On this page we present 20 practice problems on subsequences, explained step by step and designed to consolidate both the formal definition and the conceptual meaning of this fundamental tool in the study of numerical sequences.
The exercises lead the reader gradually from recognising a subsequence to choosing the indices correctly, and finally to the study of limiting behaviour. In particular, we shall see how to check whether one sequence is a subsequence of another, how to compute prescribed subsequences, and how to use two subsequences with different limits to prove that a sequence fails to converge.
Recall that a subsequence of \((a_n)\) is a sequence of the form
\[ (a_{k_n}), \]
where \((k_n)\) is a strictly increasing sequence of natural number indices, that is,
\[ k_n<k_{n+1} \]
for every \(n\in\mathbb{N}\). As in the theory, we shall assume that \(\mathbb{N}=\{0,1,2,\dots\}\).
The aim is not merely to apply a definition mechanically, but to understand why the order of the indices is essential, why every subsequence of a convergent sequence has the same limit as the original sequence, and why subsequences often allow one to detect oscillations, divergence and failure of convergence in a rigorous way.
Exercise 1 — level ★☆☆☆☆
Consider the sequence
\[ a_n=n^2. \]
Determine the subsequence obtained by selecting the even indices, that is, by setting
\[ k_n=2n. \]
Result
The subsequence is
\[ a_{2n}=4n^2. \]
Hence
\[ (a_{2n})=(0,4,16,36,\dots). \]
Solution
A subsequence is obtained by choosing a strictly increasing sequence of natural number indices \((k_n)\) and considering the corresponding terms \(a_{k_n}\).
In this case the chosen indices are
\[ k_n=2n. \]
Since \(n\in\mathbb{N}\), the indices are
\[ 0,2,4,6,\dots \]
and they are strictly increasing. Indeed, for every \(n\in\mathbb{N}\),
\[ 2n<2n+2. \]
Hence we can indeed construct a subsequence.
The original sequence is
\[ a_n=n^2. \]
To obtain the subsequence corresponding to the indices \(2n\), we substitute \(2n\) for \(n\):
\[ a_{2n}=(2n)^2. \]
Expanding the square, we obtain
\[ a_{2n}=4n^2. \]
Let us write out the first few terms to interpret the result:
\[ a_0=0^2=0,\qquad a_2=2^2=4,\qquad a_4=4^2=16,\qquad a_6=6^2=36. \]
Hence the subsequence is
\[ (a_{2n})=(0,4,16,36,\dots). \]
Conceptually, we have not built a new, arbitrary sequence: we have simply observed the original sequence along the even indices only.
Exercise 2 — level ★☆☆☆☆
Consider the sequence
\[ a_n=3n+1. \]
Determine the subsequence obtained by selecting the odd indices, that is, by setting
\[ k_n=2n+1. \]
Result
The subsequence is
\[ a_{2n+1}=6n+4. \]
Hence
\[ (a_{2n+1})=(4,10,16,22,\dots). \]
Solution
The odd indices are described by the formula
\[ k_n=2n+1. \]
Indeed, as \(n\) ranges over \(\mathbb{N}\), we obtain
\[ 1,3,5,7,\dots \]
Before computing the subsequence, let us check that these indices are strictly increasing. We have
\[ k_{n+1}=2(n+1)+1=2n+3. \]
Since
\[ 2n+1<2n+3, \]
it follows that
\[ k_n<k_{n+1}. \]
Hence the chosen indices are suitable for defining a subsequence.
The original sequence is
\[ a_n=3n+1. \]
The subsequence corresponding to the indices \(2n+1\) is
\[ a_{2n+1}. \]
We therefore substitute \(2n+1\) for \(n\) in the formula for \(a_n\):
\[ a_{2n+1}=3(2n+1)+1. \]
Carrying out the computation, we obtain
\[ a_{2n+1}=6n+3+1=6n+4. \]
Let us write out the first few terms:
\[ a_1=4,\qquad a_3=10,\qquad a_5=16,\qquad a_7=22. \]
Hence
\[ (a_{2n+1})=(4,10,16,22,\dots). \]
Exercise 3 — level ★☆☆☆☆
Let \((a_n)\) be an arbitrary sequence. Determine whether the terms
\[ a_5,\ a_2,\ a_8 \]
can form the beginning of a subsequence of \((a_n)\).
Result
No. The terms
\[ a_5,\ a_2,\ a_8 \]
cannot form the beginning of a subsequence, because the indices \(5,2,8\) are not strictly increasing.
Solution
To form a subsequence it is not enough to select some terms of the original sequence. The chosen indices must be strictly increasing.
In this case the terms given are
\[ a_5,\ a_2,\ a_8. \]
The corresponding indices are
\[ 5,\ 2,\ 8. \]
In order to be the beginning of a subsequence, we would need
\[ 5<2<8. \]
But the first inequality is false, since \(5\) is not less than \(2\).
Hence the indices do not respect the natural order in which the terms appear in the original sequence.
Consequently,
\[ a_5,\ a_2,\ a_8 \]
cannot form the beginning of a subsequence.
The conceptual point is crucial: a subsequence may skip some terms, but it may never go back in the indices. The order of the terms of the original sequence must be preserved.
For instance, by contrast,
\[ a_2,\ a_5,\ a_8 \]
could form the beginning of a subsequence, because
\[ 2<5<8. \]
Exercise 4 — level ★★☆☆☆
Consider the sequence
\[ a_n=(-1)^n. \]
Determine the subsequence of even indices and the subsequence of odd indices.
Result
The subsequence of even indices is
\[ a_{2n}=1. \]
The subsequence of odd indices is
\[ a_{2n+1}=-1. \]
Solution
Let us study the even and odd indices separately.
The even indices are given by
\[ k_n=2n. \]
The corresponding subsequence is
\[ a_{2n}=(-1)^{2n}. \]
Since \(2n\) is always even, the power \((-1)^{2n}\) always equals \(1\). Hence
\[ a_{2n}=1 \]
for every \(n\in\mathbb{N}\).
Hence the subsequence of even indices is
\[ (a_{2n})=(1,1,1,1,\dots). \]
The odd indices, on the other hand, are given by
\[ h_n=2n+1. \]
The corresponding subsequence is
\[ a_{2n+1}=(-1)^{2n+1}. \]
Since \(2n+1\) is always odd, the power \((-1)^{2n+1}\) always equals \(-1\). Hence
\[ a_{2n+1}=-1 \]
for every \(n\in\mathbb{N}\).
Hence the subsequence of odd indices is
\[ (a_{2n+1})=(-1,-1,-1,-1,\dots). \]
This exercise illustrates a very important fact: one and the same sequence may have subsequences with different behaviours. Here one subsequence is constantly equal to \(1\), while the other is constantly equal to \(-1\).
Exercise 5 — level ★★☆☆☆
Consider the sequence
\[ a_n=n^2. \]
Determine whether the sequence
\[ b_n=(n+2)^2 \]
is a subsequence of \((a_n)\).
Result
Yes. The sequence
\[ b_n=(n+2)^2 \]
is a subsequence of
\[ a_n=n^2. \]
Indeed,
\[ b_n=a_{n+2}. \]
Solution
To determine whether \((b_n)\) is a subsequence of \((a_n)\), we must check whether there exists a strictly increasing sequence of natural number indices \((k_n)\) such that
\[ b_n=a_{k_n} \]
for every \(n\in\mathbb{N}\).
The original sequence is
\[ a_n=n^2. \]
The sequence we wish to recognise as a subsequence is
\[ b_n=(n+2)^2. \]
We observe that \((n+2)^2\) is obtained from the formula \(a_n=n^2\) by substituting \(n+2\) for \(n\). We therefore choose
\[ k_n=n+2. \]
Then
\[ a_{k_n}=a_{n+2}=(n+2)^2. \]
But
\[ b_n=(n+2)^2. \]
Hence
\[ b_n=a_{k_n}. \]
It remains to check that \((k_n)\) is strictly increasing. We have
\[ k_n=n+2 \]
and
\[ k_{n+1}=n+3. \]
Since
\[ n+2<n+3, \]
it follows that
\[ k_n<k_{n+1}. \]
Hence \((k_n)\) is a strictly increasing sequence of natural number indices.
Therefore \((b_n)\) is a subsequence of \((a_n)\).
Intuitively, the sequence \((b_n)\) is obtained from \((a_n)\) by discarding the first two terms and keeping all the others in the same order.
Exercise 6 — level ★★☆☆☆
Consider the sequence
\[ a_n=n. \]
Determine whether the sequence
\[ b_n=-n \]
is a subsequence of \((a_n)\).
Result
No. The sequence
\[ b_n=-n \]
is not a subsequence of
\[ a_n=n. \]
Indeed, there is no strictly increasing sequence of natural number indices \((k_n)\) such that
\[ b_n=a_{k_n} \]
for every \(n\in\mathbb{N}\).
Solution
To determine whether \((b_n)\) is a subsequence of \((a_n)\), we must ask whether there exists a strictly increasing sequence of natural number indices \((k_n)\) such that
\[ b_n=a_{k_n} \]
for every \(n\in\mathbb{N}\).
The original sequence is
\[ a_n=n. \]
Since \(n\in\mathbb{N}\), the terms of \((a_n)\) are
\[ 0,1,2,3,\dots. \]
Hence all the terms of the sequence \((a_n)\) are natural numbers.
The proposed sequence, by contrast, is
\[ b_n=-n. \]
Its first few terms are
\[ 0,-1,-2,-3,\dots. \]
For every \(n\ge 1\), the term \(b_n\) is negative.
But no term of the sequence \((a_n)\) is negative. Indeed, for any natural number index \(k\), we have
\[ a_k=k\ge 0. \]
Consequently, for example, the term \(b_1=-1\) cannot equal any term of the sequence \((a_n)\).
Hence there can be no sequence of natural number indices \((k_n)\) such that
\[ b_n=a_{k_n} \]
for every \(n\in\mathbb{N}\).
Therefore \((b_n)\) is not a subsequence of \((a_n)\).
The conceptual point is this: a subsequence must be formed by selecting terms already present in the original sequence. Here, by contrast, \((b_n)\) contains negative terms that never appear in the sequence \((a_n)\).
Exercise 7 — level ★★☆☆☆
Consider the sequence
\[ a_n=\frac{1}{n+1}. \]
Determine whether the sequence
\[ b_n=\frac{1}{n^2+1} \]
is a subsequence of \((a_n)\).
Result
Yes. The sequence
\[ b_n=\frac{1}{n^2+1} \]
is a subsequence of
\[ a_n=\frac{1}{n+1}. \]
Indeed,
\[ b_n=a_{n^2}. \]
Solution
To determine whether \((b_n)\) is a subsequence of \((a_n)\), we look for a strictly increasing sequence of natural number indices \((k_n)\) such that
\[ b_n=a_{k_n}. \]
The original sequence is
\[ a_n=\frac{1}{n+1}. \]
If we substitute a generic index \(k_n\) for \(n\), we obtain
\[ a_{k_n}=\frac{1}{k_n+1}. \]
We want this term to coincide with
\[ b_n=\frac{1}{n^2+1}. \]
We therefore require
\[ \frac{1}{k_n+1}=\frac{1}{n^2+1}. \]
Since the denominators are positive, this equality is equivalent to
\[ k_n+1=n^2+1. \]
Subtracting \(1\) from both sides, we obtain
\[ k_n=n^2. \]
Hence the natural candidate is
\[ k_n=n^2. \]
Let us check that \((k_n)\) is strictly increasing. For every \(n\in\mathbb{N}\), we have
\[ k_{n+1}-k_n=(n+1)^2-n^2. \]
Expanding,
\[ k_{n+1}-k_n=n^2+2n+1-n^2=2n+1. \]
Since
\[ 2n+1>0 \]
for every \(n\in\mathbb{N}\), it follows that
\[ k_{n+1}>k_n. \]
Hence \((k_n)\) is strictly increasing.
Moreover, \(k_n=n^2\in\mathbb{N}\) for every \(n\in\mathbb{N}\), so the chosen indices are indeed natural number indices.
We then have
\[ a_{k_n}=a_{n^2}=\frac{1}{n^2+1}=b_n. \]
Therefore \((b_n)\) is a subsequence of \((a_n)\).
Exercise 8 — level ★★☆☆☆
Let \((k_n)\) be defined by
\[ k_n=n^2+n. \]
Verify that \((k_n)\) can be used as a sequence of indices to construct a subsequence.
Result
Yes. The sequence
\[ k_n=n^2+n \]
can be used as a sequence of indices, because it consists of natural numbers and is strictly increasing.
Solution
A sequence \((k_n)\) can be used as a sequence of indices to construct a subsequence if it satisfies two conditions.
The first condition is that each \(k_n\) be a natural number.
The second condition is that \((k_n)\) be strictly increasing, that is,
\[ k_n<k_{n+1} \]
for every \(n\in\mathbb{N}\).
In our case
\[ k_n=n^2+n. \]
Since \(n\in\mathbb{N}\), we also have \(n^2+n\in\mathbb{N}\). Hence each \(k_n\) is a natural number.
Let us now check that \((k_n)\) is strictly increasing. We compute \(k_{n+1}\):
\[ k_{n+1}=(n+1)^2+(n+1). \]
Expanding,
\[ k_{n+1}=n^2+2n+1+n+1=n^2+3n+2. \]
We compute the difference:
\[ k_{n+1}-k_n=(n^2+3n+2)-(n^2+n). \]
Simplifying,
\[ k_{n+1}-k_n=2n+2. \]
Since
\[ 2n+2>0 \]
for every \(n\in\mathbb{N}\), we obtain
\[ k_{n+1}>k_n. \]
Hence \((k_n)\) is strictly increasing.
Therefore the sequence
\[ k_n=n^2+n \]
can be used to construct a subsequence \((a_{k_n})\) of any sequence \((a_n)\).
Conceptually, the indices \(0,2,6,12,20,\dots\) select some terms of the original sequence, skipping others, but never going back.
Exercise 9 — level ★★★☆☆
Let \((a_n)\) be an arbitrary sequence. Determine whether the terms
\[ a_1,\ a_3,\ a_6,\ a_{10},\dots \]
can form a subsequence of \((a_n)\).
Result
Yes. The terms
\[ a_1,\ a_3,\ a_6,\ a_{10},\dots \]
can form a subsequence, because the indices
\[ 1,3,6,10,\dots \]
are strictly increasing.
Solution
To check whether the given terms can form a subsequence, we must look at the indices.
The terms are
\[ a_1,\ a_3,\ a_6,\ a_{10},\dots \]
so the indices are
\[ 1,\ 3,\ 6,\ 10,\dots. \]
These indices are arranged in strictly increasing order, because
\[ 1<3<6<10<\dots. \]
This is already enough to conclude that the given terms can form a subsequence.
We can also recognise an explicit formula for the indices. They are given by
\[ k_n=\frac{(n+1)(n+2)}{2}. \]
Indeed:
\[ k_0=1,\qquad k_1=3,\qquad k_2=6,\qquad k_3=10. \]
Let us check that this sequence of indices is strictly increasing. We compute
\[ k_{n+1}-k_n = \frac{(n+2)(n+3)}{2}-\frac{(n+1)(n+2)}{2}. \]
Factoring out the common factor \(\displaystyle\frac{n+2}{2}\), we obtain
\[ k_{n+1}-k_n = \frac{n+2}{2}\bigl((n+3)-(n+1)\bigr). \]
Since
\[ (n+3)-(n+1)=2, \]
we have
\[ k_{n+1}-k_n=\frac{n+2}{2}\cdot 2=n+2. \]
Now
\[ n+2>0 \]
for every \(n\in\mathbb{N}\). Hence
\[ k_{n+1}>k_n. \]
Hence the indices are strictly increasing.
Therefore the terms
\[ a_1,\ a_3,\ a_6,\ a_{10},\dots \]
can form a subsequence of \((a_n)\).
Exercise 10 — level ★★★☆☆
Consider the sequence
\[ a_n=\frac{n}{n+1}. \]
Determine the limit of the subsequence \((a_{2n})\).
Result
We have
\[ a_{2n}=\frac{2n}{2n+1} \]
and hence
\[ a_{2n}\to 1. \]
Solution
The subsequence \((a_{2n})\) is obtained by selecting the even indices, that is, by setting
\[ k_n=2n. \]
Before proceeding, we observe that the indices \(2n\) are strictly increasing, because
\[ 2n<2n+2. \]
Hence \((a_{2n})\) is indeed a subsequence of \((a_n)\).
The original sequence is
\[ a_n=\frac{n}{n+1}. \]
Substituting \(2n\) for \(n\), we obtain
\[ a_{2n}=\frac{2n}{2n+1}. \]
We must compute
\[ \lim_{n\to+\infty}\frac{2n}{2n+1}. \]
To study this limit, we may assume \(n\ge 1\) and divide numerator and denominator by \(n\):
\[ \frac{2n}{2n+1} = \frac{2}{2+\frac{1}{n}}. \]
Since
\[ \frac{1}{n}\to 0, \]
we obtain
\[ \frac{2}{2+\frac{1}{n}}\to \frac{2}{2+0}=1. \]
Hence
\[ a_{2n}\to 1. \]
This result is consistent with the general theorem on subsequences: indeed, the original sequence \((a_n)\) converges to \(1\), and every subsequence of a convergent sequence converges to the same limit.
Exercise 11 — level ★★★☆☆
Consider the sequence
\[ a_n=\frac{1}{n+1}. \]
Prove that every subsequence of \((a_n)\) converges to \(0\).
Result
Every subsequence of
\[ a_n=\frac{1}{n+1} \]
converges to \(0\).
Solution
Let \((a_{k_n})\) be an arbitrary subsequence of \((a_n)\). By the definition of subsequence, \((k_n)\) is a strictly increasing sequence of natural number indices.
Since
\[ a_n=\frac{1}{n+1}, \]
substituting \(k_n\) for \(n\) we obtain
\[ a_{k_n}=\frac{1}{k_n+1}. \]
Since \((k_n)\) is strictly increasing and takes values in \(\mathbb{N}\), we have
\[ k_n\ge n \]
for every \(n\in\mathbb{N}\). Indeed, \(k_0\ge 0\), and since the indices are natural numbers that increase strictly, from \(k_n<k_{n+1}\) it follows that \(k_{n+1}\ge k_n+1\). By induction we therefore obtain \(k_n\ge n\).
From
\[ k_n\ge n \]
it follows that
\[ k_n+1\ge n+1. \]
Since \(k_n+1\) and \(n+1\) are positive numbers, passing to reciprocals reverses the inequality. Hence
\[ \frac{1}{k_n+1}\le \frac{1}{n+1}. \]
Moreover,
\[ \frac{1}{k_n+1}>0. \]
We thus have the estimate
\[ 0<\frac{1}{k_n+1}\le \frac{1}{n+1}. \]
Now we know that
\[ \frac{1}{n+1}\to 0. \]
By the squeeze theorem, it follows that
\[ \frac{1}{k_n+1}\to 0. \]
But
\[ a_{k_n}=\frac{1}{k_n+1}. \]
Hence
\[ a_{k_n}\to 0. \]
Since \((a_{k_n})\) was an arbitrary subsequence, we have proved that every subsequence of \((a_n)\) converges to \(0\).
The conceptual meaning is the following: a subsequence may skip some terms, but it cannot escape the eventual behaviour of the sequence. Since the terms of \((a_n)\) come ever closer to \(0\), every subsequence too must come ever closer to \(0\).
Exercise 12 — level ★★★☆☆
Consider the sequence
\[ a_n=2+\frac{1}{n+1}. \]
Prove that the subsequence \((a_{n^2})\) converges to \(2\).
Result
We have
\[ a_{n^2}=2+\frac{1}{n^2+1} \]
and hence
\[ a_{n^2}\to 2. \]
Solution
The subsequence \((a_{n^2})\) is obtained by choosing the indices
\[ k_n=n^2. \]
Before computing the limit, let us check that these indices do define a subsequence.
For every \(n\in\mathbb{N}\), \(n^2\in\mathbb{N}\). Moreover
\[ k_{n+1}-k_n=(n+1)^2-n^2=2n+1. \]
Since
\[ 2n+1>0 \]
for every \(n\in\mathbb{N}\), it follows that
\[ k_{n+1}>k_n. \]
Hence \((n^2)\) is a strictly increasing sequence of natural number indices.
We now compute the subsequence. Since
\[ a_n=2+\frac{1}{n+1}, \]
substituting \(n^2\) for \(n\) we obtain
\[ a_{n^2}=2+\frac{1}{n^2+1}. \]
We must therefore compute
\[ \lim_{n\to+\infty}\left(2+\frac{1}{n^2+1}\right). \]
Since
\[ n^2+1\to+\infty, \]
we have
\[ \frac{1}{n^2+1}\to 0. \]
Hence
\[ 2+\frac{1}{n^2+1}\to 2+0=2. \]
Therefore
\[ a_{n^2}\to 2. \]
This result is consistent with the general theorem: the original sequence converges to \(2\), so every subsequence of it must converge to the same limit.
Exercise 13 — level ★★★☆☆
Consider the sequence
\[ a_n=(-1)^n. \]
Use subsequences to prove that \((a_n)\) does not converge.
Result
The sequence \((a_n)\) does not converge, because it has two subsequences converging to different limits:
\[ a_{2n}\to 1 \qquad\text{and}\qquad a_{2n+1}\to -1. \]
Solution
To prove that a sequence does not converge, we may look for two subsequences converging to different limits.
Indeed, if a sequence converged to a real limit \(\ell\), then every subsequence of it would have to converge to the same limit \(\ell\).
Consider first the even indices:
\[ k_n=2n. \]
The corresponding subsequence is
\[ a_{2n}=(-1)^{2n}. \]
Since \(2n\) is even, we have
\[ (-1)^{2n}=1. \]
Hence
\[ a_{2n}=1 \]
for every \(n\in\mathbb{N}\). Therefore
\[ a_{2n}\to 1. \]
Now consider the odd indices:
\[ h_n=2n+1. \]
The corresponding subsequence is
\[ a_{2n+1}=(-1)^{2n+1}. \]
Since \(2n+1\) is odd, we have
\[ (-1)^{2n+1}=-1. \]
Hence
\[ a_{2n+1}=-1 \]
for every \(n\in\mathbb{N}\). Therefore
\[ a_{2n+1}\to -1. \]
We have found two subsequences of the same sequence:
\[ (a_{2n}) \qquad\text{and}\qquad (a_{2n+1}), \]
such that
\[ a_{2n}\to 1 \qquad\text{and}\qquad a_{2n+1}\to -1. \]
The two limits are different, since
\[ 1\ne -1. \]
Hence the sequence \((a_n)\) cannot converge.
The conceptual reason is decisive: a convergent sequence must approach a single eventual value. Here, by contrast, along the even indices the sequence stays equal to \(1\), while along the odd indices it stays equal to \(-1\).
Exercise 14 — level ★★★☆☆
Consider the sequence
\[ a_n=\frac{1+(-1)^n}{2}. \]
Use subsequences to prove that \((a_n)\) does not converge.
Result
The sequence \((a_n)\) does not converge, because
\[ a_{2n}\to 1 \qquad\text{and}\qquad a_{2n+1}\to 0. \]
The two subsequences converge to different limits.
Solution
Let us study the sequence separately along the even and the odd indices.
Consider first the even indices. Substituting \(2n\) for \(n\), we obtain
\[ a_{2n}=\frac{1+(-1)^{2n}}{2}. \]
Since
\[ (-1)^{2n}=1, \]
we have
\[ a_{2n}=\frac{1+1}{2}=\frac{2}{2}=1. \]
Hence the subsequence of even indices is constantly equal to \(1\). Therefore
\[ a_{2n}\to 1. \]
Now consider the odd indices. Substituting \(2n+1\) for \(n\), we obtain
\[ a_{2n+1}=\frac{1+(-1)^{2n+1}}{2}. \]
Since
\[ (-1)^{2n+1}=-1, \]
we have
\[ a_{2n+1}=\frac{1-1}{2}=\frac{0}{2}=0. \]
Hence the subsequence of odd indices is constantly equal to \(0\). Therefore
\[ a_{2n+1}\to 0. \]
We have found two convergent subsequences:
\[ a_{2n}\to 1 \qquad\text{and}\qquad a_{2n+1}\to 0. \]
Since
\[ 1\ne 0, \]
the two subsequences converge to different limits.
Hence the sequence \((a_n)\) does not converge.
Conceptually, the sequence keeps alternating between the values \(1\) and \(0\). It does not settle around a single real number, and this prevents convergence.
Exercise 15 — level ★★★★☆
Consider the sequence
\[ a_n=(-1)^n+\frac{1}{n+1}. \]
Study the subsequences of even and odd indices and deduce whether \((a_n)\) converges.
Result
We have
\[ a_{2n}\to 1 \qquad\text{and}\qquad a_{2n+1}\to -1. \]
Since the two subsequences converge to different limits, the sequence \((a_n)\) does not converge.
Solution
The sequence is
\[ a_n=(-1)^n+\frac{1}{n+1}. \]
It comprises two parts: the oscillating term \((-1)^n\), which alternates between \(1\) and \(-1\), and the term \(\displaystyle\frac{1}{n+1}\), which tends to \(0\).
Let us study first the subsequence of even indices. Substituting \(2n\) for \(n\), we obtain
\[ a_{2n}=(-1)^{2n}+\frac{1}{2n+1}. \]
Since
\[ (-1)^{2n}=1, \]
it follows that
\[ a_{2n}=1+\frac{1}{2n+1}. \]
Now
\[ \frac{1}{2n+1}\to 0. \]
Hence
\[ a_{2n}=1+\frac{1}{2n+1}\to 1+0=1. \]
Therefore
\[ a_{2n}\to 1. \]
Let us now study the subsequence of odd indices. Substituting \(2n+1\) for \(n\), we obtain
\[ a_{2n+1}=(-1)^{2n+1}+\frac{1}{2n+2}. \]
Since
\[ (-1)^{2n+1}=-1, \]
it follows that
\[ a_{2n+1}=-1+\frac{1}{2n+2}. \]
Now
\[ \frac{1}{2n+2}\to 0. \]
Hence
\[ a_{2n+1}=-1+\frac{1}{2n+2}\to -1+0=-1. \]
Therefore
\[ a_{2n+1}\to -1. \]
We have found two subsequences of the same sequence:
\[ a_{2n}\to 1 \qquad\text{and}\qquad a_{2n+1}\to -1. \]
Since the limits are different, the sequence \((a_n)\) does not converge.
This example is important because it shows that an infinitesimal term, such as \(\displaystyle\frac{1}{n+1}\), is not enough to remove the main oscillation produced by \((-1)^n\). The sequence keeps approaching two different values along two different subsequences.
Exercise 16 — level ★★★★☆
Consider the sequence
\[ a_n=n. \]
Prove that the subsequence \((a_{n^2+1})\) diverges to \(+\infty\).
Result
We have
\[ a_{n^2+1}=n^2+1. \]
Hence
\[ a_{n^2+1}\to+\infty. \]
Solution
The original sequence is
\[ a_n=n. \]
The subsequence \((a_{n^2+1})\) is obtained by choosing the indices
\[ k_n=n^2+1. \]
Before studying its limit, let us check that these indices do define a subsequence.
For every \(n\in\mathbb{N}\), we have \(n^2+1\in\mathbb{N}\). Moreover
\[ k_{n+1}-k_n=((n+1)^2+1)-(n^2+1). \]
Expanding,
\[ k_{n+1}-k_n=n^2+2n+1+1-n^2-1=2n+1. \]
Since
\[ 2n+1>0 \]
for every \(n\in\mathbb{N}\), it follows that
\[ k_{n+1}>k_n. \]
Hence \((k_n)\) is a strictly increasing sequence of natural number indices.
We now compute the subsequence. Since \(a_n=n\), substituting \(n^2+1\) for \(n\) we obtain
\[ a_{n^2+1}=n^2+1. \]
We must prove that
\[ n^2+1\to+\infty. \]
We use the definition of divergence to \(+\infty\). We must show that, for every \(M\in\mathbb{R}\), there exists \(N\in\mathbb{N}\) such that, for every \(n\ge N\),
\[ n^2+1>M. \]
If \(M<0\), then for every \(n\in\mathbb{N}\) we have
\[ n^2+1\ge 1>0>M. \]
Hence in this case the inequality holds for all indices.
Suppose now \(M\ge 0\). We choose \(N\in\mathbb{N}\) such that
\[ N>\sqrt{M}. \]
Then, for every \(n\ge N\), we have
\[ n\ge N>\sqrt{M}. \]
Squaring, we obtain
\[ n^2>M. \]
Consequently
\[ n^2+1>M. \]
We have thus proved that, for every \(M\in\mathbb{R}\), the terms of the subsequence eventually become greater than \(M\).
Therefore
\[ a_{n^2+1}\to+\infty. \]
Conceptually, the original sequence \(a_n=n\) diverges to \(+\infty\). A subsequence may skip some terms, but it cannot prevent the indices from going to infinity; hence the subsequence too diverges to \(+\infty\).
Exercise 17 — level ★★★★☆
Consider the sequence
\[ a_n=-n^2. \]
Prove that the subsequence \((a_{2n+1})\) diverges to \(-\infty\).
Result
We have
\[ a_{2n+1}=-(2n+1)^2. \]
Hence
\[ a_{2n+1}\to-\infty. \]
Solution
The original sequence is
\[ a_n=-n^2. \]
The subsequence \((a_{2n+1})\) is obtained by selecting the odd indices:
\[ k_n=2n+1. \]
The indices \(2n+1\) are natural numbers and strictly increasing. Indeed
\[ k_{n+1}=2(n+1)+1=2n+3, \]
and hence
\[ k_n=2n+1<2n+3=k_{n+1}. \]
Hence \((a_{2n+1})\) is indeed a subsequence of \((a_n)\).
We now compute it explicitly. Substituting \(2n+1\) for \(n\) in the formula \(a_n=-n^2\), we obtain
\[ a_{2n+1}=-(2n+1)^2. \]
Expanding the square,
\[ (2n+1)^2=4n^2+4n+1. \]
Hence
\[ a_{2n+1}=-(4n^2+4n+1)=-4n^2-4n-1. \]
This expression becomes arbitrarily large in the negative direction as \(n\) grows. Let us prove this using the definition of divergence to \(-\infty\).
We must show that, for every \(m\in\mathbb{R}\), there exists \(N\in\mathbb{N}\) such that, for every \(n\ge N\),
\[ a_{2n+1}<m. \]
Since
\[ a_{2n+1}=-(2n+1)^2, \]
we distinguish two cases.
If \(m>0\), then for every \(n\in\mathbb{N}\) we have
\[ -(2n+1)^2<0<m. \]
Hence the inequality holds for all indices.
Suppose now \(m\le 0\). We choose \(N\in\mathbb{N}\) such that
\[ 2N+1>\sqrt{-m}. \]
Then, for every \(n\ge N\),
\[ 2n+1\ge 2N+1>\sqrt{-m}. \]
Squaring, we obtain
\[ (2n+1)^2>-m. \]
Multiplying both sides by \(-1\) reverses the inequality:
\[ -(2n+1)^2<m. \]
That is,
\[ a_{2n+1}<m. \]
We have thus proved that
\[ a_{2n+1}\to-\infty. \]
This result is consistent with the general theorem: since \(a_n=-n^2\to-\infty\), every subsequence of it diverges to \(-\infty\).
Exercise 18 — level ★★★★☆
Consider the sequence
\[ a_n=\sin\left(\frac{n\pi}{2}\right). \]
Use two subsequences to prove that \((a_n)\) does not converge.
Result
The sequence \((a_n)\) does not converge, because
\[ a_{4n+1}\to 1 \qquad\text{and}\qquad a_{4n+3}\to -1. \]
Solution
Let us examine some terms of the sequence:
\[ a_0=\sin 0=0, \]
\[ a_1=\sin\left(\frac{\pi}{2}\right)=1, \]
\[ a_2=\sin(\pi)=0, \]
\[ a_3=\sin\left(\frac{3\pi}{2}\right)=-1, \]
\[ a_4=\sin(2\pi)=0. \]
We thus see that the sequence cyclically takes the values
\[ 0,\ 1,\ 0,\ -1,\ 0,\ 1,\ 0,\ -1,\dots. \]
To prove that the sequence does not converge, we look for two subsequences converging to different limits.
Consider the indices
\[ k_n=4n+1. \]
They are strictly increasing, because
\[ k_{n+1}=4(n+1)+1=4n+5 \]
and hence
\[ 4n+1<4n+5. \]
The corresponding subsequence is
\[ a_{4n+1} = \sin\left(\frac{(4n+1)\pi}{2}\right). \]
Since
\[ \frac{(4n+1)\pi}{2}=2n\pi+\frac{\pi}{2}, \]
we obtain
\[ a_{4n+1} = \sin\left(2n\pi+\frac{\pi}{2}\right)=1. \]
Hence
\[ a_{4n+1}\to 1. \]
Now consider the indices
\[ h_n=4n+3. \]
These too are strictly increasing, because
\[ h_{n+1}=4(n+1)+3=4n+7 \]
and hence
\[ 4n+3<4n+7. \]
The corresponding subsequence is
\[ a_{4n+3} = \sin\left(\frac{(4n+3)\pi}{2}\right). \]
Since
\[ \frac{(4n+3)\pi}{2}=2n\pi+\frac{3\pi}{2}, \]
we obtain
\[ a_{4n+3} = \sin\left(2n\pi+\frac{3\pi}{2}\right)=-1. \]
Hence
\[ a_{4n+3}\to -1. \]
We have found two subsequences of the same sequence such that
\[ a_{4n+1}\to 1 \qquad\text{and}\qquad a_{4n+3}\to -1. \]
Since
\[ 1\ne -1, \]
the sequence \((a_n)\) does not converge.
Conceptually, the sequence does not approach a single eventual value: instead, it keeps repeating different values cyclically.
Exercise 19 — level ★★★★☆
Consider the sequence
\[ a_n=(-1)^n n. \]
Study the subsequences of even and odd indices and deduce the behaviour of the sequence.
Result
We have
\[ a_{2n}=2n\to+\infty \]
and
\[ a_{2n+1}=-(2n+1)\to-\infty. \]
Hence the sequence has no limit, either finite or infinite.
Solution
The sequence is
\[ a_n=(-1)^n n. \]
The factor \((-1)^n\) changes sign according to the parity of \(n\), while the factor \(n\) grows without bound.
Let us study first the even indices:
\[ k_n=2n. \]
The corresponding subsequence is
\[ a_{2n}=(-1)^{2n}\cdot 2n. \]
Since
\[ (-1)^{2n}=1, \]
we obtain
\[ a_{2n}=2n. \]
Hence
\[ a_{2n}\to+\infty. \]
Let us now study the odd indices:
\[ h_n=2n+1. \]
The corresponding subsequence is
\[ a_{2n+1}=(-1)^{2n+1}(2n+1). \]
Since
\[ (-1)^{2n+1}=-1, \]
we obtain
\[ a_{2n+1}=-(2n+1). \]
Hence
\[ a_{2n+1}\to-\infty. \]
At this point we can deduce the behaviour of the original sequence.
The sequence \((a_n)\) does not converge to a real limit, because it has a subsequence diverging to \(+\infty\) and a subsequence diverging to \(-\infty\).
Moreover, \((a_n)\) does not diverge to \(+\infty\). Indeed, if \(a_n\to+\infty\), then every subsequence of it would have to diverge to \(+\infty\). But the subsequence \((a_{2n+1})\) diverges to \(-\infty\).
Similarly, \((a_n)\) does not diverge to \(-\infty\). Indeed, if \(a_n\to-\infty\), then every subsequence of it would have to diverge to \(-\infty\). But the subsequence \((a_{2n})\) diverges to \(+\infty\).
Therefore the sequence \((a_n)\) has no limit, either finite or infinite.
The conceptual point is that the terms not only oscillate in sign but also move ever further apart: the even-indexed ones grow towards \(+\infty\), while the odd-indexed ones decrease towards \(-\infty\).
Exercise 20 — level ★★★★★
Let \((a_n)\) be a real sequence converging to a real number \(\ell\). Let \((a_{k_n})\) be a subsequence of it. Prove directly, using the definition of limit, that
\[ a_{k_n}\to \ell. \]
Result
Every subsequence of a convergent sequence converges to the same limit as the original sequence. Hence
\[ a_{k_n}\to \ell. \]
Solution
We know that the sequence \((a_n)\) converges to \(\ell\). This means that
\[ a_n\to \ell. \]
By the definition of limit, for every \(\varepsilon>0\) there exists \(N\in\mathbb{N}\) such that, for every index \(m\ge N\), we have
\[ |a_m-\ell|<\varepsilon. \]
We use the letter \(m\) to denote a generic index of the original sequence, so as to avoid confusing it with the index \(n\) of the subsequence.
Now consider an arbitrary subsequence \((a_{k_n})\). By the definition of subsequence, \((k_n)\) is a strictly increasing sequence of natural number indices.
From the fundamental property of the indices of a subsequence we know that
\[ k_n\ge n \]
for every \(n\in\mathbb{N}\).
Now take \(n\ge N\). Then, using \(k_n\ge n\), we obtain
\[ k_n\ge n\ge N. \]
Hence the index \(k_n\) is large enough for the definition of the limit of the original sequence to apply.
Indeed, since the definition of limit tells us that
\[ |a_m-\ell|<\varepsilon \]
for every \(m\ge N\), we may in particular choose
\[ m=k_n. \]
Since \(k_n\ge N\), we obtain
\[ |a_{k_n}-\ell|<\varepsilon. \]
We have thus proved that, for every \(\varepsilon>0\), there exists \(N\in\mathbb{N}\) such that, for every \(n\ge N\),
\[ |a_{k_n}-\ell|<\varepsilon. \]
This is precisely the definition of convergence of the subsequence \((a_{k_n})\) to the limit \(\ell\).
Therefore
\[ a_{k_n}\to \ell. \]
The decisive conceptual point is this: a subsequence may skip some terms, but its indices \(k_n\) nevertheless grow to infinity. Hence, once the original sequence is eventually close to \(\ell\), the subsequence too is eventually close to \(\ell\).