sequence. For example, for
\begin{equation*} \nseq{s} = 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, \dotsc \end{equation*}
we could take
\begin{align*} \nseq{a} \amp = 1, 1, 3, 3, 5, 5, 7, 7, \dotsc \\ \nseq{b} \amp = 2, 2, 3, 4, 4, 5, 6, 6, 7, \dotsc \end{align*}
and then we could recover \(\nseq{s} \) by discarding the duplication of the odd numbers in \(\nseq{b} \) and interleaving terms:
\begin{equation*} \nseq{s} = a_0, a_1, b_0, b_1, a_2, a_3, b_3, b_4, a_4, a_5, b_6, b_7, \dotsc \text{.} \end{equation*}
However, in general we should be careful not to do things like
\begin{align*} \nseq{a} \amp = 1, 3, 5, 7, \dotsc \\ \nseq{b} \amp = 2, 4, 6, 8, \dotsc \end{align*}
as we cannot simply interleave these two sequences back together to obtain \(\nseq{s} \text{.}\)

Subsection 23.3.3 Convergence versus boundedness

We think of convergent sequences as clustering around their limit values, which puts bounds on tails of the sequence. So there must be connections between boundedness and convergence.

Justification.

Suppose \(\nseq{s} \to L \text{,}\) and consider the open range \((L - 1, L + 1) \) around \(L \text{.}\) There there is some tail \(\nseqtail{s} \) entirely contained within that range, and so that tail is bounded above and below by \(L \pm 1 \text{,}\) respectively. The excluded sequence terms
\begin{equation*} s_0, s_1, s_2, \dotsc, s_{N-1} \end{equation*}
are a finite collection of numbers, and so have a maximum value and a minimum value. Let \(M \) represent that maximum excluded term value, unless that maximum is less than \(L + 1 \text{,}\) in which case let \(M = L + 1 \text{.}\) Similarly, let \(m \) represent either the minimum excluded term value or \(L - 1 \text{,}\) whichever is less. Then the full sequence satisfies \(m \le s_n \le M \text{.}\)
Be careful that a bounded sequence does not have to converge. For example, the sequence \(\bbseq{ \cos(n\pi) } \) is bounded by does not converge. However, a bounded sequence that doesn’t bounce around like this example must converge.

Justification.

We will consider only the case of a non-decreasing sequence; other cases are similar.
Suppose \(\nseq{s} \) is non-decreasing. We also assume that this sequence is bounded above, so let \(L \) be its least upper bound. Consider the open range \((a,b) \) containing \(L \text{.}\) Since \(L \) bounds the sequence above, for all \(n \) we have
\begin{equation*} s_n \le L \lt b\text{.} \end{equation*}
On the other hand, it cannot be that \(s_n \le a \) for all \(n \text{,}\) since then \(a \) would be an upper bound for \(\nseq{s} \) that is less than \(L \text{,}\) contradicting the assumption that \(L \) is the least upper bound. So suppose \(s_N \) is a particular term with \(s_N \gt a \text{.}\) As the sequence is non-decreasing, every term in the tail \(\nseqtail{s} \) must also be greater than \(a \text{.}\) Combining with \(s_n \lt b \) from above, we may conclude that the tail \(\nseqtail{s} \) is entirely contained in the range \((a,b) \text{.}\)

Example 23.3.25.

The sequence \(\lrseq{\ln\left(\frac{n+2}{n+1}\right)} \) is bounded below by \(0 \) because the ratio inside the logarithm is always greater than \(1 \text{.}\) And the sequence is decreasing because the associated function
\begin{equation*} f(x) = \ln\left(\frac{x+2}{x+1}\right) \end{equation*}
has negative derivative for positive values of \(x \text{.}\) Therefore, it must be convergent, even though we might not be sure of the value of its limit.
The following pattern is useful for comparing the convergence of one sequence to the known convergence of others.

Justification.

Consider open range \((a,b) \) containing \(L \text{.}\) Then there are tails \(\nseqtail[N_1]{r} \) and \(\nseqtail[n \ge N_2]{t} \) that are entirely contained within that range. Let \(N' \) represent the largest of \(N, N_1, N_2 \text{.}\) Then
\begin{equation*} a \lt r_n \le s_n \le t_n \lt b \qquad \text{for all } n \ge N' \text{,} \end{equation*}
so the tail \(\nseqtail[N']{s} \) is entirely contained within the range \((a,b) \text{.}\)

Example 23.3.27. Applying the Sequence Squeeze Theorem.

Consider
\begin{align*} r_n \amp = \frac{2}{n} \text{,} \amp s_n \amp = \frac{3 + {(-1)}^2}{n} \text{,} \amp t_n \amp = \frac{4}{n} \text{.} \end{align*}
We have
\begin{gather*} r_n \lt s_n = t_n \qquad \text{for even } n \text{,} \\ r_n = s_n \lt t_n \qquad \text{for odd } n \text{,} \end{gather*}
so for all \(n \) we can say \(r_n \le s_n \le t_n \text{.}\) Since both \(\nseq{r} \to 0 \) and \(\nseq{t} \to 0 \text{,}\) we can conclude \(\nseq{s} \to 0 \) as well.
Using the Sequence Squeeze Theorem, we can say more about bounded sequences even in the case that they are not monotonic.

Justification.

Suppose \(\nseq{s} \) can be bounded within the closed range \(R_0 = [a,b] \text{.}\) We will create a convergent subsequence \(\nseq{u} \) of \(\nseq{s} \text{,}\) one term at a time. First, set \(u_0 = s_0 \text{.}\)
Split \(R_0 \) in half to create two subintervals. Since \(R_0 \) contains every of the infinite number of terms in \(\nseq{s} \text{,}\) at least one of these two halves of \(R_0 \) must also contain an infinite number of terms. (But note that neither half necessarily has to contain a tail of \(\nseq{s} \text{.}\)) Let \(R_1 \) represent one of the halves of \(R_0 \) that contains an infinite number of terms, and set \(u_1 \) to be the term \(s_{k_1} \) of \(\nseq{s} \) of lowest index contained in \(R_1 \text{,}\) other than \(s_0 \) if that term happens to be \(R_1 \text{.}\)
Now split \(R_1 \) in two halves, and let \(R_2 \) represent one of the halves that contains an infinite number of terms. Then set \(u_2 \) to be the term \(s_{k_2} \) of \(\nseq{s} \) of lowest index contained in \(R_2 \text{,}\) but so that the index \(k_2 \) is greater than the previous index \(k_1 \text{.}\)
Continue in this fashion, splitting each interval in halves, choosing a half that contains an infinite number of terms of \(\nseq{s} \text{,}\) and then setting the next term in \(\nseq{u} \) to be a term of \(\nseq{s} \) contained in that half that is further along in \(\nseq{s} \) than any of the other terms already collected in \(\nseq{u} \text{.}\) As we do this, we have actually created three sequences:
  1. The sequence \(\nseq{a} \) of left endpoints of the closed intervals \(R_n \text{.}\)
  2. The sequence \(\nseq{b} \) of right endpoints of the closed intervals \(R_n \text{.}\)
  3. The sequence \(\nseq{u} \) of terms from \(\nseq{s} \text{.}\)
Then each \(R_n \) is the interval \([a_n, b_n] \) and contains \(u_n \text{,}\) so that
\begin{equation*} a_n \le u_n \le b_n \qquad \text{for all } n\text{.} \end{equation*}
At each step, either \(a_{n+1} = a_n \) and \(b_{n+1} \lt b_n \) (if \(R_{n+1} \) was chosen to be the left half of \(R_n \)), or \(a_{n+1} \gt a_n \) and \(b_{n+1} = b_n \) (if \(R_{n+1} \) was chosen to be the right half of \(R_n \)). So \(\nseq{a} \) is non-decreasing and \(\nseq{b} \) is non-increasing. Furthermore, both sequences are bounded within \(R_0 \text{.}\) By the Monotone Convergence Theorem, both sequences converge. Since the \(R_n \) intervals were created by this splitting-in-half procedure, the distance between an \(a_n \) term and the corresponding \(b_n \) term is
\begin{equation*} \frac{b - a}{2^n} \text{.} \end{equation*}
(Recall that \(a \) and \(b \) are the endpoints of the first range \(R_0 \text{.}\)) This distance gets very small for large \(n \text{,}\) so we may conclude that these two convergent sequences must have the same limit. (Can you verify this?)
Finally, we may apply the Sequence Squeeze Theorem to conclude that the constructed subsequence \(\nseq{u} \) converges to that common limit value of the two endpoint sequences.

Subsection 23.3.4 Sequence limit laws

The next law allows us to swap out sequence terms in a convergent sequence for the limit value.

Justification.

Suppose \((a,b) \) is an open range containing \(L \text{.}\) Then there is a tail \(\nseqtail{s} \) entirely contained within that range, so that
\begin{equation*} a \lt s_n \lt b \qquad \text{for all } n \ge N \text{.} \end{equation*}
But \(a \lt L \lt b \text{,}\) and so since each \(t_n \) is equal to either \(s_n \) or \(L \text{,}\) then also
\begin{equation*} a \lt t_n \lt b \qquad \text{for all } n \ge N \text{.} \end{equation*}
In other words, the tail \(\nseq{t} \) is also entirely contained within that range.
Convergence to \(0 \) is an important special case.

Justification.

Suppose it is true that every symmetric open range \((-c,c) \) containing zero contains some tail of the sequence. We must verify that it is then true that every open range \((a,b) \) containing zero, symmetric or not, contains some tail of the sequence. If this range contains zero, then necessarily \(a \lt 0 \) and \(b \gt 0 \text{.}\) Take \(c \) to be the lesser of \(\abs{a} \) and \(b \text{.}\) Then \(c \) is a positive value and the range \((-c,c) \) is contained entirely within \((a,b) \text{,}\) so that if \((-c,c) \) contains a tail of the sequence, then \((a,b) \) contains that same tail.
This next fact lets us to convert every question of convergence into a question of convergence to \(0 \text{.}\)

Justification.

Condition 1 implies Condition 2.
Assume \(\nseq{s} \to L \text{.}\) For this, we will not bother to apply Fact 23.3.30, and will simply consider arbitrary open range \((a,b) \) containing \(0 \text{.}\) Then \((a + L, b + L) \) is an open range containing \(L \text{,}\) and so contains a tail \(\nseqtail{s} \text{,}\) so that
\begin{equation*} a + L \le s_n \le b + L \qquad \text{for all } n \ge N \text{.} \end{equation*}
But then
\begin{equation*} a \le s_n - L \le b \qquad \text{for all } n \ge N \text{,} \end{equation*}
so that \((a,b) \) contains the tail \(\seqtail{s_n - L} \text{,}\) confirming that \(\seq{s_n - L} \to 0 \text{.}\)
Condition 2 implies Condition 1.
Assume \(\seq{s_n - L} \to 0 \text{.}\) Then, arguing similarly to the previous case, the open range \((a,b) \) around \(L \) will contain the tail \(\nseqtail{s} \) as long as the open range \((a - L, b - L) \) around \(0 \) contains the tail \(\seqtail{s_n - L} \text{.}\) This confirms that \(\nseq{s} \to L \text{.}\)
Condition 2 is equivalent to Condition 3.
Considering symmetric open range \((-c,c) \) containing \(0 \text{,}\) then for every number \(x \) we have \(-c \lt x \lt c \) if and only if \(0 \le \abs{x} \lt c \text{.}\) So a tail \(\seqtail{s_n - L} \) is entirely contained within \((-c, c) \) if and only if the corresponding tail \(\absseqtail{s_n - L} \) is entirely contained within \([0,c) \text{,}\) which is itself entirely contained within \((-c,c) \text{.}\) So if one of the two sequences converges, so does the other.
The following combines Fact 23.3.30 with the case of \(L = 0 \) in Fact 23.3.31, along with the fact that an absolute sequence \(\nabsseq{s} \) never contains negative terms.
Convergence of a sequence lets us bound the magnitude of the terms.

Justification.

Fact 23.3.31 implies that the statement is true in the case of \(L = 0 \text{.}\) So assume \(L \neq 0 \text{,}\) and assume \((a,b) \) is an open range around \(\abs{L} \text{.}\) We may assume that \(a \ge 0 \text{,}\) since if we can determine a tail of \(\nabsseq{s} \) that is contained in a smaller range with \(a \ge 0 \text{,}\) that same tail will also be contained in any larger range with \(a \lt 0 \text{.}\)
If \(L \gt 0 \) then \((a,b) \) also contains \(L \text{,}\) and so some tail \(\nseqtail{s} \) is entirely contained in \((a,b) \text{.}\) But since \(a \ge 0 \text{,}\) \((a,b) \) is a range of positive numbers, and so \(\nabsseqtail{s} \) and \(\nseqtail{s} \) are identical sequences. Therefore, we may conclude that \(\nabsseqtail{s} \) is entirely contained in \((a,b) \text{,}\) as required.
If \(L \lt 0 \) then \((-b,-a) \) contains \(L \text{,}\) and so some tail \(\nseqtail{s} \) is entirely contained in \((-b,-a) \text{.}\) But \((-b,-a) \) is a range of negative numbers, and so \(s_n \) is negative for all \(n \ge N \text{.}\) Therefore, we can turn
\begin{equation*} -b \lt s_n \lt -a \qquad\text{for all } n \ge N \end{equation*}
into
\begin{equation*} a \lt \abs{s_n} \lt b \qquad\text{for all } n \ge N \text{,} \end{equation*}
which says that the tail \(\nabsseqtail{s} \) is entirely contained in \((a,b) \text{,}\) as required.

Warning 23.3.34.

While we have an equivalence of statements in Fact 23.3.31, the converse of Fact 23.3.33 is false when \(L \neq 0 \text{.}\) For example, the alternating sequence defined by \(s_n = (-1)^n \) is divergent, but its sequence of absolute values is constant and so converges.
The following limit laws allow us to convert the question of convergence of a sequence into questions about the convergence of its “parts”. The proofs of the individual laws will make frequent use of Fact 23.3.31, Corollary 23.3.32, and Fact 23.3.33 without comment, and the reader is encouraged to be on the lookout for their applications.

Justification for Statement 1.

We will verify that
\begin{equation*} \lrabsseq{(s_n + t_n) - (L + M)} \to 0\text{,} \end{equation*}
so consider haf-open range \([0,c) \text{.}\) As \(\nseq{s} \to L \) and \(\nseq{t} \to M \text{,}\) there are absolute sequence tails \(\absseqtail[n \ge N_1]{s_n - L} \) and \(\absseqtail[n \ge N_2]{t_n - M} \) that are both entirely contained within \([0, c/2) \text{.}\) Let \(N \) represent the larger of the two tail boundaries \(N_1 \) and \(N_2 \text{.}\) Then for \(n \ge N \) we can say both
\begin{align*} \abs{s_n - L} \amp \lt \frac{c}{2} \text{,} \amp \abs{t_n - M} \amp \lt \frac{c}{2} \text{.} \end{align*}
Then
\begin{align*} \abs{(s _n + t_n) - (L + M)} \amp = \abs{(s_n - L) + (t_n - M)} \\ \amp \le \abs{s_n - L} + \abs{t_n - M} \amp \amp\text{(triangle inequality)} \\ \amp \lt \frac{c}{2} + \frac{c}{2} \\ \amp = c \text{,} \end{align*}
and so the absolute sequence tail \(\lrabsseqtail{(s_n + t_n) - (L + M)} \) is entirely contained within \([0,c) \text{,}\) as required.

Justification for Statement 2.

First, in the case that \(k = 0 \text{,}\) then \(\seq{k s_n} \) is a constant sequence and converges to that constant value \(k L = 0 \text{.}\)
Next, assume \(k \gt 0 \text{,}\) and consider an open range \((a,b) \) around \(k L \text{.}\) Then \((a/k, b/k) \) is an open range around \(L \text{,}\) and therefore contains some tail \(\nseqtail{s} \text{.}\) Scaling the terms of this tail creates a tail \(\seqtail{k s_n} \) that is entirely contained within \((a,b) \text{,}\) as required.
Finally, the case \(k \lt 0 \) is handled similarly as above, with the only difference that “scaling down” the range \((a,b) \) will also reflect it, since the inequality \(a \lt b \) gets flipped to \(b/k \lt a/k \) when dividing by a negative.

Justification for Statement 3.

We will handle this in a number of cases, beginning with the simplest.
Case \(L = M = 0 \).
Here we wish to verify \(\seq{s_n t_n} \to 0 \text{.}\) Consider half-open range \([0,c) \text{.}\) As both \(\nseq{s} \) and \(\nseq{t} \) converge to zero in the present case, there are tails \(\absseqtail[n \ge N_1]{s_n} \) and \(\absseqtail[n \ge N_2]{t_n} \) that are both entirely contained within \([0, \sqrt{c}) \text{.}\) Let \(N \) represent the larger of the two tail boundaries \(N_1 \) and \(N_2 \text{.}\) Then for \(n \ge N \) we have
\begin{equation*} \abs{s_n t_n} = \abs{s_n} \abs{t_n} \lt \sqrt{c} \cdot \sqrt{c} = c \text{,} \end{equation*}
so that the tail \(\absseqtail{s_n t_n} \) is entirely contained within \([0, c) \text{,}\) as required.
Case \(M = 0 \text{,}\) \(L \neq 0 \).
Again we wish to verify \(\seq{s_n t_n} \to 0 \) in this case. So consider half-open range \([0,c) \text{.}\) Since \(L \neq 0 \text{,}\) \(\bbrac{0, 2\abs{L}} \) is an open range containing \(\abs{L} \text{,}\) and so from \(\abs{s_n} \to L \) we may obtain a tail \(\absseqtail[n \ge N_1]{s_n} \) that is entirely contained within that range. There is also a tail \(\absseqtail[n \ge N_2]{t_n} \) entirely contained within half-open range \([0,d) \) for \(d = c/2\abs{L} \text{.}\) Let \(N \) represent the larger of the two tail boundaries \(N_1 \) and \(N_2 \text{.}\) Then for \(n \ge N \) we have
\begin{equation*} \abs{s_n t_n} = \abs{s_n} \abs{t_n} \lt 2 \abs{L} \cdot \frac{c}{2 \abs{L}} = c \text{,} \end{equation*}
so that the tail \(\absseqtail{s_n t_n} \) is entirely contained within \([0, c) \text{,}\) as required.
Case \(L = 0 \text{,}\) \(M \neq 0 \).
This is exactly as in the previous case, just with the roles of the two sequences reversed.
Case \(L \neq 0 \text{,}\) \(M \neq 0 \).
We will demonstrate that \(\absseq{s_n t_n - L M} \to 0 \text{.}\) So consider half-open range \([0, c/2) \text{.}\) Since \(\absseq{s_n - L} \to 0 \text{,}\) there is a tail \(\absseqtail[n \ge N_1]{s_n - L} \) that is entirely contained within the range \([0,d) \) for \(d = \abs{c/2M} \text{.}\) (Note that we would not be able to do this when \(M = 0 \text{.}\)) So for \(n \ge N_1 \) we have
\begin{align*} \abs{s_n t_n - L M} \amp = \abs{s_n t_n - s_n M + s_n M - L M} \amp \text{(adding 0, no change)}\\ \amp \le \abs{s_n t_n - s_n M} + \abs{s_n M - L M} \amp \text{(triangle inequality)}\\ \amp = \abs{s_n} \abs{t_n - M} + \abs{s_n - L} \abs{M} \\ \amp \lt \abs{s_n} \abs{t_n - M} + d \abs{M} \\ \amp \lt \abs{s_n} \abs{t_n - M} + \frac{c}{2} \text{.} \end{align*}
Since \(\nabsseq{s} \to \abs{L} \) and \(L \neq 0 \text{,}\) there is also a tail \(\absseqtail[n \ge N_2]{s_n} \) that is entirely contained within the range \(\bigl[0,2 \abs{L}\bigr) \text{.}\) And \(\absseq{t_n - M} \to 0 \text{,}\) so there is a tail \(\absseqtail[n \ge N_3]{t_n - M} \) that is entirely contained within the range \([0,d) \) for \(d = c / 4 \abs{L} \text{.}\) Let \(N \) represent the greater of the three tail boundaries \(N_1 \text{,}\) \(N_2 \text{,}\) and \(N_3 \text{.}\) Then for \(n \ge N \) we may continue the calculation above:
\begin{align*} \abs{s_n t_n - L M} \amp \lt \abs{s_n} \abs{t_n - M} + \frac{c}{2} \\ \amp \lt 2 \abs{L} \cdot \frac{c}{4 \abs{L}} + \frac{c}{2} \\ \amp = \frac{c}{2} + \frac{c}{2} \\ \amp = c \text{.} \end{align*}
That is, the tail \(\absseqtail{s_n t_n - L M} \) is entirely contained in the range \([0, c) \text{,}\) as required.

Justification for Statement 4.

Again we will verify \(\absseq{1/t_n - 1/M} \to 0 \text{,}\) so consider a half-open range \([0,c) \text{.}\) Without loss of generality, we may assume \(c \lt \abs{M} \text{,}\) since if we obtain a tail contained in a small range then that same tail will be contained in every larger half-open range beginning at \(0 \text{.}\) Let \(A \) represent the difference \(\abs{M} - c \text{;}\) under our assumption on \(c \text{,}\) \(A \) is positive. Since \(\nabsseq{t} \to \abs{M} \text{,}\) there is some tail \(\absseqtail[n \ge N_1]{t_n} \) entirely contained in the open range \(\bbrac{A, \abs{M} + c} \text{,}\) and we can say that all terms \(t_n \) in this tail satisfy \(\abs{t_n} \gt A \text{.}\)
So for \(n \ge N_1 \text{,}\) we have
\begin{align*} \abs{\frac{1}{t_n} - \frac{1}{M}} \amp = \abs{\frac{M - t_n}{t_n M}} \\ \amp = \frac{\abs{M - t_n}}{\abs{t_n} \abs{M}} \\ \amp = \frac{\abs{t_n - M}}{\abs{t_n} \abs{M}} \\ \amp \lt \frac{\abs{t_n - M}}{A \abs{M}} \text{.} \end{align*}
Because \(\abs{t_n - M} \to 0 \text{,}\) there is some tail \(\absseqtail[n \ge N_2]{t_n - M} \) entirely contained within \(\bigl[0, c A \abs{M}\bigr) \text{.}\) If we take \(N \) to be the greater of \(N_1 \) and \(N_2 \text{,}\) we may continue the above calculation, so that for \(n \ge N \) we have
\begin{align*} \abs{\frac{1}{t_n} - \frac{1}{M}} \amp \lt \frac{\abs{t_n - M}}{A \abs{M}} \\ \amp \lt \frac{c A \abs{M}}{A \abs{M}} \\ \amp = c \text{.} \end{align*}
This says that the tail \(\absseqtail{1/t_n - 1/M} \) is entirely contained within \([0, c) \text{,}\) as required.

Justification for Statement 6.

We assume \(L \neq 0 \text{,}\) so there is a tail \(\absseqtail[n \ge N_1]{s_n} \) that is entirely contained in the open range \(\bbrac{\abs{L}/2, 2 \abs{L}} \text{.}\) Given a large, positive number \(K \text{,}\) there is some tail \(\absseqtail[n \ge N_2]{t_n} \) that is entirely contained in the range \([0, d) \) for \(d = \abs{L}/2K \text{.}\) Take \(N \) to be the greater of the two tail boundaries \(N_1 \) and \(N_2 \text{.}\) But then for \(n \ge N \) we have
\begin{equation*} \abs{s_n} \gt \frac{\abs{L}}{2} \end{equation*}
and
\begin{equation*} \abs{t_n} \lt \frac{\abs{L}}{2 K} \qquad\implies\qquad \frac{1}{\abs{t_n}} \gt \frac{2 K}{\abs{L}} \text{.} \end{equation*}
But then
\begin{equation*} \abs{\frac{s_n}{t_n}} = \abs{s_n} \cdot \frac{1}{\abs{t_n}} \gt \frac{\abs{L}}{2} \cdot \frac{2 K}{\abs{L}} = K \text{.} \end{equation*}
We have proved that every unbounded open range \((K,\infty) \) with \(K \gt 0 \) contains a tail of \(\absseq{s_n / t_n} \text{,}\) so it is impossible that tails of that absolute sequence will ever become bounded around some potential limit value. Therefore, \(\absseq{s_n / t_n} \) diverges, and hence so must \(\seq{s_n / t_n} \text{.}\)
The following example effectively justifies our dominant terms approach to long-term behaviour in Chapter 4.

Example 23.3.36. Dominant terms in a rational sequence.

Consider \(\nseq{s} \) for
\begin{equation*} s_n = \frac{n^2 - 1}{2 n^2 + 1} \text{.} \end{equation*}
The numerator and denominator sequences both diverge to infinity, but for \(n \gt 0 \) we may manipulate algebraically to obtain convergent numerator and denominator:
\begin{equation*} s_n = \frac{n^2 - 1}{2 n^2 + 1} \cdot \frac{1/n^2}{1/n^2} = \frac{1 - 1/n^2}{2 + 1/n^2}\text{.} \end{equation*}
Since \(\seqtail[n \gt 0]{1/n^2} \to 0 \text{,}\) we may apply Statement 1 of the Sequence limit laws to both numerator and denominator to conclude
\begin{align*} \seqtail[n \gt 0]{1 - 1/n^2} \amp \to 1 \amp \seqtail[n \gt 0]{2 + 1/n^2} \amp \to 2\text{.} \end{align*}
Then apply Statement 5 of the Sequence limit laws to see that the tail \(\seqtail[n \gt 0]{s_n} \to 1/2 \text{.}\) Since this tail converges to \(1/2 \text{,}\) we must have \(\nseq{s} \to 1/2 \) as well.
Finally, we record some patterns about how limits compare if we can compare their terms.

Justification for Statement 1.

First we consider the case \(C = 0 \text{,}\) We assume \(s_n \ge 0 \) for all \(n \text{,}\) and wish to prove that \(L \ge 0 \text{.}\) If it were possible that \(L \lt 0 \text{,}\) then the open range \((2L, 0) \) that contains \(L \) would also contain some tail of \(\nseq{s} \text{.}\) But that range only contains negative numbers, and no term of \(\nseq{s} \) is negative. Since \(L \lt 0 \) is impossible, we must have \(L \ge 0 \text{.}\)
For \(C \neq 0 \text{,}\) simply apply the argument above to the sequence \(\seq{s_n - C} \text{,}\) which converges to \(L - C \text{.}\)

Justification for Statement 2.

The sequence \(\seq{-s_n} \) converges to \(-L \) and satisfies \(-s_n \ge -C \) for all \(n \text{,}\) so we may apply Statement 1 to conclude that \(-L \ge -C \text{.}\)

Justification for Statement 3.

The sequence \(\seq{s_n - t_n} \) converges to \(L - M \) and satisfies \(s_n - t_n \ge 0 \) for all \(n \text{,}\) so we may apply Statement 1 with \(C = 0 \) to conclude that \(L - M \ge 0 \text{.}\)

Remark 23.3.38.

The conditions in Pattern 23.3.37 do not need to hold for all \(n \) in order for the conclusions to be true — it is sufficient for them to hold for some tail of the sequence involved.
PrevTopNext