The exercises below consolidate the basic notions concerning sequences: definition, notation, general term, recurrence, monotonicity, boundedness, arithmetic and geometric progressions, graphical representation, and a first look at subsequences.
In each exercise we apply the definitions explicitly, making clear not only the final answer but also the correct line of reasoning.
Exercise 1 — level ★☆☆☆☆
Write the first five terms of the sequence defined by
\[ a_n=2n-1,\qquad n\ge 1. \]
Answer
The first five terms are
\[ 1,\ 3,\ 5,\ 7,\ 9. \]
Solution
The sequence is given by its general term
\[ a_n=2n-1. \]
To find the first five terms we substitute the values \(1,2,3,4,5\) for \(n\).
For \(n=1\) we obtain
\[ a_1=2\cdot 1-1=1. \]
For \(n=2\) we obtain
\[ a_2=2\cdot 2-1=3. \]
For \(n=3\) we obtain
\[ a_3=2\cdot 3-1=5. \]
Similarly,
\[ a_4=2\cdot 4-1=7, \qquad a_5=2\cdot 5-1=9. \]
Hence the first five terms of the sequence are
\[ 1,\ 3,\ 5,\ 7,\ 9. \]
Notice that these are the first odd natural numbers. Nevertheless, the sequence is not merely the set of odd numbers: it is an ordered list, in which the first term is \(1\), the second is \(3\), the third is \(5\), and so on.
Exercise 2 — level ★☆☆☆☆
Determine the first four terms of the sequence
\[ a_n=\frac{1}{n+1}, \]
in the following two cases:
- \(n\ge 0\);
- \(n\ge 1\).
Answer
If \(n\ge 0\), the first four terms are
\[ 1,\ \frac12,\ \frac13,\ \frac14. \]
If \(n\ge 1\), the first four terms are
\[ \frac12,\ \frac13,\ \frac14,\ \frac15. \]
Solution
The formula for the general term is the same in both cases:
\[ a_n=\frac{1}{n+1}. \]
What changes, however, is the starting index of the sequence.
If \(n\ge 0\), the first index is \(0\). Hence the first four terms correspond to \(n=0,1,2,3\).
We compute:
\[ a_0=\frac{1}{0+1}=1, \]
\[ a_1=\frac{1}{1+1}=\frac12, \]
\[ a_2=\frac{1}{2+1}=\frac13, \]
\[ a_3=\frac{1}{3+1}=\frac14. \]
Thus, if \(n\ge 0\), the sequence begins with
\[ 1,\ \frac12,\ \frac13,\ \frac14,\ldots \]
If instead \(n\ge 1\), the first index is \(1\). The first four terms then correspond to \(n=1,2,3,4\).
We compute:
\[ a_1=\frac12,\qquad a_2=\frac13,\qquad a_3=\frac14,\qquad a_4=\frac15. \]
Thus, if \(n\ge 1\), the sequence begins with
\[ \frac12,\ \frac13,\ \frac14,\ \frac15,\ldots \]
This exercise shows that one and the same formula can generate different sequences when the index set changes.
Exercise 3 — level ★☆☆☆☆
Decide whether the formula
\[ a_n=\frac{1}{n-2} \]
defines a real-valued sequence for every \(n\ge 1\). If it does not, indicate from which index onward it may be considered.
Answer
The formula does not define a real-valued sequence for every \(n\ge 1\), since for \(n=2\) the denominator vanishes. It may be considered, for instance, for \(n\ge 3\).
Solution
In order to define a real-valued sequence, the term \(a_n\) must be a real number for every admissible index.
The formula is
\[ a_n=\frac{1}{n-2}. \]
The denominator is
\[ n-2. \]
This denominator vanishes when
\[ n-2=0. \]
Hence
\[ n=2. \]
For \(n=2\) we would have
\[ a_2=\frac{1}{2-2}=\frac10, \]
which is undefined.
Hence the formula does not define a real-valued sequence for all indices \(n\ge 1\).
To avoid this difficulty, we may consider the sequence starting from \(n=3\). In that case we obtain
\[ a_3=1,\qquad a_4=\frac12,\qquad a_5=\frac13,\qquad a_6=\frac14,\ldots \]
Thus the formula correctly defines a real-valued sequence if we take, for instance,
\[ n\ge 3. \]
Exercise 4 — level ★★☆☆☆
Given the sequence
\[ a_n=\frac{n+1}{n}, \qquad n\ge 1, \]
write its first four terms and rewrite the general term in the form \(1+\displaystyle \frac1n\).
Answer
The first four terms are
\[ 2,\ \frac32,\ \frac43,\ \frac54. \]
Moreover,
\[ a_n=1+\frac1n. \]
Solution
We compute the first terms by substituting \(n=1,2,3,4\).
For \(n=1\),
\[ a_1=\frac{1+1}{1}=2. \]
For \(n=2\),
\[ a_2=\frac{2+1}{2}=\frac32. \]
For \(n=3\),
\[ a_3=\frac{3+1}{3}=\frac43. \]
For \(n=4\),
\[ a_4=\frac{4+1}{4}=\frac54. \]
Hence the first four terms are
\[ 2,\ \frac32,\ \frac43,\ \frac54. \]
Now we rewrite the general term:
\[ \frac{n+1}{n}=\frac{n}{n}+\frac{1}{n}. \]
Since
\[ \frac{n}{n}=1, \]
we obtain
\[ a_n=1+\frac1n. \]
This form is often more telling than the original one, since it shows that each term is obtained by adding to \(1\) the quantity \(\displaystyle \frac1n\).
Exercise 5 — level ★★☆☆☆
Consider the sequence defined recursively by
\[ a_1=4,\qquad a_{n+1}=a_n+5\quad \text{for every } n\ge 1. \]
Write the first five terms and find an explicit formula for \(a_n\).
Answer
The first five terms are
\[ 4,\ 9,\ 14,\ 19,\ 24. \]
The explicit formula is
\[ a_n=4+5(n-1). \]
Equivalently,
\[ a_n=5n-1. \]
Solution
The sequence is defined recursively. This means that each term is obtained from the preceding one.
We know that
\[ a_1=4. \]
Moreover,
\[ a_{n+1}=a_n+5. \]
Hence each subsequent term is obtained by adding \(5\) to the previous one.
We compute:
\[ a_2=a_1+5=4+5=9, \]
\[ a_3=a_2+5=9+5=14, \]
\[ a_4=a_3+5=14+5=19, \]
\[ a_5=a_4+5=19+5=24. \]
The first five terms are therefore
\[ 4,\ 9,\ 14,\ 19,\ 24. \]
To find the explicit formula, observe that in passing from \(a_1\) to \(a_n\) we add \(5\) exactly \(n-1\) times.
Therefore
\[ a_n=4+5(n-1). \]
Expanding,
\[ a_n=4+5n-5=5n-1. \]
Hence an explicit formula for the sequence is
\[ a_n=5n-1. \]
Exercise 6 — level ★★☆☆☆
Consider the sequence defined by
\[ b_1=3,\qquad b_{n+1}=2b_n\quad \text{for every } n\ge 1. \]
Write the first five terms and identify the type of sequence.
Answer
The first five terms are
\[ 3,\ 6,\ 12,\ 24,\ 48. \]
This is a geometric progression with first term \(3\) and common ratio \(2\).
Solution
The sequence is defined recursively:
\[ b_1=3, \qquad b_{n+1}=2b_n. \]
This means that each subsequent term is obtained by multiplying the previous one by \(2\).
We compute:
\[ b_2=2b_1=2\cdot 3=6, \]
\[ b_3=2b_2=2\cdot 6=12, \]
\[ b_4=2b_3=2\cdot 12=24, \]
\[ b_5=2b_4=2\cdot 24=48. \]
Hence the first five terms are
\[ 3,\ 6,\ 12,\ 24,\ 48. \]
Since each term is obtained from the preceding one by multiplying always by the same number, the sequence is geometric.
The first term is
\[ b_1=3, \]
while the common ratio is
\[ q=2. \]
The explicit formula is therefore
\[ b_n=3\cdot 2^{n-1}. \]
Exercise 7 — level ★★☆☆☆
Decide whether the sequence
\[ a_n=7,\qquad n\ge 1, \]
is increasing, decreasing, and bounded.
Answer
The sequence is constant. Hence it is both non-strictly increasing and non-strictly decreasing. Moreover, it is bounded.
Solution
The sequence is
\[ a_n=7\quad \text{for every } n\ge 1. \]
This means that all its terms are equal to \(7\):
\[ 7,\ 7,\ 7,\ 7,\ldots \]
To check whether it is increasing, we must verify whether
\[ a_n\le a_{n+1}\quad \text{for every } n\ge 1. \]
In this case
\[ a_n=7 \qquad \text{and} \qquad a_{n+1}=7. \]
Hence
\[ a_n=a_{n+1}. \]
In particular,
\[ a_n\le a_{n+1}. \]
Hence the sequence is non-strictly increasing.
Similarly, since
\[ a_n=a_{n+1}, \]
we also have
\[ a_n\ge a_{n+1}. \]
Hence the sequence is also non-strictly decreasing.
Finally, the sequence is bounded, since all its terms coincide with \(7\). For example, we may write
\[ 6\le a_n\le 8\quad \text{for every } n\ge 1. \]
In fact, the set of values taken by the sequence is simply
\[ \{7\}. \]
This is enough to conclude that the sequence is bounded: indeed, all its terms remain equal to \(7\).
Exercise 8 — level ★★☆☆☆
Prove that the sequence
\[ a_n=n^2+1, \qquad n\ge 1, \]
is strictly increasing.
Answer
The sequence is strictly increasing.
Solution
To prove that a sequence is strictly increasing, we must show that
\[ a_n<a_{n+1}\quad \text{for every } n\ge 1. \]
Equivalently, we may show that
\[ a_{n+1}-a_n>0\quad \text{for every } n\ge 1. \]
In our case
\[ a_n=n^2+1. \]
We compute the next term:
\[ a_{n+1}=(n+1)^2+1. \]
Hence
\[ a_{n+1}-a_n=\bigl((n+1)^2+1\bigr)-(n^2+1). \]
We expand:
\[ (n+1)^2+1=n^2+2n+1+1=n^2+2n+2. \]
Then
\[ a_{n+1}-a_n=(n^2+2n+2)-(n^2+1)=2n+1. \]
Since \(n\ge 1\), we have
\[ 2n+1>0. \]
Therefore
\[ a_{n+1}-a_n>0\quad \text{for every } n\ge 1. \]
Consequently,
\[ a_n<a_{n+1}\quad \text{for every } n\ge 1, \]
and the sequence is strictly increasing.
Exercise 9 — level ★★☆☆☆
Prove that the sequence
\[ a_n=\frac1n, \qquad n\ge 1, \]
is strictly decreasing.
Answer
The sequence is strictly decreasing.
Solution
To prove that a sequence is strictly decreasing, we must show that
\[ a_{n+1}<a_n\quad \text{for every } n\ge 1. \]
In our case
\[ a_n=\frac1n. \]
The next term is
\[ a_{n+1}=\frac{1}{n+1}. \]
Since
\[ n+1>n \]
and since \(n\) and \(n+1\) are positive, on passing to reciprocals the direction of the inequality is reversed:
\[ \frac{1}{n+1}<\frac1n. \]
That is,
\[ a_{n+1}<a_n\quad \text{for every } n\ge 1. \]
Hence the sequence
\[ \left(\frac1n\right)_{n\ge 1} \]
is strictly decreasing.
This example is important because it shows that a decreasing sequence need not become negative: indeed, all the terms of the sequence are positive.
Exercise 10 — level ★★☆☆☆
Decide whether the sequence
\[ a_n=(-1)^n, \qquad n\ge 1, \]
is increasing, decreasing, or monotone.
Answer
The sequence is neither increasing nor decreasing. Consequently, it is not monotone.
Solution
We compute the first terms of the sequence:
\[ a_1=(-1)^1=-1, \]
\[ a_2=(-1)^2=1, \]
\[ a_3=(-1)^3=-1, \]
\[ a_4=(-1)^4=1. \]
Hence the sequence is
\[ -1,\ 1,\ -1,\ 1,\ldots \]
For it to be increasing, we would need
\[ a_n\le a_{n+1}\quad \text{for every } n\ge 1. \]
However, for \(n=2\) we have
\[ a_2=1 \qquad \text{and} \qquad a_3=-1. \]
Hence
\[ a_2>a_3. \]
This is enough to conclude that the sequence is not increasing.
For it to be decreasing, we would need
\[ a_n\ge a_{n+1}\quad \text{for every } n\ge 1. \]
However, for \(n=1\) we have
\[ a_1=-1 \qquad \text{and} \qquad a_2=1. \]
Hence
\[ a_1<a_2. \]
This is enough to conclude that the sequence is not decreasing.
Since a monotone sequence is one that is either increasing or decreasing, the given sequence is not monotone.
Exercise 11 — level ★★★☆☆
Investigate the boundedness of the sequence
\[ a_n=\frac{n}{n+1}, \qquad n\ge 1. \]
Determine a lower bound, an upper bound, the infimum, and the supremum of the set of values taken.
Answer
The sequence is bounded. We have
\[ 0<a_n<1\quad \text{for every } n\ge 1. \]
A lower bound is \(0\), an upper bound is \(1\). Moreover,
\[ \inf\{a_n:n\ge 1\}=\frac12, \qquad \sup\{a_n:n\ge 1\}=1. \]
Solution
Consider the sequence
\[ a_n=\frac{n}{n+1}. \]
Since \(n\ge 1\), both \(n\) and \(n+1\) are positive. Hence
\[ \frac{n}{n+1}>0. \]
Hence \(0\) is a lower bound of the sequence.
Moreover, since
\[ n<n+1, \]
dividing by \(n+1>0\) we obtain
\[ \frac{n}{n+1}<1. \]
Hence \(1\) is an upper bound of the sequence.
We therefore have
\[ 0<a_n<1\quad \text{for every } n\ge 1. \]
In particular, the sequence is bounded.
We now determine the infimum and the supremum of the set of values taken.
We compute the first terms:
\[ a_1=\frac12,\qquad a_2=\frac23,\qquad a_3=\frac34,\qquad a_4=\frac45. \]
The sequence is increasing, because
\[ a_n=1-\frac{1}{n+1}. \]
As \(n\) increases, the quantity \(\frac{1}{n+1}\) decreases; hence \(1-\frac{1}{n+1}\) increases.
The first term is
\[ a_1=\frac12. \]
Since the sequence is increasing, the smallest value it attains is \(\displaystyle \frac12\). Therefore
\[ \min\{a_n:n\ge 1\}=\inf\{a_n:n\ge 1\}=\frac12. \]
Notice that the supremum is not a term of the sequence, since there is no \(n\ge 1\) such that
On the other hand, all the terms are less than \(1\), yet they come arbitrarily close to \(1\). Hence \(1\) is the least of all upper bounds.
Therefore
\[ \sup\{a_n:n\ge 1\}=1. \]
Notice that the supremum is not a term of the sequence, since there is no \(n\ge 1\) such that
\[ \frac{n}{n+1}=1. \]
Exercise 12 — level ★★★☆☆
Prove that the sequence
\[ a_n=\frac{(-1)^n}{n}, \qquad n\ge 1, \]
is bounded.
Answer
The sequence is bounded. Indeed,
\[ |a_n|\le 1\quad \text{for every } n\ge 1. \]
Solution
To prove that a sequence is bounded, we may use the criterion involving the absolute value.
A sequence \((a_n)\) is bounded if there exists a real number \(K>0\) such that
\[ |a_n|\le K\quad \text{for every } n\ge 1. \]
In our case
\[ a_n=\frac{(-1)^n}{n}. \]
We compute the absolute value:
\[ |a_n|=\left|\frac{(-1)^n}{n}\right|. \]
Since
\[ |(-1)^n|=1 \]
for every \(n\ge 1\), we obtain
\[ |a_n|=\frac1n. \]
Since \(n\ge 1\), we have
\[ \frac1n\le 1. \]
Hence
\[ |a_n|\le 1\quad \text{for every } n\ge 1. \]
Choosing \(K=1\), we conclude that the sequence is bounded.
The first terms are
\[ -1,\ \frac12,\ -\frac13,\ \frac14,\ldots \]
They change sign, but all remain between \(-1\) and \(1\).
Exercise 13 — level ★★★☆☆
Decide whether the sequence
\[ a_n=n^2-3n, \qquad n\ge 1, \]
is bounded above, bounded below, or bounded.
Answer
The sequence is bounded below but not bounded above. Consequently, it is not bounded.
Solution
Consider
\[ a_n=n^2-3n. \]
We compute a few terms:
\[ a_1=1-3=-2, \]
\[ a_2=4-6=-2, \]
\[ a_3=9-9=0, \]
\[ a_4=16-12=4. \]
The terms thus begin as follows:
\[ -2,\ -2,\ 0,\ 4,\ldots \]
To study boundedness, we rewrite the general term by completing the square:
\[ n^2-3n=\left(n-\frac32\right)^2-\frac94. \]
Since a square is always nonnegative, we have
\[ \left(n-\frac32\right)^2\ge 0. \]
Hence
\[ a_n=\left(n-\frac32\right)^2-\frac94\ge -\frac94. \]
This shows that the sequence is bounded below. However, since \(n\) is restricted to natural-number values, this lower bound need not be the least value actually attained by the sequence.
In fact, since \(n\) is a natural number, the smallest value attained is \(-2\), reached for \(n=1\) and \(n=2\). Indeed,
\[ a_1=a_2=-2. \]
Let us now ask whether the sequence is bounded above.
For large values of \(n\), the dominant term is \(n^2\). The term \(-3n\) grows in absolute value much more slowly than \(n^2\).
We can make this rigorous by observing that, for \(n\ge 6\), we have
\[ 3n\le \frac{n^2}{2}. \]
Indeed, this inequality is equivalent to
\[ 6n\le n^2, \]
that is,
\[ 6\le n. \]
Hence, for \(n\ge 6\),
\[ a_n=n^2-3n\ge n^2-\frac{n^2}{2}=\frac{n^2}{2}. \]
The quantity \(\displaystyle \frac{n^2}{2}\) exceeds any prescribed real number, provided \(n\) is chosen large enough.
Therefore the sequence is not bounded above.
We conclude that the sequence is bounded below but not above. Hence it is not bounded.
Exercise 14 — level ★★☆☆☆
Verify that the sequence
\[ 5,\ 8,\ 11,\ 14,\ldots \]
is an arithmetic progression and determine its general term.
Answer
The sequence is an arithmetic progression with first term \(a_1=5\) and common difference \(d=3\). The general term is
\[ a_n=5+3(n-1). \]
Equivalently,
\[ a_n=3n+2. \]
Solution
A sequence is an arithmetic progression if the difference between two consecutive terms is constant.
We compute the differences:
\[ 8-5=3, \]
\[ 11-8=3, \]
\[ 14-11=3. \]
The difference between consecutive terms is always \(3\). Hence the sequence is an arithmetic progression.
The first term is
\[ a_1=5, \]
and the common difference is
\[ d=3. \]
The general term of an arithmetic progression is
\[ a_n=a_1+(n-1)d. \]
Substituting \(a_1=5\) and \(d=3\), we obtain
\[ a_n=5+3(n-1). \]
Expanding,
\[ a_n=5+3n-3=3n+2. \]
Hence
\[ a_n=3n+2. \]
Exercise 15 — level ★★☆☆☆
Verify that the sequence
\[ 2,\ 6,\ 18,\ 54,\ldots \]
is a geometric progression and determine its general term.
Answer
The sequence is a geometric progression with first term \(a_1=2\) and common ratio \(q=3\). The general term is
\[ a_n=2\cdot 3^{n-1}. \]
Solution
A sequence is a geometric progression if each term is obtained from the preceding one by multiplying always by the same number.
We compute the ratios of consecutive terms:
\[ \frac62=3, \]
\[ \frac{18}{6}=3, \]
\[ \frac{54}{18}=3. \]
The ratio is constant and equal to \(3\). Hence the sequence is a geometric progression.
The first term is
\[ a_1=2, \]
and the common ratio is
\[ q=3. \]
The general term of a geometric progression is
\[ a_n=a_1q^{n-1}. \]
Substituting \(a_1=2\) and \(q=3\), we obtain
\[ a_n=2\cdot 3^{n-1}. \]
We check this against the first terms:
\[ a_1=2\cdot 3^0=2, \]
\[ a_2=2\cdot 3^1=6, \]
\[ a_3=2\cdot 3^2=18. \]
The formula is therefore consistent with the given terms.
Exercise 16 — level ★★★☆☆
Study the sign of the sequence
\[ a_n=(-1)^{n+1}\frac{n}{n+1}, \qquad n\ge 1. \]
Decide whether it is positive, negative, or of alternating sign.
Answer
The sequence is of alternating sign. The terms with odd index are positive, while the terms with even index are negative.
Solution
The sequence is
\[ a_n=(-1)^{n+1}\frac{n}{n+1}. \]
The factor
\[ \frac{n}{n+1} \]
is always positive, since \(n\ge 1\) and \(n+1>0\). Hence the sign of \(a_n\) depends solely on the factor
\[ (-1)^{n+1}. \]
If \(n\) is odd, then \(n+1\) is even. Therefore
\[ (-1)^{n+1}=1. \]
In this case
\[ a_n=\frac{n}{n+1}>0. \]
If instead \(n\) is even, then \(n+1\) is odd. Therefore
\[ (-1)^{n+1}=-1. \]
In this case
\[ a_n=-\frac{n}{n+1}<0. \]
We compute the first terms:
\[ a_1=\frac12, \]
\[ a_2=-\frac23, \]
\[ a_3=\frac34, \]
\[ a_4=-\frac45. \]
Hence the sequence is
\[ \frac12,\ -\frac23,\ \frac34,\ -\frac45,\ldots \]
The terms change sign at every step. Consequently, the sequence is of alternating sign.
Exercise 17 — level ★★☆☆☆
Consider the sequence
\[ a_n=\frac1n, \qquad n\ge 1. \]
State which points appear in its graphical representation and explain why the graph of a sequence is not a continuous curve.
Answer
The first points of the graph are
\[ \left(1,1\right),\ \left(2,\frac12\right),\ \left(3,\frac13\right),\ \left(4,\frac14\right),\ldots \]
The graph is not a continuous curve because the sequence is defined only for natural-number values of the index.
Solution
A real-valued sequence is a function defined on the natural numbers.
In our case
\[ a_n=\frac1n. \]
To represent it graphically, we associate with each index \(n\) the point of the plane
\[ (n,a_n). \]
For \(n=1\) we obtain
\[ a_1=1, \]
so the first point is
\[ (1,1). \]
For \(n=2\) we obtain
\[ a_2=\frac12, \]
so the second point is
\[ \left(2,\frac12\right). \]
For \(n=3\) we obtain
\[ a_3=\frac13, \]
so the third point is
\[ \left(3,\frac13\right). \]
Similarly, for \(n=4\) we obtain
\[ \left(4,\frac14\right). \]
Hence the first points of the graph are
\[ \left(1,1\right),\ \left(2,\frac12\right),\ \left(3,\frac13\right),\ \left(4,\frac14\right),\ldots \]
The graph of a sequence is not a continuous curve, because the index \(n\) does not take all real values, but only natural-number values.
Thus, between the point corresponding to \(n=1\) and the one corresponding to \(n=2\) there are no points of the sequence. The graphical representation consists of isolated points, not of a continuous line.
Exercise 18 — level ★★★☆☆
Let
\[ a_n=(-1)^n, \qquad n\ge 1. \]
Write the subsequence formed by the terms of even index and the subsequence formed by the terms of odd index.
Answer
The subsequence of even indices is
\[ a_{2k}=1,\qquad k\ge 1. \]
The subsequence of odd indices is
\[ a_{2k-1}=-1,\qquad k\ge 1. \]
Solution
The sequence is
\[ a_n=(-1)^n. \]
Its first terms are
\[ -1,\ 1,\ -1,\ 1,\ldots \]
We first consider the even indices. An even index can be written in the form
\[ n=2k, \qquad k\ge 1. \]
The corresponding subsequence is
\[ a_{2k}=(-1)^{2k}. \]
Since \(2k\) is even, we have
\[ (-1)^{2k}=1. \]
Hence
\[ a_{2k}=1\quad \text{for every } k\ge 1. \]
The subsequence of even indices is therefore
\[ 1,\ 1,\ 1,\ 1,\ldots \]
We now consider the odd indices. An odd index can be written in the form
\[ n=2k-1, \qquad k\ge 1. \]
The corresponding subsequence is
\[ a_{2k-1}=(-1)^{2k-1}. \]
Since \(2k-1\) is odd, we have
\[ (-1)^{2k-1}=-1. \]
Hence
\[ a_{2k-1}=-1\quad \text{for every } k\ge 1. \]
The subsequence of odd indices is therefore
\[ -1,\ -1,\ -1,\ -1,\ldots \]
This exercise shows that a non-constant sequence may contain constant subsequences.
Exercise 19 — level ★★★☆☆
Let
\[ a_n=n, \qquad n\ge 1. \]
Decide whether the list
\[ a_3,\ a_5,\ a_8,\ a_{10},\ldots \]
can be a subsequence. Then decide whether the list
\[ a_5,\ a_3,\ a_8,\ a_{10},\ldots \]
can be a subsequence.
Answer
The first list can be a subsequence, because the indices are strictly increasing. The second list cannot be a subsequence, because the indices are not strictly increasing.
Solution
A subsequence of \((a_n)\) is obtained by choosing a strictly increasing sequence of natural-number indices
\[ n_1<n_2<n_3<\cdots. \]
The subsequence is then
\[ a_{n_1},\ a_{n_2},\ a_{n_3},\ldots \]
Consider the first list:
\[ a_3,\ a_5,\ a_8,\ a_{10},\ldots \]
The indices are
\[ 3,\ 5,\ 8,\ 10,\ldots \]
They are strictly increasing, because
\[ 3<5<8<10<\cdots. \]
Hence this list can be a subsequence.
Now consider the second list:
\[ a_5,\ a_3,\ a_8,\ a_{10},\ldots \]
The indices are
\[ 5,\ 3,\ 8,\ 10,\ldots \]
This sequence of indices is not strictly increasing, because
\[ 5>3. \]
Hence the second list cannot be a subsequence.
The essential point is that a subsequence may skip some terms of the original sequence, but it cannot change the order in which the terms appear.
Exercise 20 — level ★★★★☆
Consider the sequence
\[ a_n=(-1)^n+\frac1n, \qquad n\ge 1. \]
Write the first six terms, decide whether the sequence is monotone, and show that it is bounded.
Answer
The first six terms are
\[ 0,\ \frac32,\ -\frac23,\ \frac54,\ -\frac45,\ \frac76. \]
The sequence is not monotone. Moreover, it is bounded, since
\[ -1\le a_n\le 2\quad \text{for every } n\ge 1. \]
Solution
The sequence is
\[ a_n=(-1)^n+\frac1n. \]
We compute the first six terms.
For \(n=1\),
\[ a_1=(-1)^1+\frac11=-1+1=0. \]
For \(n=2\),
\[ a_2=(-1)^2+\frac12=1+\frac12=\frac32. \]
For \(n=3\),
\[ a_3=(-1)^3+\frac13=-1+\frac13=-\frac23. \]
For \(n=4\),
\[ a_4=(-1)^4+\frac14=1+\frac14=\frac54. \]
For \(n=5\),
\[ a_5=(-1)^5+\frac15=-1+\frac15=-\frac45. \]
For \(n=6\),
\[ a_6=(-1)^6+\frac16=1+\frac16=\frac76. \]
Hence the first six terms are
\[ 0,\ \frac32,\ -\frac23,\ \frac54,\ -\frac45,\ \frac76. \]
We now study monotonicity. Observe that
\[ a_1=0 \qquad \text{and} \qquad a_2=\frac32. \]
Hence
\[ a_1<a_2. \]
However,
\[ a_2=\frac32 \qquad \text{and} \qquad a_3=-\frac23. \]
Hence
\[ a_2>a_3. \]
The sequence first increases and then decreases. Therefore it is not increasing.
Moreover, since \(a_1<a_2\), it is not decreasing either.
Consequently, the sequence is not monotone.
Finally, we show that it is bounded.
We know that
\[ -1\le (-1)^n\le 1 \]
for every \(n\ge 1\). Moreover,
\[ 0<\frac1n\le 1. \]
Adding these facts, we obtain on the one hand
\[ (-1)^n+\frac1n\ge -1+0=-1. \]
On the other hand,
\[ (-1)^n+\frac1n\le 1+1=2. \]
Hence
\[ -1\le a_n\le 2\quad \text{for every } n\ge 1. \]
This proves that the sequence is bounded.
This exercise is instructive because it exhibits a sequence that is bounded but not monotone: boundedness and monotonicity are distinct and independent properties.