In this collection we present 20 step-by-step practice problems on the comparison theorem for sequences. The aim is not merely to compute limits, but to learn how to recognise which form of the theorem to apply: comparison between convergent sequences, the squeeze theorem, or comparison with sequences diverging to \(+\infty\) or to \(-\infty\).
In each problem we shall pay close attention to the direction of the inequalities and to the requirement that they hold eventually, that is, from some index onwards. This point is essential: to apply a theorem on limits correctly, an inequality need not hold for every index, but one must show that it holds for all sufficiently large indices.
The problems are arranged in order of increasing difficulty. In the first ones we shall use immediate estimates, such as \(-1\le \sin n\le 1\) or \(-1\le \cos n\le 1\). In the later ones, however, we shall need to construct sharper estimates and to choose with care the sequences to be compared.
Exercise 1 — level ★☆☆☆☆
Compute the limit
\[ \lim_{n\to+\infty}\frac{\sin n}{n}. \]
Answer
\[ \lim_{n\to+\infty}\frac{\sin n}{n}=0. \]
Solution
The difficulty of this exercise lies in the fact that the sequence \(\sin n\) has no limit. Indeed, the values of \(\sin n\) oscillate and do not approach a single real number. Nevertheless, we know that the sine is always bounded between \(-1\) and \(1\). Hence, for every \(n\in\mathbb{N}\),
\[ -1\le \sin n\le 1. \]
Since we wish to study \(\displaystyle \frac{\sin n}{n}\), we divide all sides by \(n\). For \(n\ge 1\) the number \(n\) is positive, so the direction of the inequalities is unchanged:
\[ -\frac{1}{n}\le \frac{\sin n}{n}\le \frac{1}{n}. \]
We now examine the two outer sequences. We have
\[ \lim_{n\to+\infty}\left(-\frac{1}{n}\right)=0 \qquad\text{and}\qquad \lim_{n\to+\infty}\frac{1}{n}=0. \]
Thus the sequence \(\displaystyle \frac{\sin n}{n}\) lies, for every \(n\ge 1\), between two sequences both tending to \(0\). By the squeeze theorem it follows that the sequence in the middle also tends to \(0\). Therefore
\[ \lim_{n\to+\infty}\frac{\sin n}{n}=0. \]
Exercise 2 — level ★☆☆☆☆
Compute the limit
\[ \lim_{n\to+\infty}\left(4+\frac{(-1)^n}{n}\right). \]
Answer
\[ \lim_{n\to+\infty}\left(4+\frac{(-1)^n}{n}\right)=4. \]
Solution
The sequence \(((-1)^n)\) oscillates between \(1\) and \(-1\). Indeed, for every \(n\),
\[ -1\le (-1)^n\le 1. \]
For \(n\ge 1\) we divide all sides of the inequality by \(n\). Since \(n\) is positive, the direction of the inequalities is unchanged:
\[ -\frac{1}{n}\le \frac{(-1)^n}{n}\le \frac{1}{n}. \]
We now add \(4\) to all sides. Adding the same number does not alter the direction of the inequalities, so we obtain
\[ 4-\frac{1}{n}\le 4+\frac{(-1)^n}{n}\le 4+\frac{1}{n}. \]
We examine the limits of the two outer sequences:
\[ \lim_{n\to+\infty}\left(4-\frac{1}{n}\right)=4 \qquad\text{and}\qquad \lim_{n\to+\infty}\left(4+\frac{1}{n}\right)=4. \]
The given sequence therefore lies, for every \(n\ge 1\), between two sequences both tending to \(4\). By the squeeze theorem, the sequence in the middle also tends to \(4\). Therefore
\[ \lim_{n\to+\infty}\left(4+\frac{(-1)^n}{n}\right)=4. \]
Exercise 3 — level ★☆☆☆☆
Compute the limit
\[ \lim_{n\to+\infty}\frac{\cos n}{n+1}. \]
Answer
\[ \lim_{n\to+\infty}\frac{\cos n}{n+1}=0. \]
Solution
Here too the sequence \(\cos n\) has no limit, since it oscillates. Nevertheless it is always bounded between \(-1\) and \(1\). Indeed, for every \(n\in\mathbb{N}\),
\[ -1\le \cos n\le 1. \]
We must divide by \(n+1\). For every \(n\in\mathbb{N}\) we have \(n+1>0\). Hence we may divide all sides of the inequality by \(n+1\) without changing the direction of the inequalities:
\[ -\frac{1}{n+1}\le \frac{\cos n}{n+1}\le \frac{1}{n+1}. \]
The two outer sequences both have limit \(0\); indeed,
\[ \lim_{n\to+\infty}\left(-\frac{1}{n+1}\right)=0 \qquad\text{and}\qquad \lim_{n\to+\infty}\frac{1}{n+1}=0. \]
The sequence \(\displaystyle \frac{\cos n}{n+1}\) thus lies between two sequences tending to the same limit. By the squeeze theorem we conclude that
\[ \lim_{n\to+\infty}\frac{\cos n}{n+1}=0. \]
Exercise 4 — level ★★☆☆☆
Compute the limit
\[ \lim_{n\to+\infty}\left(1+\frac{\sin n}{n}\right). \]
Answer
\[ \lim_{n\to+\infty}\left(1+\frac{\sin n}{n}\right)=1. \]
Solution
The given sequence is made up of a constant part, equal to \(1\), and an oscillating part, equal to \(\displaystyle \frac{\sin n}{n}\). To determine the limit, we must show that the oscillating part tends to \(0\).
For every \(n\in\mathbb{N}\) we know that
\[ -1\le \sin n\le 1. \]
For \(n\ge 1\), dividing all sides by \(n\), we obtain
\[ -\frac{1}{n}\le \frac{\sin n}{n}\le \frac{1}{n}. \]
We now add \(1\) to all sides of the inequality. Adding the same number to all sides does not change the direction of the inequalities:
\[ 1-\frac{1}{n}\le 1+\frac{\sin n}{n}\le 1+\frac{1}{n}. \]
We examine the limits of the two outer sequences:
\[ \lim_{n\to+\infty}\left(1-\frac{1}{n}\right)=1 \qquad\text{and}\qquad \lim_{n\to+\infty}\left(1+\frac{1}{n}\right)=1. \]
The given sequence therefore lies eventually between two sequences both tending to \(1\). By the squeeze theorem,
\[ \lim_{n\to+\infty}\left(1+\frac{\sin n}{n}\right)=1. \]
Exercise 5 — level ★★☆☆☆
Compute the limit
\[ \lim_{n\to+\infty}\frac{\sqrt{n^2+n}}{n}. \]
Answer
\[ \lim_{n\to+\infty}\frac{\sqrt{n^2+n}}{n}=1. \]
Solution
We first observe that, for \(n\ge 1\), the denominator \(n\) is positive. We may therefore rewrite the sequence by taking \(n^2\) out of the square root:
\[ \frac{\sqrt{n^2+n}}{n} = \frac{\sqrt{n^2\left(1+\frac{1}{n}\right)}}{n}. \]
Since \(n\ge 1\), we have \(\sqrt{n^2}=n\). Hence
\[ \frac{\sqrt{n^2\left(1+\frac{1}{n}\right)}}{n} = \frac{n\sqrt{1+\frac{1}{n}}}{n} = \sqrt{1+\frac{1}{n}}. \]
We must therefore study the limit
\[ \lim_{n\to+\infty}\sqrt{1+\frac{1}{n}}. \]
To apply the squeeze theorem, we construct a two-sided estimate. Since \(\displaystyle \frac{1}{n}\ge 0\), we have
\[ 1\le 1+\frac{1}{n}. \]
The square-root function is increasing on the non-negative reals, so
\[ 1\le \sqrt{1+\frac{1}{n}}. \]
Moreover, for every \(x\ge 0\) the inequality
\[ \sqrt{1+x}\le 1+x \]
holds. Indeed, both sides are non-negative, and on squaring, the inequality becomes
\[ 1+x\le (1+x)^2. \]
This holds for \(x\ge 0\), because
\[ (1+x)^2-(1+x)=x+x^2\ge 0. \]
Applying this inequality with \(\displaystyle x=\frac{1}{n}\), we obtain
\[ \sqrt{1+\frac{1}{n}}\le 1+\frac{1}{n}. \]
We thus have the two-sided estimate
\[ 1\le \sqrt{1+\frac{1}{n}}\le 1+\frac{1}{n}. \]
Now the two outer sequences both tend to \(1\):
\[ \lim_{n\to+\infty}1=1 \qquad\text{and}\qquad \lim_{n\to+\infty}\left(1+\frac{1}{n}\right)=1. \]
By the squeeze theorem it follows that
\[ \lim_{n\to+\infty}\sqrt{1+\frac{1}{n}}=1. \]
Since the original sequence coincides with \(\displaystyle \sqrt{1+\frac{1}{n}}\) for \(n\ge 1\), we finally obtain
\[ \lim_{n\to+\infty}\frac{\sqrt{n^2+n}}{n}=1. \]
Exercise 6 — level ★★☆☆☆
Compute the limit
\[ \lim_{n\to+\infty}\frac{n+\cos n}{n}. \]
Answer
\[ \lim_{n\to+\infty}\frac{n+\cos n}{n}=1. \]
Solution
The sequence contains the oscillating term \(\cos n\). On its own, \(\cos n\) has no limit, but it always lies between \(-1\) and \(1\). Indeed, for every \(n\in\mathbb{N}\),
\[ -1\le \cos n\le 1. \]
We add \(n\) to all sides of the inequality. We obtain
\[ n-1\le n+\cos n\le n+1. \]
Now we divide all sides by \(n\). For \(n\ge 1\) the number \(n\) is positive, so the direction of the inequalities is unchanged:
\[ \frac{n-1}{n}\le \frac{n+\cos n}{n}\le \frac{n+1}{n}. \]
We simplify the two outer sequences:
\[ \frac{n-1}{n}=1-\frac{1}{n}, \qquad \frac{n+1}{n}=1+\frac{1}{n}. \]
Hence, for every \(n\ge 1\), we have
\[ 1-\frac{1}{n}\le \frac{n+\cos n}{n}\le 1+\frac{1}{n}. \]
The two outer sequences both tend to \(1\); indeed,
\[ \lim_{n\to+\infty}\left(1-\frac{1}{n}\right)=1 \qquad\text{and}\qquad \lim_{n\to+\infty}\left(1+\frac{1}{n}\right)=1. \]
By the squeeze theorem, the sequence in the middle also tends to \(1\). Therefore
\[ \lim_{n\to+\infty}\frac{n+\cos n}{n}=1. \]
Exercise 7 — level ★★☆☆☆
Compute the limit
\[ \lim_{n\to+\infty}\frac{(-1)^n+\sin n}{\sqrt{n}}. \]
Answer
\[ \lim_{n\to+\infty}\frac{(-1)^n+\sin n}{\sqrt{n}}=0. \]
Solution
The numerator contains two oscillating terms: \((-1)^n\) and \(\sin n\). Neither has a limit, but both are bounded. Indeed, for every \(n\),
\[ |(-1)^n|=1 \]
and
\[ |\sin n|\le 1. \]
We now use the triangle inequality:
\[ |(-1)^n+\sin n|\le |(-1)^n|+|\sin n|. \]
Since \(|(-1)^n|=1\) and \(|\sin n|\le 1\), we obtain
\[ |(-1)^n+\sin n|\le 2. \]
Dividing by \(\sqrt{n}\), which is positive for every \(n\ge 1\), it follows that
\[ \left|\frac{(-1)^n+\sin n}{\sqrt{n}}\right| = \frac{|(-1)^n+\sin n|}{\sqrt{n}} \le \frac{2}{\sqrt{n}}. \]
Moreover, an absolute value is always non-negative, so
\[ 0\le \left|\frac{(-1)^n+\sin n}{\sqrt{n}}\right| \le \frac{2}{\sqrt{n}}. \]
Now the two outer sequences tend to \(0\):
\[ \lim_{n\to+\infty}0=0 \qquad\text{and}\qquad \lim_{n\to+\infty}\frac{2}{\sqrt{n}}=0. \]
By the squeeze theorem,
\[ \lim_{n\to+\infty} \left|\frac{(-1)^n+\sin n}{\sqrt{n}}\right| =0. \]
If the absolute value of a sequence tends to \(0\), then the sequence itself tends to \(0\) as well. Indeed, the distance of the sequence from \(0\) tends to \(0\). Hence
\[ \lim_{n\to+\infty}\frac{(-1)^n+\sin n}{\sqrt{n}}=0. \]
Exercise 8 — level ★★☆☆☆
Determine the limit of the sequence
\[ a_n=n^2+\sin n. \]
Answer
\[ \lim_{n\to+\infty}\left(n^2+\sin n\right)=+\infty. \]
Solution
The sequence \(n^2+\sin n\) is the sum of a term tending to \(+\infty\), namely \(n^2\), and an oscillating term, namely \(\sin n\). Since \(\sin n\) always remains between \(-1\) and \(1\), it cannot offset the growth of \(n^2\).
To make this argument rigorous, we use the estimate
\[ \sin n\ge -1. \]
Adding \(n^2\) to both sides, we obtain
\[ n^2+\sin n\ge n^2-1. \]
We now observe that
\[ \lim_{n\to+\infty}(n^2-1)=+\infty. \]
We have thus found a sequence, \(n^2-1\), which tends to \(+\infty\) and is less than or equal to the given sequence:
\[ n^2-1\le n^2+\sin n. \]
By comparison with sequences diverging to \(+\infty\), if a sequence is eventually bounded below by a sequence tending to \(+\infty\), then it too tends to \(+\infty\).
Therefore
\[ \lim_{n\to+\infty}\left(n^2+\sin n\right)=+\infty. \]
Exercise 9 — level ★★☆☆☆
Determine the limit of the sequence
\[ b_n=-n+\cos n. \]
Answer
\[ \lim_{n\to+\infty}\left(-n+\cos n\right)=-\infty. \]
Solution
The sequence \(-n+\cos n\) contains the term \(-n\), which tends to \(-\infty\), and the term \(\cos n\), which instead oscillates while remaining bounded between \(-1\) and \(1\). Intuitively, the oscillation of \(\cos n\) cannot prevent the sequence from descending towards \(-\infty\).
To prove this rigorously, we use the upper estimate
\[ \cos n\le 1. \]
Adding \(-n\) to both sides, we obtain
\[ -n+\cos n\le -n+1. \]
We now observe that
\[ \lim_{n\to+\infty}(-n+1)=-\infty. \]
Hence the given sequence is bounded above by a sequence tending to \(-\infty\):
\[ -n+\cos n\le -n+1. \]
In the comparison with sequences diverging to \(-\infty\), the direction is crucial: to conclude that a sequence tends to \(-\infty\), one must bound it above by a sequence tending to \(-\infty\). And that is precisely what we have done.
Therefore
\[ \lim_{n\to+\infty}\left(-n+\cos n\right)=-\infty. \]
Exercise 10 — level ★★★☆☆
Compute the limit
\[ \lim_{n\to+\infty}\frac{\sqrt{n^2+\sin n}}{n}. \]
Answer
\[ \lim_{n\to+\infty}\frac{\sqrt{n^2+\sin n}}{n}=1. \]
Solution
The presence of \(\sin n\) inside the root may make the limit less immediate. Nevertheless \(\sin n\) is a bounded term, whereas \(n^2\) grows without bound. We therefore expect the dominant behaviour of the root to be that of \(\sqrt{n^2}=n\), and hence the ratio to tend to \(1\).
To make the argument rigorous, we begin with the basic estimate
\[ -1\le \sin n\le 1. \]
Adding \(n^2\) to all sides, we obtain
\[ n^2-1\le n^2+\sin n\le n^2+1. \]
For \(n\ge 1\) the terms involved are non-negative. Since the square-root function is increasing on the non-negative reals, we may take square roots throughout:
\[ \sqrt{n^2-1}\le \sqrt{n^2+\sin n}\le \sqrt{n^2+1}. \]
We now divide by \(n\). For \(n\ge 1\) the number \(n\) is positive, so the direction of the inequalities is unchanged:
\[ \frac{\sqrt{n^2-1}}{n} \le \frac{\sqrt{n^2+\sin n}}{n} \le \frac{\sqrt{n^2+1}}{n}. \]
We rewrite the two outer sequences. Since \(n\ge 1\), we have \(\sqrt{n^2}=n\), so
\[ \frac{\sqrt{n^2-1}}{n} = \sqrt{\frac{n^2-1}{n^2}} = \sqrt{1-\frac{1}{n^2}}, \]
and
\[ \frac{\sqrt{n^2+1}}{n} = \sqrt{\frac{n^2+1}{n^2}} = \sqrt{1+\frac{1}{n^2}}. \]
We have thus obtained the estimate
\[ \sqrt{1-\frac{1}{n^2}} \le \frac{\sqrt{n^2+\sin n}}{n} \le \sqrt{1+\frac{1}{n^2}}. \]
Now
\[ \lim_{n\to+\infty}\left(1-\frac{1}{n^2}\right)=1 \qquad\text{and}\qquad \lim_{n\to+\infty}\left(1+\frac{1}{n^2}\right)=1. \]
Since the square root is continuous at positive points, it follows that
\[ \lim_{n\to+\infty}\sqrt{1-\frac{1}{n^2}}=1 \qquad\text{and}\qquad \lim_{n\to+\infty}\sqrt{1+\frac{1}{n^2}}=1. \]
The given sequence therefore lies eventually between two sequences both tending to \(1\). By the squeeze theorem,
\[ \lim_{n\to+\infty}\frac{\sqrt{n^2+\sin n}}{n}=1. \]
Exercise 11 — level ★★★☆☆
Compute the limit
\[ \lim_{n\to+\infty}\frac{n+\sin n}{n+\cos n}. \]
Answer
\[ \lim_{n\to+\infty}\frac{n+\sin n}{n+\cos n}=1. \]
Solution
The sequence contains two oscillating terms, \(\sin n\) and \(\cos n\). However, both are bounded, while the term \(n\) grows without bound. For this reason we expect the ratio to behave like
\[ \frac{n}{n}=1. \]
To prove this rigorously, we examine the distance of the sequence from the candidate limit \(1\):
\[ \left|\frac{n+\sin n}{n+\cos n}-1\right|. \]
We put the terms over a common denominator:
\[ \left|\frac{n+\sin n}{n+\cos n}-1\right| = \left|\frac{n+\sin n-(n+\cos n)}{n+\cos n}\right|. \]
Simplifying the numerator, we obtain
\[ \left|\frac{n+\sin n}{n+\cos n}-1\right| = \left|\frac{\sin n-\cos n}{n+\cos n}\right|. \]
We must now estimate the numerator and the denominator separately. For the numerator, using the triangle inequality, we have
\[ |\sin n-\cos n|\le |\sin n|+|\cos n|\le 2. \]
For the denominator, since \(\cos n\ge -1\), we have
\[ n+\cos n\ge n-1. \]
In particular, for \(n\ge 2\) the denominator is positive and
\[ |n+\cos n|=n+\cos n\ge n-1. \]
Hence, for every \(n\ge 2\),
\[ 0\le \left|\frac{n+\sin n}{n+\cos n}-1\right| \le \frac{2}{n-1}. \]
Since
\[ \lim_{n\to+\infty}\frac{2}{n-1}=0, \]
by the squeeze theorem we obtain
\[ \lim_{n\to+\infty} \left|\frac{n+\sin n}{n+\cos n}-1\right|=0. \]
This means that the distance between the given sequence and \(1\) tends to \(0\). Therefore
\[ \lim_{n\to+\infty}\frac{n+\sin n}{n+\cos n}=1. \]
Exercise 12 — level ★★★☆☆
Compute the limit
\[ \lim_{n\to+\infty}\frac{2n+(-1)^n}{3n+\sin n}. \]
Answer
\[ \lim_{n\to+\infty}\frac{2n+(-1)^n}{3n+\sin n}=\frac{2}{3}. \]
Solution
The dominant term in the numerator is \(2n\), while the dominant term in the denominator is \(3n\). The terms \((-1)^n\) and \(\sin n\), on the other hand, are bounded. For this reason the candidate limit is
\[ \frac{2n}{3n}=\frac{2}{3}. \]
To prove this by means of the comparison theorem, we consider the distance of the sequence from the candidate limit:
\[ \left|\frac{2n+(-1)^n}{3n+\sin n}-\frac{2}{3}\right|. \]
We put the terms over a common denominator:
\[ \left|\frac{2n+(-1)^n}{3n+\sin n}-\frac{2}{3}\right| = \left|\frac{3(2n+(-1)^n)-2(3n+\sin n)}{3(3n+\sin n)}\right|. \]
We expand the numerator:
\[ 3(2n+(-1)^n)-2(3n+\sin n) = 6n+3(-1)^n-6n-2\sin n. \]
Hence
\[ 3(2n+(-1)^n)-2(3n+\sin n) = 3(-1)^n-2\sin n. \]
We therefore obtain
\[ \left|\frac{2n+(-1)^n}{3n+\sin n}-\frac{2}{3}\right| = \frac{|3(-1)^n-2\sin n|}{3|3n+\sin n|}. \]
We estimate the numerator. By the triangle inequality,
\[ |3(-1)^n-2\sin n| \le 3|(-1)^n|+2|\sin n|. \]
Since \(|(-1)^n|=1\) and \(|\sin n|\le 1\), it follows that
\[ |3(-1)^n-2\sin n|\le 5. \]
We now estimate the denominator. Since \(\sin n\ge -1\), we have
\[ 3n+\sin n\ge 3n-1. \]
For every \(n\ge 1\) the number \(3n-1\) is positive, hence \(3n+\sin n\) is positive as well. Consequently
\[ |3n+\sin n|=3n+\sin n\ge 3n-1. \]
Therefore, for every \(n\ge 1\),
\[ 0\le \left|\frac{2n+(-1)^n}{3n+\sin n}-\frac{2}{3}\right| \le \frac{5}{3(3n-1)}. \]
Since
\[ \lim_{n\to+\infty}\frac{5}{3(3n-1)}=0, \]
by the squeeze theorem we have
\[ \lim_{n\to+\infty} \left|\frac{2n+(-1)^n}{3n+\sin n}-\frac{2}{3}\right|=0. \]
This means that the given sequence tends to \(\displaystyle \frac{2}{3}\). Hence
\[ \lim_{n\to+\infty}\frac{2n+(-1)^n}{3n+\sin n}=\frac{2}{3}. \]
Exercise 13 — level ★★★☆☆
Determine the limit of the sequence
\[ a_n=n+(-1)^n\sqrt{n}. \]
Answer
\[ \lim_{n\to+\infty}\left(n+(-1)^n\sqrt{n}\right)=+\infty. \]
Solution
The sequence contains a leading term, \(n\), which tends to \(+\infty\), and an oscillating term, \((-1)^n\sqrt{n}\), which may be positive or negative. We cannot therefore simply say that every term is greater than \(n\), because when \((-1)^n=-1\) the second term subtracts \(\sqrt{n}\).
To show that the sequence nonetheless tends to \(+\infty\), we must find a lower estimate that tends to \(+\infty\).
Since \((-1)^n\ge -1\), multiplying by \(\sqrt{n}\ge 0\) we obtain
\[ (-1)^n\sqrt{n}\ge -\sqrt{n}. \]
Adding \(n\) to both sides, it follows that
\[ n+(-1)^n\sqrt{n}\ge n-\sqrt{n}. \]
We must now show that \(n-\sqrt{n}\to+\infty\). For \(n\ge 4\), we have
\[ \sqrt{n}\le \frac{n}{2}. \]
Indeed, since both sides are non-negative, we may square them and obtain an equivalent inequality:
\[ n\le \frac{n^2}{4}. \]
For \(n>0\) this is equivalent to
\[ 4\le n, \]
which holds for \(n\ge 4\).
Hence, for \(n\ge 4\),
\[ n-\sqrt{n}\ge n-\frac{n}{2}=\frac{n}{2}. \]
Since
\[ \lim_{n\to+\infty}\frac{n}{2}=+\infty, \]
we also have \(n-\sqrt{n}\to+\infty\). Moreover, we have shown that eventually
\[ n+(-1)^n\sqrt{n}\ge n-\sqrt{n}. \]
By comparison with sequences diverging to \(+\infty\), the given sequence tends to \(+\infty\). Therefore
\[ \lim_{n\to+\infty}\left(n+(-1)^n\sqrt{n}\right)=+\infty. \]
Exercise 14 — level ★★★☆☆
Determine the limit of the sequence
\[ b_n=-n^2+n\sin n. \]
Answer
\[ \lim_{n\to+\infty}\left(-n^2+n\sin n\right)=-\infty. \]
Solution
The sequence contains the term \(-n^2\), which tends to \(-\infty\), and the term \(n\sin n\), which may be positive or negative. Although \(n\sin n\) is not bounded, it grows at most like \(n\), whereas \(n^2\) grows much more rapidly.
To prove rigorously that the sequence tends to \(-\infty\), we must find an upper estimate that tends to \(-\infty\).
Since \(\sin n\le 1\), multiplying by \(n\ge 0\) we obtain
\[ n\sin n\le n. \]
Adding \(-n^2\) to both sides, it follows that
\[ -n^2+n\sin n\le -n^2+n. \]
We now show that the sequence on the right tends to \(-\infty\). We have
\[ -n^2+n=-n(n-1). \]
For \(n\ge 2\) we have \(n-1\ge \frac{n}{2}\). Indeed,
\[ n-1\ge \frac{n}{2} \]
is equivalent to
\[ \frac{n}{2}\ge 1, \]
that is, \(n\ge 2\). Hence, for \(n\ge 2\),
\[ n(n-1)\ge \frac{n^2}{2}. \]
Multiplying by \(-1\), the direction of the inequality is reversed:
\[ -n(n-1)\le -\frac{n^2}{2}. \]
Therefore, for \(n\ge 2\),
\[ -n^2+n\le -\frac{n^2}{2}. \]
Since
\[ \lim_{n\to+\infty}\left(-\frac{n^2}{2}\right)=-\infty, \]
we also have
\[ \lim_{n\to+\infty}(-n^2+n)=-\infty. \]
We have thus found a sequence which tends to \(-\infty\) and which eventually bounds the given sequence above:
\[ -n^2+n\sin n\le -n^2+n. \]
By comparison with sequences diverging to \(-\infty\), it follows that
\[ \lim_{n\to+\infty}\left(-n^2+n\sin n\right)=-\infty. \]
Exercise 15 — level ★★★★☆
Compute the limit
\[ \lim_{n\to+\infty}\left(\sqrt{n^2+\sin n}-n\right). \]
Answer
\[ \lim_{n\to+\infty}\left(\sqrt{n^2+\sin n}-n\right)=0. \]
Solution
The expression contains a difference between two quantities both large: \(\sqrt{n^2+\sin n}\) and \(n\). At first sight the behaviour of the difference is not immediately clear. For this reason it is convenient to rationalise.
We multiply and divide by the conjugate:
\[ \sqrt{n^2+\sin n}-n = \frac{\left(\sqrt{n^2+\sin n}-n\right)\left(\sqrt{n^2+\sin n}+n\right)} {\sqrt{n^2+\sin n}+n}. \]
In the numerator we use the formula
\[ (A-B)(A+B)=A^2-B^2. \]
With
\[ A=\sqrt{n^2+\sin n} \qquad\text{and}\qquad B=n, \]
we obtain
\[ \left(\sqrt{n^2+\sin n}\right)^2-n^2 = n^2+\sin n-n^2 = \sin n. \]
Hence
\[ \sqrt{n^2+\sin n}-n = \frac{\sin n}{\sqrt{n^2+\sin n}+n}. \]
We now estimate the absolute value:
\[ \left|\sqrt{n^2+\sin n}-n\right| = \left|\frac{\sin n}{\sqrt{n^2+\sin n}+n}\right|. \]
Since \(|\sin n|\le 1\), we have
\[ \left|\sqrt{n^2+\sin n}-n\right| \le \frac{1}{\sqrt{n^2+\sin n}+n}. \]
We must now bound the denominator below. Since \(\sin n\ge -1\), we have
\[ n^2+\sin n\ge n^2-1. \]
For \(n\ge 1\) both sides are non-negative, so, taking square roots,
\[ \sqrt{n^2+\sin n}\ge \sqrt{n^2-1}. \]
Therefore
\[ \sqrt{n^2+\sin n}+n \ge \sqrt{n^2-1}+n. \]
In particular, since \(\sqrt{n^2-1}\ge 0\), for \(n\ge 1\) we have
\[ \sqrt{n^2+\sin n}+n\ge n. \]
Hence
\[ 0\le \left|\sqrt{n^2+\sin n}-n\right| \le \frac{1}{n}. \]
Since
\[ \lim_{n\to+\infty}\frac{1}{n}=0, \]
by the squeeze theorem it follows that
\[ \lim_{n\to+\infty} \left|\sqrt{n^2+\sin n}-n\right|=0. \]
If the distance of the sequence from \(0\) tends to \(0\), then the sequence itself tends to \(0\). Therefore
\[ \lim_{n\to+\infty}\left(\sqrt{n^2+\sin n}-n\right)=0. \]
Exercise 16 — level ★★★★☆
Compute the limit
\[ \lim_{n\to+\infty} n\left(\sqrt{1+\frac{1}{n}}-1\right). \]
Answer
\[ \lim_{n\to+\infty} n\left(\sqrt{1+\frac{1}{n}}-1\right)=\frac{1}{2}. \]
Solution
The expression contains a difference between two quantities both tending to \(1\):
\[ \sqrt{1+\frac{1}{n}} \qquad\text{and}\qquad 1. \]
To avoid an unwieldy form, we rationalise the difference. We multiply and divide by the conjugate:
\[ n\left(\sqrt{1+\frac{1}{n}}-1\right) = n\cdot \frac{\left(\sqrt{1+\frac{1}{n}}-1\right)\left(\sqrt{1+\frac{1}{n}}+1\right)} {\sqrt{1+\frac{1}{n}}+1}. \]
In the numerator we use the formula
\[ (A-B)(A+B)=A^2-B^2. \]
With
\[ A=\sqrt{1+\frac{1}{n}} \qquad\text{and}\qquad B=1, \]
we obtain
\[ \left(\sqrt{1+\frac{1}{n}}\right)^2-1 = 1+\frac{1}{n}-1 = \frac{1}{n}. \]
Hence
\[ n\left(\sqrt{1+\frac{1}{n}}-1\right) = n\cdot \frac{\frac{1}{n}}{\sqrt{1+\frac{1}{n}}+1}. \]
Cancelling \(n\) against \(\displaystyle \frac{1}{n}\), we obtain
\[ n\left(\sqrt{1+\frac{1}{n}}-1\right) = \frac{1}{\sqrt{1+\frac{1}{n}}+1}. \]
We now construct an estimate. Since \(\displaystyle \frac{1}{n}\ge 0\), we have
\[ 1\le \sqrt{1+\frac{1}{n}}. \]
Moreover, for every \(x\ge 0\) we have
\[ \sqrt{1+x}\le 1+x. \]
Applying this inequality with \(\displaystyle x=\frac{1}{n}\), we obtain
\[ \sqrt{1+\frac{1}{n}}\le 1+\frac{1}{n}. \]
Hence
\[ 1\le \sqrt{1+\frac{1}{n}}\le 1+\frac{1}{n}. \]
Adding \(1\) to all sides,
\[ 2\le \sqrt{1+\frac{1}{n}}+1\le 2+\frac{1}{n}. \]
All sides are positive. Passing to reciprocals, the direction of the inequalities is reversed:
\[ \frac{1}{2+\frac{1}{n}} \le \frac{1}{\sqrt{1+\frac{1}{n}}+1} \le \frac{1}{2}. \]
The two outer sequences both tend to \(\displaystyle \frac{1}{2}\); indeed,
\[ \lim_{n\to+\infty}\frac{1}{2+\frac{1}{n}}=\frac{1}{2} \qquad\text{and}\qquad \lim_{n\to+\infty}\frac{1}{2}=\frac{1}{2}. \]
By the squeeze theorem,
\[ \lim_{n\to+\infty} \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \frac{1}{2}. \]
Since the original sequence coincides with this expression, we conclude that
\[ \lim_{n\to+\infty} n\left(\sqrt{1+\frac{1}{n}}-1\right)=\frac{1}{2}. \]
Exercise 17 — level ★★★★☆
Compute the limit
\[ \lim_{n\to+\infty}\frac{n^2}{n^3+(-1)^n}. \]
Answer
\[ \lim_{n\to+\infty}\frac{n^2}{n^3+(-1)^n}=0. \]
Solution
The denominator contains the dominant term \(n^3\) and the oscillating term \((-1)^n\). Since \((-1)^n\) takes only the values \(1\) and \(-1\), its effect is negligible compared with \(n^3\).
To apply the comparison theorem, we must estimate the sequence in absolute value. For every \(n\),
\[ (-1)^n\ge -1. \]
Hence
\[ n^3+(-1)^n\ge n^3-1. \]
For \(n\ge 2\) we have
\[ n^3-1\ge \frac{n^3}{2}. \]
Indeed, this inequality is equivalent to
\[ \frac{n^3}{2}\ge 1, \]
which holds for \(n\ge 2\). Hence, for \(n\ge 2\),
\[ n^3+(-1)^n\ge \frac{n^3}{2}. \]
In particular the denominator is eventually positive. Therefore, for \(n\ge 2\),
\[ 0\le \frac{n^2}{n^3+(-1)^n} \le \frac{n^2}{\frac{n^3}{2}}. \]
Simplifying the term on the right,
\[ \frac{n^2}{\frac{n^3}{2}} = \frac{2n^2}{n^3} = \frac{2}{n}. \]
We thus have
\[ 0\le \frac{n^2}{n^3+(-1)^n}\le \frac{2}{n} \]
eventually. Since
\[ \lim_{n\to+\infty}0=0 \qquad\text{and}\qquad \lim_{n\to+\infty}\frac{2}{n}=0, \]
by the squeeze theorem it follows that
\[ \lim_{n\to+\infty}\frac{n^2}{n^3+(-1)^n}=0. \]
Exercise 18 — level ★★★★☆
Determine the limit of the sequence
\[ a_n=n^2+n(-1)^n. \]
Answer
\[ \lim_{n\to+\infty}\left(n^2+n(-1)^n\right)=+\infty. \]
Solution
The sequence contains a leading term \(n^2\), which tends to \(+\infty\), and an oscillating term \(n(-1)^n\), which may equal \(n\) or \(-n\). The least favourable case, for proving divergence to \(+\infty\), occurs when the oscillating term is negative.
Since
\[ (-1)^n\ge -1, \]
multiplying by \(n\ge 0\) we obtain
\[ n(-1)^n\ge -n. \]
Adding \(n^2\) to both sides,
\[ n^2+n(-1)^n\ge n^2-n. \]
We now show that the sequence \(n^2-n\) tends to \(+\infty\). We have
\[ n^2-n=n(n-1). \]
For \(n\ge 2\) we have
\[ n-1\ge \frac{n}{2}. \]
Hence, for \(n\ge 2\),
\[ n(n-1)\ge \frac{n^2}{2}. \]
Since
\[ \lim_{n\to+\infty}\frac{n^2}{2}=+\infty, \]
we also have \(n^2-n\to+\infty\). We have thus found a lower estimate:
\[ n^2+n(-1)^n\ge n^2-n, \]
where the sequence on the right tends to \(+\infty\).
By comparison with sequences diverging to \(+\infty\), we conclude that
\[ \lim_{n\to+\infty}\left(n^2+n(-1)^n\right)=+\infty. \]
Exercise 19 — level ★★★★☆
Compute the limit
\[ \lim_{n\to+\infty}\frac{\sqrt{n^4+n^2\sin n}}{n^2}. \]
Answer
\[ \lim_{n\to+\infty}\frac{\sqrt{n^4+n^2\sin n}}{n^2}=1. \]
Solution
The leading term inside the root is \(n^4\), while \(n^2\sin n\) is an oscillating term of lower order. We therefore expect the root to behave like
\[ \sqrt{n^4}=n^2. \]
To prove this rigorously, we begin with the estimate
\[ -1\le \sin n\le 1. \]
Multiplying all sides by \(n^2\), which is non-negative, we obtain
\[ -n^2\le n^2\sin n\le n^2. \]
Adding \(n^4\) to all sides,
\[ n^4-n^2\le n^4+n^2\sin n\le n^4+n^2. \]
For \(n\ge 1\) the three sides are non-negative. Since the square-root function is increasing on the non-negative reals, we may take square roots throughout:
\[ \sqrt{n^4-n^2} \le \sqrt{n^4+n^2\sin n} \le \sqrt{n^4+n^2}. \]
We now divide by \(n^2\), which is positive for \(n\ge 1\):
\[ \frac{\sqrt{n^4-n^2}}{n^2} \le \frac{\sqrt{n^4+n^2\sin n}}{n^2} \le \frac{\sqrt{n^4+n^2}}{n^2}. \]
We rewrite the outer sequences:
\[ \frac{\sqrt{n^4-n^2}}{n^2} = \sqrt{\frac{n^4-n^2}{n^4}} = \sqrt{1-\frac{1}{n^2}}, \]
and
\[ \frac{\sqrt{n^4+n^2}}{n^2} = \sqrt{\frac{n^4+n^2}{n^4}} = \sqrt{1+\frac{1}{n^2}}. \]
Hence
\[ \sqrt{1-\frac{1}{n^2}} \le \frac{\sqrt{n^4+n^2\sin n}}{n^2} \le \sqrt{1+\frac{1}{n^2}}. \]
The two outer sequences both tend to \(1\):
\[ \lim_{n\to+\infty}\sqrt{1-\frac{1}{n^2}}=1 \qquad\text{and}\qquad \lim_{n\to+\infty}\sqrt{1+\frac{1}{n^2}}=1. \]
By the squeeze theorem, the sequence in the middle also tends to \(1\). Therefore
\[ \lim_{n\to+\infty}\frac{\sqrt{n^4+n^2\sin n}}{n^2}=1. \]
Exercise 20 — level ★★★★★
Determine whether, from the mere fact that a sequence \((b_n)\) eventually satisfies
\[ 0\le b_n\le 1, \]
one may conclude that \((b_n)\) is convergent.
Answer
No. From the mere fact that \(0\le b_n\le 1\) eventually, one cannot conclude that \((b_n)\) is convergent.
Solution
The squeeze theorem allows one to conclude the convergence of a middle sequence when it lies eventually between two sequences tending to the same limit. Here, however, we know only that
\[ 0\le b_n\le 1 \]
eventually. The two outer sequences are the constant sequences \(0\) and \(1\), which have different limits:
\[ \lim_{n\to+\infty}0=0 \qquad\text{and}\qquad \lim_{n\to+\infty}1=1. \]
Since the two outer limits do not coincide, the squeeze theorem is not applicable. The inequality merely tells us that the terms of \((b_n)\) remain in the interval \([0,1]\), but this is not enough to guarantee the existence of the limit.
To show that convergence cannot be concluded, it suffices to give a counterexample. Consider the sequence
\[ b_n=\frac{1+(-1)^n}{2}. \]
If \(n\) is even, then \((-1)^n=1\), so
\[ b_n=\frac{1+1}{2}=1. \]
If \(n\) is odd, then \((-1)^n=-1\), so
\[ b_n=\frac{1-1}{2}=0. \]
Hence the sequence takes alternately the values \(1\) and \(0\). In particular, for every \(n\),
\[ 0\le b_n\le 1. \]
However, \((b_n)\) does not converge. Indeed, one subsequence of its terms is always equal to \(1\), while another subsequence is always equal to \(0\). More precisely,
\[ b_{2k}=1 \qquad\text{and}\qquad b_{2k+1}=0. \]
The subsequence \((b_{2k})\) tends to \(1\), while the subsequence \((b_{2k+1})\) tends to \(0\). Since a convergent sequence cannot have two subsequences with different limits, \((b_n)\) is not convergent.
Therefore, from the mere fact that a sequence lies eventually between \(0\) and \(1\), one cannot conclude that it is convergent. To apply the squeeze theorem it is essential that the two outer sequences tend to the same limit.