This collection offers 20 practice problems on numerical sequences and the limits of sequences, with particular attention to the distinction between convergent sequences, sequences divergent to infinity, and oscillating sequences.
Each exercise is solved step by step, recalling the fundamental definitions and showing how to recognise the behaviour of a sequence as the index \(n\) tends to \(+\infty\).
Throughout the exercises, the index \(n\) is assumed to satisfy \(n\geq 1\).
Exercise 1 — level ★☆☆☆☆
Study the behaviour of the sequence
\[ a_n=\frac{1}{n}. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is convergent, and
\[ \lim_{n\to+\infty}\frac{1}{n}=0. \]
Solution
The sequence is
\[ a_n=\frac{1}{n}. \]
Its first terms are
\[ 1,\ \frac12,\ \frac13,\ \frac14,\ \ldots \]
As \(n\) grows, the denominator becomes larger and larger while the numerator stays equal to \(1\). The terms therefore become smaller and smaller and approach \(0\).
To prove this from the definition, fix an arbitrary number
\[ \varepsilon>0. \]
We seek an index \(n_\varepsilon\) such that, for every \(n\geq n_\varepsilon\),
\[ \left|\frac1n-0\right|<\varepsilon. \]
Since
\[ \left|\frac1n-0\right|=\frac1n, \]
we must require
\[ \frac1n<\varepsilon. \]
This inequality is equivalent to
\[ n>\frac1\varepsilon. \]
We therefore choose \(n_\varepsilon\in\mathbb N\) such that
\[ n_\varepsilon>\frac1\varepsilon. \]
Then, for every \(n\geq n_\varepsilon\),
\[ n\geq n_\varepsilon>\frac1\varepsilon, \]
and hence
\[ \frac1n<\varepsilon. \]
We have shown that, for every \(\varepsilon>0\), from some index onwards all the terms of the sequence lie within \(\varepsilon\) of \(0\).
Therefore
\[ \lim_{n\to+\infty}\frac1n=0. \]
The sequence is thus convergent.
Exercise 2 — level ★☆☆☆☆
Study the behaviour of the sequence
\[ a_n=\frac{n}{n+1}. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is convergent, and
\[ \lim_{n\to+\infty}\frac{n}{n+1}=1. \]
Solution
Consider the sequence
\[ a_n=\frac{n}{n+1}. \]
We rewrite the general term as follows:
\[ \frac{n}{n+1}=\frac{n+1-1}{n+1}=1-\frac{1}{n+1}. \]
Since
\[ \frac{1}{n+1}\to0 \]
as \(n\to+\infty\), we expect that
\[ 1-\frac{1}{n+1}\to1. \]
Let us verify this from the definition. We need to examine the distance between \(a_n\) and \(1\):
\[ |a_n-1|=\left|\frac{n}{n+1}-1\right|. \]
We compute:
\[ \frac{n}{n+1}-1=\frac{n-(n+1)}{n+1}=-\frac{1}{n+1}. \]
Hence
\[ |a_n-1|=\left|-\frac{1}{n+1}\right|=\frac{1}{n+1}. \]
Given \(\varepsilon>0\), we want
\[ \frac{1}{n+1}<\varepsilon. \]
This inequality holds when
\[ n+1>\frac1\varepsilon, \]
that is, when
\[ n>\frac1\varepsilon-1. \]
Choosing \(n_\varepsilon\in\mathbb N\) such that
\[ n_\varepsilon>\frac1\varepsilon-1, \]
for every \(n\geq n_\varepsilon\) we obtain
\[ |a_n-1|<\varepsilon. \]
Therefore
\[ \lim_{n\to+\infty}\frac{n}{n+1}=1. \]
The sequence is convergent.
Exercise 3 — level ★☆☆☆☆
Study the behaviour of the sequence
\[ a_n=3. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is convergent, and
\[ \lim_{n\to+\infty}3=3. \]
Solution
The sequence is constant:
\[ a_n=3 \]
for every \(n\in\mathbb N\).
Every term of the sequence equals \(3\). The sequence does not merely approach \(3\): it is always exactly equal to \(3\).
Let us verify this from the definition. We must show that, for every \(\varepsilon>0\), there exists an index \(n_\varepsilon\) such that, for every \(n\geq n_\varepsilon\),
\[ |a_n-3|<\varepsilon. \]
Since \(a_n=3\), we have
\[ |a_n-3|=|3-3|=0. \]
But
\[ 0<\varepsilon \]
for every \(\varepsilon>0\).
Hence the inequality holds for every \(n\). We may choose, for instance,
\[ n_\varepsilon=1. \]
It follows that
\[ \lim_{n\to+\infty}a_n=3. \]
The sequence is thus convergent.
Exercise 4 — level ★★☆☆☆
Study the behaviour of the sequence
\[ a_n=\frac{2n+1}{n}. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is convergent, and
\[ \lim_{n\to+\infty}\frac{2n+1}{n}=2. \]
Solution
We rewrite the general term by splitting the fraction:
\[ a_n=\frac{2n+1}{n}=\frac{2n}{n}+\frac1n=2+\frac1n. \]
Since
\[ \frac1n\to0, \]
we expect that
\[ 2+\frac1n\to2. \]
We check the distance from \(2\):
\[ |a_n-2|=\left|2+\frac1n-2\right|=\frac1n. \]
Given \(\varepsilon>0\), we want
\[ |a_n-2|<\varepsilon. \]
Since
\[ |a_n-2|=\frac1n, \]
it suffices to require
\[ \frac1n<\varepsilon. \]
This inequality holds if
\[ n>\frac1\varepsilon. \]
We therefore choose \(n_\varepsilon\in\mathbb N\) such that
\[ n_\varepsilon>\frac1\varepsilon. \]
Then, for every \(n\geq n_\varepsilon\),
\[ |a_n-2|<\varepsilon. \]
Therefore
\[ \lim_{n\to+\infty}\frac{2n+1}{n}=2. \]
The sequence is convergent.
Exercise 5 — level ★★☆☆☆
Study the behaviour of the sequence
\[ a_n=\frac{3n-2}{2n+5}. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is convergent, and
\[ \lim_{n\to+\infty}\frac{3n-2}{2n+5}=\frac32. \]
Solution
Consider
\[ a_n=\frac{3n-2}{2n+5}. \]
Both numerator and denominator are first-degree polynomials in \(n\). As \(n\to+\infty\), the leading behaviour is governed by the highest-degree terms:
\[ 3n \quad \text{and} \quad 2n. \]
We divide numerator and denominator by \(n\):
\[ \frac{3n-2}{2n+5}=\frac{3-\displaystyle \frac2n}{2+\displaystyle \frac5n}. \]
Since
\[ \frac2n\to0 \qquad\text{and}\qquad \frac5n\to0, \]
we obtain
\[ \frac{3-\displaystyle \frac2n}{2+\displaystyle \frac5n}\to\frac{3-0}{2+0}=\frac32. \]
Hence
\[ \lim_{n\to+\infty}\frac{3n-2}{2n+5}=\frac32. \]
Since the limit is a finite real number, the sequence is convergent.
Exercise 6 — level ★★☆☆☆
Study the behaviour of the sequence
\[ a_n=n. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is divergent to \(+\infty\), and
\[ \lim_{n\to+\infty}n=+\infty. \]
Solution
The sequence is
\[ a_n=n. \]
Its terms are
\[ 1,\ 2,\ 3,\ 4,\ \ldots \]
and become arbitrarily large.
To prove that
\[ \lim_{n\to+\infty}n=+\infty, \]
we use the definition of divergence to \(+\infty\). We must show that, for every \(M>0\), there exists \(n_M\in\mathbb N\) such that, for every \(n\geq n_M\),
\[ a_n>M. \]
Since \(a_n=n\), we need
\[ n>M. \]
We choose \(n_M\in\mathbb N\) such that
\[ n_M>M. \]
Then, if \(n\geq n_M\),
\[ n\geq n_M>M. \]
Hence
\[ a_n=n>M. \]
This proves that the sequence diverges to \(+\infty\).
Exercise 7 — level ★★☆☆☆
Study the behaviour of the sequence
\[ a_n=n^2. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is divergent to \(+\infty\), and
\[ \lim_{n\to+\infty}n^2=+\infty. \]
Solution
The sequence is
\[ a_n=n^2. \]
Its first terms are
\[ 1,\ 4,\ 9,\ 16,\ \ldots \]
and grow without bound.
We prove that the sequence diverges to \(+\infty\). Fix an arbitrary number \(M>0\). We seek an index \(n_M\) such that, for every \(n\geq n_M\),
\[ n^2>M. \]
Since \(n\) is positive, the inequality
\[ n^2>M \]
holds when
\[ n>\sqrt{M}. \]
We therefore choose \(n_M\in\mathbb N\) such that
\[ n_M>\sqrt{M}. \]
Then, for every \(n\geq n_M\),
\[ n\geq n_M>\sqrt{M}. \]
Squaring, we obtain
\[ n^2>M. \]
Thus, for every positive threshold \(M\), from some index onwards the terms of the sequence exceed \(M\).
Therefore
\[ \lim_{n\to+\infty}n^2=+\infty. \]
The sequence is divergent to \(+\infty\).
Exercise 8 — level ★★☆☆☆
Study the behaviour of the sequence
\[ a_n=-n. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is divergent to \(-\infty\), and
\[ \lim_{n\to+\infty}(-n)=-\infty. \]
Solution
The sequence is
\[ a_n=-n. \]
Its terms are
\[ -1,\ -2,\ -3,\ -4,\ \ldots \]
and become smaller and smaller.
To prove that
\[ \lim_{n\to+\infty}(-n)=-\infty, \]
we use the definition of divergence to \(-\infty\). Fix \(M>0\). We must find \(n_M\in\mathbb N\) such that, for every \(n\geq n_M\),
\[ -n<-M. \]
Multiplying both sides by \(-1\) reverses the inequality:
\[ n>M. \]
We then choose \(n_M\in\mathbb N\) such that
\[ n_M>M. \]
If \(n\geq n_M\), then
\[ n\geq n_M>M, \]
and hence
\[ -n<-M. \]
This shows that the terms of the sequence drop below any negative threshold.
Therefore
\[ \lim_{n\to+\infty}(-n)=-\infty. \]
The sequence is divergent to \(-\infty\).
Exercise 9 — level ★★☆☆☆
Study the behaviour of the sequence
\[ a_n=-2n+5. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is divergent to \(-\infty\), and
\[ \lim_{n\to+\infty}(-2n+5)=-\infty. \]
Solution
The sequence is
\[ a_n=-2n+5. \]
The dominant term is \(-2n\), which tends to \(-\infty\). The constant term \(5\) does not affect the behaviour at infinity.
Let us prove this from the definition. Fix \(M>0\). We seek \(n_M\) such that, for every \(n\geq n_M\),
\[ -2n+5<-M. \]
We solve the inequality:
\[ -2n+5<-M. \]
Subtracting \(5\) from both sides, we obtain
\[ -2n<-M-5. \]
Dividing by \(-2\) reverses the inequality:
\[ n>\frac{M+5}{2}. \]
We choose \(n_M\in\mathbb N\) such that
\[ n_M>\frac{M+5}{2}. \]
Then, for every \(n\geq n_M\),
\[ -2n+5<-M. \]
Hence the sequence diverges to \(-\infty\).
Therefore
\[ \lim_{n\to+\infty}(-2n+5)=-\infty. \]
Exercise 10 — level ★★☆☆☆
Study the behaviour of the sequence
\[ a_n=(-1)^n. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is oscillating.
Solution
Consider the sequence
\[ a_n=(-1)^n. \]
Its terms are
\[ -1,\ 1,\ -1,\ 1,\ -1,\ 1,\ \ldots \]
The sequence oscillates between the values \(-1\) and \(1\), so it does not appear to approach any single real number.
Consider the even indices. If \(n=2k\), then
\[ a_{2k}=(-1)^{2k}=1. \]
The subsequence of even-indexed terms is therefore constant and equal to \(1\), so that
\[ \lim_{k\to+\infty}a_{2k}=1. \]
Now consider the odd indices. If \(n=2k-1\), then
\[ a_{2k-1}=(-1)^{2k-1}=-1. \]
Hence
\[ \lim_{k\to+\infty}a_{2k-1}=-1. \]
We have found two subsequences converging to different limits:
\[ 1 \qquad\text{and}\qquad -1. \]
The sequence therefore cannot be convergent.
Moreover, it is bounded, since for every \(n\in\mathbb N\)
\[ -1\leq (-1)^n\leq 1. \]
Being bounded, it can diverge neither to \(+\infty\) nor to \(-\infty\).
The sequence is thus neither convergent nor divergent. It is therefore oscillating.
Exercise 11 — level ★★☆☆☆
Study the behaviour of the sequence
\[ a_n=1+(-1)^n. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is oscillating.
Solution
The sequence is
\[ a_n=1+(-1)^n. \]
We examine the even and odd indices separately.
If \(n=2k\), then
\[ a_{2k}=1+(-1)^{2k}=1+1=2. \]
Hence
\[ \lim_{k\to+\infty}a_{2k}=2. \]
If instead \(n=2k-1\), then
\[ a_{2k-1}=1+(-1)^{2k-1}=1-1=0. \]
Hence
\[ \lim_{k\to+\infty}a_{2k-1}=0. \]
The sequence has two subsequences with different limits:
\[ 2 \qquad\text{and}\qquad 0. \]
Consequently the sequence is not convergent.
Moreover, its terms take only the values \(0\) and \(2\). The sequence is therefore bounded:
\[ 0\leq a_n\leq 2 \]
for every \(n\in\mathbb N\).
Being bounded, it can diverge neither to \(+\infty\) nor to \(-\infty\).
The sequence is therefore oscillating.
Exercise 12 — level ★★★☆☆
Study the behaviour of the sequence
\[ a_n=\frac{(-1)^n}{n}. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is convergent, and
\[ \lim_{n\to+\infty}\frac{(-1)^n}{n}=0. \]
Solution
The sequence is
\[ a_n=\frac{(-1)^n}{n}. \]
The factor \((-1)^n\) makes the sign of the terms alternate, while the denominator \(n\) grows ever larger.
We study the absolute value:
\[ |a_n|=\left|\frac{(-1)^n}{n}\right|=\frac{|(-1)^n|}{n}. \]
Since
\[ |(-1)^n|=1, \]
we have
\[ |a_n|=\frac1n. \]
Since
\[ \frac1n\to0, \]
\(a_n\) tends to \(0\) as well.
Let us verify this directly. Given \(\varepsilon>0\), we want
\[ |a_n-0|<\varepsilon. \]
But
\[ |a_n-0|=|a_n|=\frac1n. \]
It therefore suffices to require
\[ \frac1n<\varepsilon. \]
As already seen, this condition holds for
\[ n>\frac1\varepsilon. \]
Choosing \(n_\varepsilon\in\mathbb N\) such that
\[ n_\varepsilon>\frac1\varepsilon, \]
for every \(n\geq n_\varepsilon\) we have
\[ |a_n-0|<\varepsilon. \]
Therefore
\[ \lim_{n\to+\infty}\frac{(-1)^n}{n}=0. \]
The sequence is convergent. This example shows that a sequence may alternate in sign and yet converge, provided the amplitude of the oscillation tends to zero.
Exercise 13 — level ★★★☆☆
Study the behaviour of the sequence
\[ a_n=\frac{n^2+1}{n}. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is divergent to \(+\infty\), and
\[ \lim_{n\to+\infty}\frac{n^2+1}{n}=+\infty. \]
Solution
We rewrite the general term:
\[ a_n=\frac{n^2+1}{n}=n+\frac1n. \]
The term \(n\) tends to \(+\infty\), while the term \(\frac1n\) tends to \(0\). The dominant behaviour is therefore that of \(n\).
We prove that \(a_n\to+\infty\). Given \(M>0\), we want
\[ n+\frac1n>M. \]
Since
\[ \frac1n>0 \]
for every \(n\in\mathbb N\), we have
\[ n+\frac1n>n. \]
Hence, if we require
\[ n>M, \]
we automatically obtain
\[ n+\frac1n>M. \]
We choose \(n_M\in\mathbb N\) such that
\[ n_M>M. \]
Then, for every \(n\geq n_M\),
\[ a_n=n+\frac1n>n\geq n_M>M. \]
Hence the sequence diverges to \(+\infty\).
Exercise 14 — level ★★★☆☆
Study the behaviour of the sequence
\[ a_n=\frac{n}{n^2+1}. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is convergent, and
\[ \lim_{n\to+\infty}\frac{n}{n^2+1}=0. \]
Solution
Consider
\[ a_n=\frac{n}{n^2+1}. \]
The denominator has degree \(2\), whereas the numerator has degree \(1\). As \(n\to+\infty\), the denominator grows faster than the numerator.
We divide numerator and denominator by \(n^2\):
\[ \frac{n}{n^2+1} = \frac{\frac1n}{1+\frac1{n^2}}. \]
Since
\[ \frac1n\to0 \qquad\text{and}\qquad \frac1{n^2}\to0, \]
we obtain
\[ \frac{\frac1n}{1+\frac1{n^2}}\to\frac0{1+0}=0. \]
Therefore
\[ \lim_{n\to+\infty}\frac{n}{n^2+1}=0. \]
The limit is finite, so the sequence is convergent.
Exercise 15 — level ★★★☆☆
Study the behaviour of the sequence
\[ a_n=\sin\left(\frac{n\pi}{2}\right). \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is oscillating.
Solution
Consider
\[ a_n=\sin\left(\frac{n\pi}{2}\right). \]
We compute a few terms:
\[ a_1=\sin\left(\frac{\pi}{2}\right)=1, \]
\[ a_2=\sin(\pi)=0, \]
\[ a_3=\sin\left(\frac{3\pi}{2}\right)=-1, \]
\[ a_4=\sin(2\pi)=0. \]
Thus the terms repeat according to the pattern
\[ 1,\ 0,\ -1,\ 0,\ 1,\ 0,\ -1,\ 0,\ \ldots \]
The sequence does not approach any single real number.
Indeed, taking the indices of the form \(4k+1\), we obtain
\[ a_{4k+1}=1. \]
Hence
\[ \lim_{k\to+\infty}a_{4k+1}=1. \]
Taking instead the indices of the form \(4k+2\), we obtain
\[ a_{4k+2}=0. \]
Hence
\[ \lim_{k\to+\infty}a_{4k+2}=0. \]
The sequence has two subsequences converging to different limits. Consequently it is not convergent.
Moreover, for every \(n\in\mathbb N\),
\[ -1\leq a_n\leq 1. \]
The sequence is therefore bounded and can diverge neither to \(+\infty\) nor to \(-\infty\).
It is thus oscillating.
Exercise 16 — level ★★★★☆
Study the behaviour of the sequence
\[ a_n=n(-1)^n. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is oscillating.
Solution
The sequence is
\[ a_n=n(-1)^n. \]
We examine the even and odd indices separately.
If \(n=2k\), then
\[ a_{2k}=2k(-1)^{2k}=2k. \]
Hence
\[ a_{2k}\to+\infty \]
as \(k\to+\infty\).
If instead \(n=2k-1\), then
\[ a_{2k-1}=(2k-1)(-1)^{2k-1}=-(2k-1). \]
Hence
\[ a_{2k-1}\to-\infty \]
as \(k\to+\infty\).
The sequence cannot converge to a real number, since some of its terms increase without upper bound while others decrease without lower bound.
It does not diverge to \(+\infty\), because the odd-indexed terms become ever more negative and so cannot, from any point onwards, exceed every positive threshold.
Nor does it diverge to \(-\infty\), because the even-indexed terms become ever more positive and so cannot, from any point onwards, fall below every negative threshold.
Thus the sequence is not convergent and diverges neither to \(+\infty\) nor to \(-\infty\).
It is therefore oscillating.
Exercise 17 — level ★★★★☆
Study the behaviour of the sequence
\[ a_n=\frac{(-1)^n n}{n+1}. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is oscillating.
Solution
Consider
\[ a_n=\frac{(-1)^n n}{n+1}. \]
The factor
\[ \frac{n}{n+1} \]
tends to \(1\), while the factor \((-1)^n\) alternates the sign.
We study the even and odd subsequences.
If \(n=2k\), then
\[ a_{2k}=\frac{(-1)^{2k}\,2k}{2k+1}=\frac{2k}{2k+1}. \]
Since
\[ \frac{2k}{2k+1}\to1, \]
we have
\[ \lim_{k\to+\infty}a_{2k}=1. \]
If instead \(n=2k-1\), then
\[ a_{2k-1}=\frac{(-1)^{2k-1}(2k-1)}{2k}=-\frac{2k-1}{2k}. \]
Since
\[ \frac{2k-1}{2k}\to1, \]
we obtain
\[ \lim_{k\to+\infty}a_{2k-1}=-1. \]
The sequence thus has two subsequences converging to different limits:
\[ 1 \qquad\text{and}\qquad -1. \]
The sequence is therefore not convergent.
Moreover, it is bounded, since
\[ \left|\frac{(-1)^n n}{n+1}\right|=\frac{n}{n+1}<1. \]
Being bounded, it diverges neither to \(+\infty\) nor to \(-\infty\).
The sequence is thus oscillating.
Exercise 18 — level ★★★★☆
Study the behaviour of the sequence
\[ a_n=\frac{n^2+3n}{2n^2-1}. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is convergent, and
\[ \lim_{n\to+\infty}\frac{n^2+3n}{2n^2-1}=\frac12. \]
Solution
Consider
\[ a_n=\frac{n^2+3n}{2n^2-1}. \]
Numerator and denominator are polynomials of the same degree, namely degree \(2\).
In such cases the limit equals the ratio of the leading coefficients. Let us verify this by dividing numerator and denominator by \(n^2\):
\[ \frac{n^2+3n}{2n^2-1} = \frac{1+\frac3n}{2-\frac1{n^2}}. \]
Now
\[ \frac3n\to0 \qquad\text{and}\qquad \frac1{n^2}\to0. \]
Hence
\[ \frac{1+\frac3n}{2-\frac1{n^2}}\to\frac{1+0}{2-0}=\frac12. \]
Therefore
\[ \lim_{n\to+\infty}\frac{n^2+3n}{2n^2-1}=\frac12. \]
Since the limit is real and finite, the sequence is convergent.
Exercise 19 — level ★★★★☆
Study the behaviour of the sequence
\[ a_n=\frac{n^3+1}{n^2+1}. \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is divergent to \(+\infty\).
Solution
Consider
\[ a_n=\frac{n^3+1}{n^2+1}. \]
The numerator has degree \(3\), whereas the denominator has degree \(2\). The numerator therefore grows faster than the denominator.
We divide numerator and denominator by \(n^2\):
\[ \frac{n^3+1}{n^2+1} = \frac{n+\frac1{n^2}}{1+\frac1{n^2}}. \]
As \(n\to+\infty\),
\[ n+\frac1{n^2}\to+\infty \]
and
\[ 1+\frac1{n^2}\to1. \]
The sequence therefore tends to \(+\infty\).
Let us also establish a simple estimate. For every \(n\geq1\),
\[ n^2+1\leq 2n^2. \]
Furthermore
\[ n^3+1\geq n^3. \]
Hence
\[ a_n=\frac{n^3+1}{n^2+1}\geq\frac{n^3}{2n^2}=\frac n2. \]
Since
\[ \frac n2\to+\infty, \]
\(a_n\to+\infty\) as well.
Therefore
\[ \lim_{n\to+\infty}\frac{n^3+1}{n^2+1}=+\infty. \]
The sequence is divergent to \(+\infty\).
Exercise 20 — level ★★★★★
Study the behaviour of the sequence
\[ a_n= \begin{cases} \dfrac{1}{n} & \text{if } n \text{ is even},\\[6pt] 2+\dfrac{1}{n} & \text{if } n \text{ is odd}. \end{cases} \]
Determine whether it is convergent, divergent, or oscillating.
Answer
The sequence is oscillating.
Solution
The sequence is defined differently according to whether the index \(n\) is even or odd:
\[ a_n= \begin{cases} \dfrac{1}{n} & \text{if } n \text{ is even},\\[6pt] 2+\dfrac{1}{n} & \text{if } n \text{ is odd}. \end{cases} \]
We study the subsequence of even indices. If \(n=2k\), then
\[ a_{2k}=\frac{1}{2k}. \]
Since
\[ \frac{1}{2k}\to0 \]
as \(k\to+\infty\), we obtain
\[ \lim_{k\to+\infty}a_{2k}=0. \]
Now we study the subsequence of odd indices. If \(n=2k-1\), then
\[ a_{2k-1}=2+\frac{1}{2k-1}. \]
Since
\[ \frac{1}{2k-1}\to0, \]
it follows that
\[ 2+\frac{1}{2k-1}\to2. \]
Hence
\[ \lim_{k\to+\infty}a_{2k-1}=2. \]
The sequence therefore has two subsequences converging to different limits:
\[ 0 \qquad\text{and}\qquad 2. \]
For this reason the sequence cannot be convergent.
Moreover, the sequence is bounded. Indeed, for even \(n\) we have
\[ a_n=\frac1n, \]
so that \(0<a_n\leq \frac12\), while for odd \(n\) we have
\[ a_n=2+\frac1n, \]
so that \(2<a_n\leq3\).
In either case the terms remain within a bounded interval. For instance,
\[ 0<a_n\leq3 \]
for every \(n\geq1\).
Being bounded, the sequence can diverge neither to \(+\infty\) nor to \(-\infty\).
The sequence is thus neither convergent nor divergent.
It is therefore oscillating.