This page offers 20 step-by-step practice problems on bounded sequences. They are designed to help you learn to recognize when a sequence is bounded above, bounded below, or bounded both below and above.
In every exercise we shall work with real sequences defined on \(\mathbb{N}\), where
\[ \mathbb{N}=\{0,1,2,\dots\}. \]
Recall that a sequence \((a_n)\) is bounded above if there exists \(M\in\mathbb{R}\) such that
\[ a_n\le M \]
for every \(n\in\mathbb{N}\). Likewise, it is bounded below if there exists \(m\in\mathbb{R}\) such that
\[ a_n\ge m \]
for every \(n\in\mathbb{N}\).
Finally, a sequence is bounded if it is both bounded below and bounded above. Equivalently, \((a_n)\) is bounded if there exists \(K>0\) such that
\[ |a_n|\le K \]
for every \(n\in\mathbb{N}\).
Exercise 1 — level ★☆☆☆☆
Determine whether the sequence
\[ a_n=\frac{1}{n+1} \]
is bounded above, bounded below, and bounded.
Answer
The sequence is bounded. More precisely,
\[ 0\le a_n\le 1 \]
for every \(n\in\mathbb{N}\).
Solution
The sequence is given by
\[ a_n=\frac{1}{n+1}, \qquad n\in\mathbb{N}. \]
Since throughout this collection of exercises we assume
\[ \mathbb{N}=\{0,1,2,\dots\}, \]
we have
\[ n\ge 0. \]
Adding \(1\) to both sides, we obtain
\[ n+1\ge 1. \]
Hence the denominator \(n+1\) is always positive and at least \(1\).
From \(n+1\ge 1\) it follows that
\[ 0<\frac{1}{n+1}\le 1. \]
Since
\[ a_n=\frac{1}{n+1}, \]
we may write
\[ 0<a_n\le 1. \]
In particular, from \(a_n\le 1\) it follows that \(1\) is an upper bound for the sequence. Thus the sequence is bounded above.
Moreover, from \(a_n>0\) it also follows that
\[ a_n\ge 0. \]
Hence \(0\) is a lower bound for the sequence, so the sequence is bounded below.
The sequence is therefore bounded.
Exercise 2 — level ★☆☆☆☆
Determine whether the sequence
\[ a_n=\frac{n}{n+1} \]
is bounded above, bounded below, and bounded.
Answer
The sequence is bounded. More precisely,
\[ 0\le a_n<1 \]
for every \(n\in\mathbb{N}\). In particular, \(0\) is a lower bound and \(1\) is an upper bound.
Solution
Consider the sequence
\[ a_n=\frac{n}{n+1}. \]
Since \(n\in\mathbb{N}\), we have
\[ n\ge 0. \]
Furthermore
\[ n+1>0. \]
The numerator is therefore nonnegative, while the denominator is positive. Consequently
\[ \frac{n}{n+1}\ge 0. \]
Hence
\[ a_n\ge 0 \]
for every \(n\in\mathbb{N}\). This shows that the sequence is bounded below.
We now examine boundedness above. Since
\[ n<n+1 \]
and since \(n+1>0\), dividing both sides by \(n+1\) we obtain
\[ \frac{n}{n+1}<1. \]
Hence
\[ a_n<1. \]
In particular, \(1\) is an upper bound for the sequence, since every term is less than \(1\). Thus the sequence is bounded above.
We have shown that
\[ 0\le a_n<1 \]
for every \(n\in\mathbb{N}\).
The sequence is therefore bounded.
Exercise 3 — level ★☆☆☆☆
Determine whether the sequence
\[ a_n=\frac{(-1)^n}{n+1} \]
is bounded.
Answer
The sequence is bounded. Indeed
\[ |a_n|\le 1 \]
for every \(n\in\mathbb{N}\). Hence
\[ -1\le a_n\le 1. \]
Solution
The sequence is
\[ a_n=\frac{(-1)^n}{n+1}. \]
To decide whether it is bounded, it is convenient to estimate the absolute value of its terms.
We compute:
\[ |a_n| = \left|\frac{(-1)^n}{n+1}\right|. \]
The absolute value of a quotient equals the quotient of the absolute values, so
\[ |a_n| = \frac{|(-1)^n|}{|n+1|}. \]
For every \(n\in\mathbb{N}\), we have
\[ |(-1)^n|=1. \]
Moreover, since \(n+1>0\), we have
\[ |n+1|=n+1. \]
Hence
\[ |a_n|=\frac{1}{n+1}. \]
Since \(n+1\ge 1\), we have
\[ \frac{1}{n+1}\le 1. \]
Hence
\[ |a_n|\le 1 \]
for every \(n\in\mathbb{N}\).
By the characterisation via the absolute value, a real sequence is bounded if there exists \(K>0\) such that
\[ |a_n|\le K \]
for every \(n\in\mathbb{N}\). Here we may take \(K=1\).
The sequence is therefore bounded.
Exercise 4 — level ★☆☆☆☆
Determine whether the sequence
\[ a_n=n+3 \]
is bounded above, bounded below, and bounded.
Answer
The sequence is bounded below but not bounded above. Hence it is not bounded.
Solution
Consider the sequence
\[ a_n=n+3. \]
Since \(n\in\mathbb{N}\), we have
\[ n\ge 0. \]
Adding \(3\) to both sides, we obtain
\[ n+3\ge 3. \]
Since \(a_n=n+3\), it follows that
\[ a_n\ge 3 \]
for every \(n\in\mathbb{N}\). Hence \(3\) is a lower bound for the sequence, so the sequence is bounded below.
We now check whether the sequence is bounded above.
For it to be bounded above, there would have to exist a real number \(M\) such that
\[ a_n\le M \]
for every \(n\in\mathbb{N}\). We show that this fails.
Let \(M\in\mathbb{R}\) be an arbitrary real number. Since the natural numbers are not bounded above, we may choose \(n\in\mathbb{N}\) such that
\[ n>M-3. \]
Adding \(3\) to both sides, we obtain
\[ n+3>M. \]
But \(a_n=n+3\), so
\[ a_n>M. \]
We have shown that, whatever \(M\in\mathbb{R}\) may be, there is a term of the sequence greater than \(M\). Hence the sequence is not bounded above.
Since it is bounded below but not above, the sequence is not bounded.
Exercise 5 — level ★☆☆☆☆
Determine whether the sequence
\[ a_n=-n^2 \]
is bounded above, bounded below, and bounded.
Answer
The sequence is bounded above but not bounded below. Hence it is not bounded.
Solution
Consider the sequence
\[ a_n=-n^2. \]
Since the square of a real number is always nonnegative, for every \(n\in\mathbb{N}\) we have
\[ n^2\ge 0. \]
Multiplying both sides by \(-1\) reverses the inequality. We thus obtain
\[ -n^2\le 0. \]
Since \(a_n=-n^2\), it follows that
\[ a_n\le 0 \]
for every \(n\in\mathbb{N}\). Hence \(0\) is an upper bound for the sequence, so the sequence is bounded above.
We now examine boundedness below.
For it to be bounded below, there would have to exist a real number \(m\) such that
\[ a_n\ge m \]
for every \(n\in\mathbb{N}\). We show that no real number \(m\) can be a lower bound.
Let \(m\in\mathbb{R}\). We seek an index \(n\in\mathbb{N}\) such that
\[ a_n<m. \]
It suffices to choose \(n\in\mathbb{N}\) large enough that
\[ n^2>|m|+1. \]
This is possible because \(n^2\) grows without bound as \(n\) increases.
From
\[ n^2>|m|+1 \]
it follows in particular that
\[ n^2>|m|. \]
Since \(|m|\ge -m\) holds for every \(m\in\mathbb{R}\), we obtain
\[ n^2>-m. \]
Multiplying by \(-1\) reverses the inequality:
\[ -n^2<m. \]
Since \(a_n=-n^2\), it follows that
\[ a_n<m. \]
We have shown that, for every \(m\in\mathbb{R}\), there is a term of the sequence less than \(m\). Hence the sequence is not bounded below.
Since it is bounded above but not below, the sequence is not bounded.
Exercise 6 — level ★☆☆☆☆
Determine whether the sequence
\[ a_n=\frac{n^2}{n^2+1} \]
is bounded above, bounded below, and bounded.
Answer
The sequence is bounded. More precisely,
\[ 0\le a_n<1 \]
for every \(n\in\mathbb{N}\). In particular, \(0\) is a lower bound and \(1\) is an upper bound.
Solution
Consider the sequence
\[ a_n=\frac{n^2}{n^2+1}. \]
For every \(n\in\mathbb{N}\), we have
\[ n^2\ge 0. \]
Furthermore
\[ n^2+1>0. \]
The numerator is therefore nonnegative, while the denominator is positive. Consequently
\[ \frac{n^2}{n^2+1}\ge 0. \]
Hence
\[ a_n\ge 0 \]
for every \(n\in\mathbb{N}\). This shows that the sequence is bounded below.
We now examine boundedness above. Since
\[ n^2<n^2+1 \]
and since \(n^2+1>0\), dividing both sides by \(n^2+1\) we obtain
\[ \frac{n^2}{n^2+1}<1. \]
Hence
\[ a_n<1 \]
for every \(n\in\mathbb{N}\). In particular, \(1\) is an upper bound for the sequence, so the sequence is bounded above.
We have shown that
\[ 0\le a_n<1 \]
for every \(n\in\mathbb{N}\).
The sequence is therefore bounded.
Exercise 7 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=(-1)^n n \]
is bounded above, bounded below, and bounded.
Answer
The sequence is bounded neither above nor below. Hence it is not bounded.
Solution
Consider the sequence
\[ a_n=(-1)^n n. \]
The factor \((-1)^n\) changes sign according to the parity of \(n\).
If \(n\) is even, then \((-1)^n=1\), and hence
\[ a_n=n. \]
If instead \(n\) is odd, then \((-1)^n=-1\), and hence
\[ a_n=-n. \]
We first examine boundedness above.
To prove that the sequence is not bounded above, we must show that, whatever \(M\in\mathbb{R}\) may be, there exists an index \(n\in\mathbb{N}\) such that
\[ a_n>M. \]
So let \(M\in\mathbb{R}\). We choose an even index \(n=2q\) large enough that
\[ 2q>M. \]
This is possible because the even natural numbers grow without bound.
For this index \(n=2q\), since \(n\) is even, we have
\[ (-1)^n=1. \]
Hence
\[ a_n=(-1)^n n=n=2q>M. \]
We have thus shown that no real number \(M\) can be an upper bound. The sequence is not bounded above.
We now examine boundedness below.
To prove that the sequence is not bounded below, we must show that, whatever \(m\in\mathbb{R}\) may be, there exists an index \(n\in\mathbb{N}\) such that
\[ a_n<m. \]
So let \(m\in\mathbb{R}\). We choose an odd index \(n=2q+1\) large enough that
\[ -(2q+1)<m. \]
This is possible because the numbers of the form \(-(2q+1)\) decrease without bound as \(q\) increases.
For this index \(n=2q+1\), since \(n\) is odd, we have
\[ (-1)^n=-1. \]
Hence
\[ a_n=(-1)^n n=-n=-(2q+1)<m. \]
We have thus shown that no real number \(m\) can be a lower bound. The sequence is not bounded below.
Since the sequence is bounded neither above nor below, it is not bounded.
Exercise 8 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=(-1)^n\frac{n}{n+1} \]
is bounded.
Answer
The sequence is bounded. Indeed
\[ |a_n|<1 \]
for every \(n\in\mathbb{N}\). Hence, in particular,
\[ -1\le a_n\le 1. \]
Solution
Consider the sequence
\[ a_n=(-1)^n\frac{n}{n+1}. \]
Since the sequence contains the alternating factor \((-1)^n\), it is natural to estimate the absolute value of its terms.
We compute:
\[ |a_n| = \left|(-1)^n\frac{n}{n+1}\right|. \]
Using the properties of the absolute value, we obtain
\[ |a_n| = |(-1)^n|\left|\frac{n}{n+1}\right|. \]
For every \(n\in\mathbb{N}\), we have
\[ |(-1)^n|=1. \]
Moreover \(n\ge 0\) and \(n+1>0\), so
\[ \left|\frac{n}{n+1}\right|=\frac{n}{n+1}. \]
Hence
\[ |a_n|=\frac{n}{n+1}. \]
Since
\[ n<n+1 \]
and \(n+1>0\), dividing by \(n+1\) we obtain
\[ \frac{n}{n+1}<1. \]
Hence
\[ |a_n|<1 \]
for every \(n\in\mathbb{N}\).
In particular, we also have
\[ |a_n|\le 1. \]
By the characterisation via the absolute value, since there exists \(K>0\), for instance \(K=1\), such that
\[ |a_n|\le K \]
for every \(n\in\mathbb{N}\), the sequence is bounded.
From the inequality \(|a_n|\le 1\) it also follows that
\[ -1\le a_n\le 1. \]
Exercise 9 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=\frac{2n+1}{n+2} \]
is bounded above, bounded below, and bounded.
Answer
The sequence is bounded. More precisely,
\[ \frac{1}{2}\le a_n<2 \]
for every \(n\in\mathbb{N}\).
Solution
Consider the sequence
\[ a_n=\frac{2n+1}{n+2}. \]
To study its boundedness, we rewrite the numerator in terms of the denominator. Observe that
\[ 2n+1=2(n+2)-3. \]
Indeed
\[ 2(n+2)-3=2n+4-3=2n+1. \]
Hence
\[ a_n=\frac{2(n+2)-3}{n+2}. \]
Splitting the fraction, we obtain
\[ a_n = \frac{2(n+2)}{n+2}-\frac{3}{n+2} = 2-\frac{3}{n+2}. \]
Since \(n\in\mathbb{N}\), we have
\[ n+2\ge 2. \]
Hence
\[ \frac{3}{n+2}>0. \]
From
\[ a_n=2-\frac{3}{n+2} \]
and from \(\displaystyle\frac{3}{n+2}>0\), it follows that
\[ a_n<2. \]
Hence \(2\) is an upper bound for the sequence, so the sequence is bounded above.
We now look for a lower bound. Since \(n+2\ge 2\), dividing \(3\) by a number greater than or equal to \(2\) we obtain
\[ \frac{3}{n+2}\le \frac{3}{2}. \]
Changing sign reverses the inequality:
\[ -\frac{3}{n+2}\ge -\frac{3}{2}. \]
Adding \(2\) to both sides, we obtain
\[ 2-\frac{3}{n+2}\ge 2-\frac{3}{2}. \]
Since
\[ 2-\frac{3}{2}=\frac{1}{2}, \]
it follows that
\[ a_n\ge \frac{1}{2}. \]
Hence \(\displaystyle\frac{1}{2}\) is a lower bound for the sequence, so the sequence is bounded below.
We have shown that
\[ \frac{1}{2}\le a_n<2 \]
for every \(n\in\mathbb{N}\). The sequence is therefore bounded.
Exercise 10 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=\frac{n^2+3}{n^2+1} \]
is bounded above, bounded below, and bounded.
Answer
The sequence is bounded. More precisely,
\[ 1<a_n\le 3 \]
for every \(n\in\mathbb{N}\).
Solution
Consider the sequence
\[ a_n=\frac{n^2+3}{n^2+1}. \]
To study its boundedness, we rewrite the numerator so as to bring out the denominator:
\[ n^2+3=(n^2+1)+2. \]
Hence
\[ a_n=\frac{(n^2+1)+2}{n^2+1}. \]
Splitting the fraction, we obtain
\[ a_n = \frac{n^2+1}{n^2+1} + \frac{2}{n^2+1} = 1+\frac{2}{n^2+1}. \]
Since \(n^2\ge 0\), we have
\[ n^2+1\ge 1. \]
Consequently
\[ \frac{2}{n^2+1}>0. \]
From
\[ a_n=1+\frac{2}{n^2+1} \]
it then follows that
\[ a_n>1. \]
In particular, \(1\) is a lower bound for the sequence, so the sequence is bounded below.
We now look for an upper bound. From \(n^2+1\ge 1\) it follows that
\[ \frac{2}{n^2+1}\le 2. \]
Adding \(1\) to both sides, we obtain
\[ 1+\frac{2}{n^2+1}\le 3. \]
Since
\[ a_n=1+\frac{2}{n^2+1}, \]
it follows that
\[ a_n\le 3. \]
Hence \(3\) is an upper bound for the sequence, so the sequence is bounded above.
We have shown that
\[ 1<a_n\le 3 \]
for every \(n\in\mathbb{N}\). The sequence is therefore bounded.
Exercise 11 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=\frac{3n^2-2}{n^2+1} \]
is bounded above, bounded below, and bounded.
Answer
The sequence is bounded. More precisely,
\[ -2\le a_n<3 \]
for every \(n\in\mathbb{N}\).
Solution
Consider the sequence
\[ a_n=\frac{3n^2-2}{n^2+1}. \]
To study its boundedness, we rewrite the numerator so as to bring out the denominator. Observe that
\[ 3n^2-2=3(n^2+1)-5. \]
Indeed
\[ 3(n^2+1)-5=3n^2+3-5=3n^2-2. \]
Hence
\[ a_n=\frac{3(n^2+1)-5}{n^2+1}. \]
Splitting the fraction, we obtain
\[ a_n = \frac{3(n^2+1)}{n^2+1}-\frac{5}{n^2+1} = 3-\frac{5}{n^2+1}. \]
Since \(n^2\ge 0\), we have
\[ n^2+1\ge 1. \]
In particular, the denominator \(n^2+1\) is always positive. Hence
\[ \frac{5}{n^2+1}>0. \]
From
\[ a_n=3-\frac{5}{n^2+1} \]
and from \(\displaystyle\frac{5}{n^2+1}>0\), it follows that
\[ a_n<3. \]
Hence \(3\) is an upper bound for the sequence, so the sequence is bounded above.
We now look for a lower bound. From \(n^2+1\ge 1\) it follows that
\[ \frac{5}{n^2+1}\le 5. \]
Changing sign reverses the inequality:
\[ -\frac{5}{n^2+1}\ge -5. \]
Adding \(3\) to both sides, we obtain
\[ 3-\frac{5}{n^2+1}\ge 3-5. \]
Since
\[ 3-5=-2, \]
it follows that
\[ a_n\ge -2. \]
Hence \(-2\) is a lower bound for the sequence, so the sequence is bounded below.
We have shown that
\[ -2\le a_n<3 \]
for every \(n\in\mathbb{N}\). The sequence is therefore bounded.
Exercise 12 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=\frac{n}{n^2+1} \]
is bounded above, bounded below, and bounded.
Answer
The sequence is bounded. More precisely,
\[ 0\le a_n\le \frac{1}{2} \]
for every \(n\in\mathbb{N}\).
Solution
Consider the sequence
\[ a_n=\frac{n}{n^2+1}. \]
Since \(n\in\mathbb{N}\), we have
\[ n\ge 0. \]
Furthermore
\[ n^2+1>0. \]
The numerator is nonnegative, while the denominator is positive. Consequently
\[ \frac{n}{n^2+1}\ge 0. \]
Hence
\[ a_n\ge 0 \]
for every \(n\in\mathbb{N}\). The sequence is therefore bounded below.
We now examine boundedness above. We wish to show that
\[ \frac{n}{n^2+1}\le \frac{1}{2}. \]
Since \(n^2+1>0\), we may multiply both sides by \(2(n^2+1)\), which is positive. The preceding inequality is equivalent to
\[ 2n\le n^2+1. \]
Moving everything to the right-hand side, we obtain
\[ 0\le n^2-2n+1. \]
But
\[ n^2-2n+1=(n-1)^2. \]
Hence the inequality becomes
\[ 0\le (n-1)^2. \]
This always holds, since the square of a real number is always nonnegative.
Hence
\[ \frac{n}{n^2+1}\le \frac{1}{2} \]
for every \(n\in\mathbb{N}\).
Therefore \(\displaystyle\frac{1}{2}\) is an upper bound for the sequence, so the sequence is bounded above.
We have shown that
\[ 0\le a_n\le \frac{1}{2} \]
for every \(n\in\mathbb{N}\). The sequence is therefore bounded.
Exercise 13 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=(-1)^n\frac{n+2}{n+1} \]
is bounded.
Answer
The sequence is bounded. Indeed
\[ |a_n|\le 2 \]
for every \(n\in\mathbb{N}\). Hence
\[ -2\le a_n\le 2. \]
Solution
Consider the sequence
\[ a_n=(-1)^n\frac{n+2}{n+1}. \]
Since the alternating factor \((-1)^n\) is present, it is convenient to study the absolute value of the terms.
We compute:
\[ |a_n| = \left|(-1)^n\frac{n+2}{n+1}\right|. \]
Using the properties of the absolute value, we obtain
\[ |a_n| = |(-1)^n|\left|\frac{n+2}{n+1}\right|. \]
For every \(n\in\mathbb{N}\), we have
\[ |(-1)^n|=1. \]
Moreover \(n+1>0\) and \(n+2>0\), so
\[ \left|\frac{n+2}{n+1}\right|=\frac{n+2}{n+1}. \]
Therefore
\[ |a_n|=\frac{n+2}{n+1}. \]
We now rewrite the fraction:
\[ \frac{n+2}{n+1} = \frac{(n+1)+1}{n+1} = 1+\frac{1}{n+1}. \]
Since \(n+1\ge 1\), we have
\[ \frac{1}{n+1}\le 1. \]
Adding \(1\) to both sides, we obtain
\[ 1+\frac{1}{n+1}\le 2. \]
Since
\[ |a_n|=1+\frac{1}{n+1}, \]
it follows that
\[ |a_n|\le 2. \]
By the characterisation via the absolute value, since there exists \(K>0\), for instance \(K=2\), such that
\[ |a_n|\le K \]
for every \(n\in\mathbb{N}\), the sequence is bounded.
In particular, from the inequality \(|a_n|\le 2\) it follows that
\[ -2\le a_n\le 2. \]
Exercise 14 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=\frac{n^3}{n^2+1} \]
is bounded above, bounded below, and bounded.
Answer
The sequence is bounded below but not bounded above. Hence it is not bounded.
Solution
Consider the sequence
\[ a_n=\frac{n^3}{n^2+1}. \]
Since \(n\in\mathbb{N}\), we have
\[ n\ge 0. \]
Hence
\[ n^3\ge 0. \]
Furthermore
\[ n^2+1>0. \]
The numerator is nonnegative, while the denominator is positive. Therefore
\[ \frac{n^3}{n^2+1}\ge 0. \]
Hence
\[ a_n\ge 0 \]
for every \(n\in\mathbb{N}\). The sequence is therefore bounded below.
We now show that the sequence is not bounded above.
For \(n\ge 1\), we have
\[ n^2+1\le 2n^2. \]
Indeed, if \(n\ge 1\), then \(1\le n^2\), and hence
\[ n^2+1\le n^2+n^2=2n^2. \]
Since \(n^2+1\le 2n^2\) and all the quantities involved are positive, taking reciprocals reverses the inequality:
\[ \frac{1}{n^2+1}\ge \frac{1}{2n^2}. \]
Multiplying by \(n^3\ge 0\), we obtain
\[ \frac{n^3}{n^2+1}\ge \frac{n^3}{2n^2}. \]
Simplifying,
\[ \frac{n^3}{2n^2}=\frac{n}{2}. \]
Hence, for every \(n\ge 1\),
\[ a_n\ge \frac{n}{2}. \]
Now let \(M\in\mathbb{R}\). We seek an index \(n\in\mathbb{N}\) such that
\[ a_n>M. \]
We choose \(n\in\mathbb{N}\) such that
\[ n\ge 1 \qquad \text{and} \qquad \frac{n}{2}>M. \]
This is possible because \(\displaystyle\frac{n}{2}\) grows without bound as \(n\) increases.
For this index, using the previous estimate, we have
\[ a_n\ge \frac{n}{2}>M. \]
We have thus shown that, whatever \(M\in\mathbb{R}\) may be, there is a term of the sequence greater than \(M\). Hence the sequence is not bounded above.
Since the sequence is bounded below but not above, it is not bounded.
Exercise 15 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=n^2-4n \]
is bounded above, bounded below, and bounded.
Answer
The sequence is bounded below but not bounded above. Hence it is not bounded.
Solution
Consider the sequence
\[ a_n=n^2-4n. \]
To study boundedness below, we complete the square:
\[ n^2-4n=n^2-4n+4-4. \]
Since
\[ n^2-4n+4=(n-2)^2, \]
we obtain
\[ a_n=(n-2)^2-4. \]
Now, for every \(n\in\mathbb{N}\), the square \((n-2)^2\) is nonnegative. Hence
\[ (n-2)^2\ge 0. \]
Subtracting \(4\) from both sides, we obtain
\[ (n-2)^2-4\ge -4. \]
Since
\[ a_n=(n-2)^2-4, \]
it follows that
\[ a_n\ge -4. \]
Hence \(-4\) is a lower bound for the sequence, so the sequence is bounded below.
We now show that the sequence is not bounded above.
For \(n\ge 8\), we have
\[ n^2-4n\ge \frac{n^2}{2}. \]
Let us verify this estimate. The inequality
\[ n^2-4n\ge \frac{n^2}{2} \]
is equivalent to
\[ \frac{n^2}{2}-4n\ge 0. \]
Factoring out \(n\), we obtain
\[ n\left(\frac{n}{2}-4\right)\ge 0. \]
If \(n\ge 8\), then
\[ \frac{n}{2}-4\ge 0, \]
and since \(n\ge 0\), the product is nonnegative. Hence, for every \(n\ge 8\),
\[ a_n=n^2-4n\ge \frac{n^2}{2}. \]
Now let \(M\in\mathbb{R}\). We seek an index \(n\in\mathbb{N}\) such that
\[ a_n>M. \]
Since \(\displaystyle\frac{n^2}{2}\) grows without bound as \(n\) increases, we may choose \(n\in\mathbb{N}\) such that
\[ n\ge 8 \qquad \text{and} \qquad \frac{n^2}{2}>M. \]
For this index, the previous estimate gives
\[ a_n\ge \frac{n^2}{2}>M. \]
Hence, whatever \(M\in\mathbb{R}\) may be, there is a term of the sequence greater than \(M\). The sequence is therefore not bounded above.
Since the sequence is bounded below but not above, it is not bounded.
Exercise 16 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=\sqrt{n+1}-\sqrt{n} \]
is bounded above, bounded below, and bounded.
Answer
The sequence is bounded. More precisely,
\[ 0<a_n\le 1 \]
for every \(n\in\mathbb{N}\).
Solution
Consider the sequence
\[ a_n=\sqrt{n+1}-\sqrt{n}. \]
Since \(n+1>n\) and the square-root function is increasing on \([0,+\infty)\), we have
\[ \sqrt{n+1}>\sqrt{n}. \]
Subtracting \(\sqrt{n}\) from both sides, we obtain
\[ \sqrt{n+1}-\sqrt{n}>0. \]
Hence
\[ a_n>0 \]
for every \(n\in\mathbb{N}\). In particular, \(0\) is a lower bound for the sequence, so the sequence is bounded below.
We now examine boundedness above. To estimate \(a_n\), we rationalize:
\[ a_n=\sqrt{n+1}-\sqrt{n} = \frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}. \]
In the numerator we use the identity
\[ (x-y)(x+y)=x^2-y^2. \]
We thus obtain
\[ a_n= \frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}} = \frac{1}{\sqrt{n+1}+\sqrt{n}}. \]
Since \(n\ge 0\), we have
\[ \sqrt{n+1}\ge 1 \qquad \text{and} \qquad \sqrt{n}\ge 0. \]
Hence
\[ \sqrt{n+1}+\sqrt{n}\ge 1. \]
Consequently
\[ \frac{1}{\sqrt{n+1}+\sqrt{n}}\le 1. \]
Since
\[ a_n=\frac{1}{\sqrt{n+1}+\sqrt{n}}, \]
it follows that
\[ a_n\le 1. \]
Hence \(1\) is an upper bound for the sequence, so the sequence is bounded above.
We have shown that
\[ 0<a_n\le 1 \]
for every \(n\in\mathbb{N}\).
The sequence is therefore bounded.
Exercise 17 — level ★★☆☆☆
Determine whether the sequence
\[ a_n=\sqrt{n^2+1}-n \]
is bounded above, bounded below, and bounded.
Answer
The sequence is bounded. More precisely,
\[ 0<a_n\le 1 \]
for every \(n\in\mathbb{N}\).
Solution
Consider the sequence
\[ a_n=\sqrt{n^2+1}-n. \]
Since
\[ n^2+1>n^2, \]
and since the square root is increasing on \([0,+\infty)\), we obtain
\[ \sqrt{n^2+1}>\sqrt{n^2}. \]
Since \(n\in\mathbb{N}\), we have \(n\ge 0\), so
\[ \sqrt{n^2}=n. \]
Therefore
\[ \sqrt{n^2+1}>n. \]
Subtracting \(n\) from both sides, it follows that
\[ \sqrt{n^2+1}-n>0. \]
Hence
\[ a_n>0 \]
for every \(n\in\mathbb{N}\). The sequence is therefore bounded below, for instance by \(0\).
We now examine boundedness above. We rationalize the expression:
\[ a_n=\sqrt{n^2+1}-n = \frac{(\sqrt{n^2+1}-n)(\sqrt{n^2+1}+n)}{\sqrt{n^2+1}+n}. \]
In the numerator we obtain
\[ (\sqrt{n^2+1})^2-n^2=n^2+1-n^2=1. \]
Hence
\[ a_n=\frac{1}{\sqrt{n^2+1}+n}. \]
Since \(n\ge 0\), we have
\[ \sqrt{n^2+1}\ge 1 \qquad \text{and} \qquad n\ge 0. \]
Hence
\[ \sqrt{n^2+1}+n\ge 1. \]
From this it follows that
\[ \frac{1}{\sqrt{n^2+1}+n}\le 1. \]
Since
\[ a_n=\frac{1}{\sqrt{n^2+1}+n}, \]
we obtain
\[ a_n\le 1. \]
Hence \(1\) is an upper bound for the sequence, so the sequence is bounded above.
We have shown that
\[ 0<a_n\le 1 \]
for every \(n\in\mathbb{N}\). The sequence is therefore bounded.
Exercise 18 — level ★★★☆☆
Determine whether the sequence
\[ a_n=(-1)^n+\frac{1}{n+1} \]
is bounded.
Answer
The sequence is bounded. For instance,
\[ -1\le a_n\le 2 \]
for every \(n\in\mathbb{N}\).
Solution
Consider the sequence
\[ a_n=(-1)^n+\frac{1}{n+1}. \]
For every \(n\in\mathbb{N}\), the term \((-1)^n\) can take only the values \(1\) and \(-1\). Hence
\[ -1\le (-1)^n\le 1. \]
Moreover, since \(n+1\ge 1\), we have
\[ 0<\frac{1}{n+1}\le 1. \]
From this inequality it follows in particular that
\[ 0\le \frac{1}{n+1}\le 1. \]
We now add the two estimates:
\[ -1\le (-1)^n\le 1 \]
and
\[ 0\le \frac{1}{n+1}\le 1. \]
Adding term by term, we obtain
\[ -1+0\le (-1)^n+\frac{1}{n+1}\le 1+1. \]
Hence
\[ -1\le (-1)^n+\frac{1}{n+1}\le 2. \]
Since
\[ a_n=(-1)^n+\frac{1}{n+1}, \]
it follows that
\[ -1\le a_n\le 2 \]
for every \(n\in\mathbb{N}\).
Hence \(-1\) is a lower bound and \(2\) is an upper bound for the sequence. The sequence is therefore bounded.
Exercise 19 — level ★★★☆☆
Determine whether the sequence
\[ a_n=\frac{(-1)^n n^2+n}{n^2+1} \]
is bounded.
Answer
The sequence is bounded. For instance,
\[ |a_n|\le 2 \]
for every \(n\in\mathbb{N}\).
Solution
Consider the sequence
\[ a_n=\frac{(-1)^n n^2+n}{n^2+1}. \]
To prove that the sequence is bounded, we estimate the absolute value:
\[ |a_n| = \left|\frac{(-1)^n n^2+n}{n^2+1}\right|. \]
Since \(n^2+1>0\), we may write
\[ |a_n| = \frac{|(-1)^n n^2+n|}{n^2+1}. \]
We now use the triangle inequality:
\[ |x+y|\le |x|+|y|. \]
In our case,
\[ |(-1)^n n^2+n| \le |(-1)^n n^2|+|n|. \]
Since
\[ |(-1)^n|=1 \]
and \(n\ge 0\), we obtain
\[ |(-1)^n n^2|=n^2 \qquad \text{and} \qquad |n|=n. \]
Hence
\[ |(-1)^n n^2+n|\le n^2+n. \]
Consequently
\[ |a_n| \le \frac{n^2+n}{n^2+1}. \]
We now wish to bound this fraction from above. Since \(n\in\mathbb{N}\), for every \(n\) we have
\[ n\le n^2+1. \]
Indeed, this inequality is equivalent to
\[ n^2-n+1\ge 0, \]
and it holds for every \(n\in\mathbb{N}\). For instance, if \(n=0\) it is immediate, while if \(n\ge 1\) then \(n^2\ge n\), so \(n^2+1\ge n\).
From \(n\le n^2+1\) it follows that
\[ n^2+n\le n^2+(n^2+1)=2n^2+1. \]
Since
\[ 2n^2+1\le 2n^2+2=2(n^2+1), \]
we obtain
\[ n^2+n\le 2(n^2+1). \]
Dividing by \(n^2+1>0\), it follows that
\[ \frac{n^2+n}{n^2+1}\le 2. \]
Therefore
\[ |a_n|\le 2. \]
Since there exists \(K>0\), for instance \(K=2\), such that
\[ |a_n|\le K \]
for every \(n\in\mathbb{N}\), the sequence is bounded.
Exercise 20 — level ★★★☆☆
Determine whether the sequence
\[ a_n=(-1)^n n+\frac{1}{n+1} \]
is bounded above, bounded below, and bounded.
Answer
The sequence is bounded neither above nor below. Hence it is not bounded.
Solution
Consider the sequence
\[ a_n=(-1)^n n+\frac{1}{n+1}. \]
The leading term is \((-1)^n n\), which takes ever larger positive values at even indices and ever smaller (more negative) values at odd indices. The term
\[ \frac{1}{n+1} \]
is, on the other hand, always positive and lies between \(0\) and \(1\). We show rigorously that the sequence is bounded neither above nor below.
We first examine boundedness above. Let \(M\in\mathbb{R}\). We seek an index \(n\in\mathbb{N}\) such that
\[ a_n>M. \]
We choose an even index \(n=2q\). Then
\[ (-1)^n=(-1)^{2q}=1. \]
For such indices, the sequence becomes
\[ a_{2q}=2q+\frac{1}{2q+1}. \]
Since
\[ \frac{1}{2q+1}>0, \]
it follows that
\[ a_{2q}=2q+\frac{1}{2q+1}>2q. \]
Now choose \(q\in\mathbb{N}\) large enough that
\[ 2q>M. \]
Then
\[ a_{2q}>2q>M. \]
We have shown that, whatever \(M\in\mathbb{R}\) may be, there is a term of the sequence greater than \(M\). Hence the sequence is not bounded above.
We now examine boundedness below. Let \(m\in\mathbb{R}\). We seek an index \(n\in\mathbb{N}\) such that
\[ a_n<m. \]
We choose an odd index \(n=2q+1\). Then
\[ (-1)^n=(-1)^{2q+1}=-1. \]
For such indices, the sequence becomes
\[ a_{2q+1}=-(2q+1)+\frac{1}{2q+2}. \]
Since
\[ 0<\frac{1}{2q+2}\le 1, \]
we obtain
\[ a_{2q+1} = -(2q+1)+\frac{1}{2q+2} \le -(2q+1)+1. \]
Hence
\[ a_{2q+1}\le -2q. \]
Now choose \(q\in\mathbb{N}\) large enough that
\[ -2q<m. \]
This is possible because \(-2q\) tends to \(-\infty\) as \(q\) increases.
For this choice of \(q\), we have
\[ a_{2q+1}\le -2q<m. \]
We have shown that, whatever \(m\in\mathbb{R}\) may be, there is a term of the sequence less than \(m\). Hence the sequence is not bounded below.
Since the sequence is bounded neither above nor below, it is not bounded.