In this collection we present 20 step-by-step practice problems on the monotone convergence theorem for sequences. The problems show how to study the monotonicity of a sequence, how to determine whether it is bounded, and how to deduce the existence of the limit using the monotone convergence theorem.
In particular, we shall use the fact that an increasing sequence tends to its supremum, possibly equal to \(+\infty\), whereas a decreasing sequence tends to its infimum, possibly equal to \(-\infty\).
Exercise 1 — level ★☆☆☆☆
Investigate the monotonicity of the sequence
\[ a_n=\frac{1}{n} \]
and deduce its limit using the monotone convergence theorem.
Answer
The sequence is decreasing, bounded below, and
\[ \lim_{n\to+\infty}\frac{1}{n}=0. \]
Solution
Consider the sequence
\[ a_n=\frac{1}{n}. \]
To study its monotonicity we compare two consecutive terms:
\[ a_{n+1}=\frac{1}{n+1}. \]
Since \(n+1>n\) and the denominators are positive, we have
\[ \frac{1}{n+1}<\frac{1}{n}. \]
Hence
\[ a_{n+1}<a_n. \]
The sequence is therefore strictly decreasing, and in particular decreasing.
Moreover, for every \(n\geq1\),
\[ \frac1n>0. \]
Thus \(0\) is a lower bound of the sequence. The sequence is decreasing and bounded below.
By the monotone convergence theorem, it converges to its infimum:
\[ \lim_{n\to+\infty}a_n=\inf\left\{\frac1n:n\in\mathbb N,\ n\geq1\right\}. \]
The infimum equals \(0\): indeed, all the terms are positive, yet they become arbitrarily small.
Therefore
\[ \lim_{n\to+\infty}\frac1n=0. \]
Exercise 2 — level ★☆☆☆☆
Investigate the monotonicity of the sequence
\[ a_n=1-\frac{1}{n} \]
and deduce its limit.
Answer
The sequence is increasing, bounded above, and
\[ \lim_{n\to+\infty}\left(1-\frac1n\right)=1. \]
Solution
The sequence is
\[ a_n=1-\frac1n. \]
We compute the next term:
\[ a_{n+1}=1-\frac{1}{n+1}. \]
Since
\[ \frac{1}{n+1}<\frac1n, \]
changing sign we obtain
\[ -\frac{1}{n+1}>-\frac1n. \]
Adding \(1\) to both sides:
\[ 1-\frac{1}{n+1}>1-\frac1n. \]
Hence
\[ a_{n+1}>a_n. \]
The sequence is strictly increasing.
Moreover, for every \(n\geq1\),
\[ 1-\frac1n<1. \]
Thus \(1\) is an upper bound of the sequence.
By the monotone convergence theorem, an increasing sequence that is bounded above converges to its supremum.
In this case
\[ \sup\left\{1-\frac1n:n\in\mathbb N,\ n\geq1\right\}=1. \]
Indeed, the terms are always less than \(1\), yet they come arbitrarily close to \(1\).
Therefore
\[ \lim_{n\to+\infty}\left(1-\frac1n\right)=1. \]
Exercise 3 — level ★☆☆☆☆
Study the sequence
\[ a_n=n. \]
Determine whether it is monotone and bounded, and find its limit.
Answer
The sequence is increasing, not bounded above, and
\[ \lim_{n\to+\infty}n=+\infty. \]
Solution
The sequence is
\[ a_n=n. \]
The next term is
\[ a_{n+1}=n+1. \]
For every \(n\geq1\),
\[ n+1>n. \]
Hence
\[ a_{n+1}>a_n. \]
The sequence is strictly increasing.
It is not bounded above. Indeed, given any real number \(M\), we can choose an integer \(n\) such that
\[ n>M. \]
Then
\[ a_n=n>M. \]
Hence the sequence grows without bound.
By the monotone convergence theorem, an increasing sequence that is not bounded above diverges to \(+\infty\).
Hence
\[ \lim_{n\to+\infty}n=+\infty. \]
Exercise 4 — level ★☆☆☆☆
Study the sequence
\[ a_n=-n. \]
Determine whether it is monotone and bounded, and find its limit.
Answer
The sequence is decreasing, not bounded below, and
\[ \lim_{n\to+\infty}(-n)=-\infty. \]
Solution
The sequence is
\[ a_n=-n. \]
The next term is
\[ a_{n+1}=-(n+1)=-n-1. \]
Since
\[ -n-1<-n, \]
we have
\[ a_{n+1}<a_n. \]
The sequence is strictly decreasing.
It is not bounded below. Indeed, given \(M>0\), we can choose \(n\) such that
\[ n>M. \]
Multiplying by \(-1\), we obtain
\[ -n<-M. \]
Hence the terms of the sequence fall below any negative threshold.
By the monotone convergence theorem, a decreasing sequence that is not bounded below diverges to \(-\infty\).
Therefore
\[ \lim_{n\to+\infty}(-n)=-\infty. \]
Exercise 5 — level ★★☆☆☆
Study the sequence
\[ a_n=\frac{n}{n+1}. \]
Determine whether it is monotone and bounded, and find its limit.
Answer
The sequence is increasing, bounded above, and
\[ \lim_{n\to+\infty}\frac{n}{n+1}=1. \]
Solution
Let us rewrite the general term:
\[ a_n=\frac{n}{n+1}=1-\frac{1}{n+1}. \]
Since the sequence
\[ \frac{1}{n+1} \]
is decreasing, the sequence
\[ 1-\frac{1}{n+1} \]
is increasing.
Let us verify this directly. We have
\[ a_{n+1}=\frac{n+1}{n+2}. \]
We compute the difference:
\[ a_{n+1}-a_n=\frac{n+1}{n+2}-\frac{n}{n+1}. \]
Putting over a common denominator:
\[ a_{n+1}-a_n= \frac{(n+1)^2-n(n+2)}{(n+2)(n+1)}. \]
Expanding the numerator:
\[ (n+1)^2-n(n+2)=n^2+2n+1-(n^2+2n)=1. \]
Hence
\[ a_{n+1}-a_n=\frac{1}{(n+1)(n+2)}>0. \]
Thus \(a_{n+1}>a_n\), that is, the sequence is strictly increasing.
Moreover
\[ \frac{n}{n+1}<1 \]
for every \(n\geq1\), so \(1\) is an upper bound.
By the monotone convergence theorem, the sequence converges to its supremum.
Since the terms approach \(1\) from below, we have
\[ \sup\left\{\frac{n}{n+1}:n\in\mathbb N,\ n\geq1\right\}=1. \]
Therefore
\[ \lim_{n\to+\infty}\frac{n}{n+1}=1. \]
Exercise 6 — level ★★☆☆☆
Study the sequence
\[ a_n=\frac{n+1}{n}. \]
Determine whether it is monotone and bounded, and find its limit.
Answer
The sequence is decreasing, bounded below, and
\[ \lim_{n\to+\infty}\frac{n+1}{n}=1. \]
Solution
Let us rewrite the sequence:
\[ a_n=\frac{n+1}{n}=1+\frac1n. \]
Since \(\displaystyle \frac1n\) is decreasing, the expression
\[ 1+\frac1n \]
is also decreasing.
Let us check this with consecutive terms:
\[ a_{n+1}=1+\frac{1}{n+1}. \]
Since
\[ \frac{1}{n+1}<\frac1n, \]
adding \(1\) to both sides we obtain
\[ 1+\frac{1}{n+1}<1+\frac1n. \]
Hence
\[ a_{n+1}<a_n. \]
The sequence is strictly decreasing.
Moreover, for every \(n\geq1\),
\[ 1+\frac1n>1. \]
Thus \(1\) is a lower bound.
A decreasing sequence that is bounded below converges to its infimum.
In this case
\[ \inf\left\{1+\frac1n:n\in\mathbb N,\ n\geq1\right\}=1. \]
Therefore
\[ \lim_{n\to+\infty}\frac{n+1}{n}=1. \]
Exercise 7 — level ★★☆☆☆
Study the sequence
\[ a_n=2-\frac{3}{n}. \]
Determine whether it is monotone and bounded, and find its limit.
Answer
The sequence is increasing, bounded above, and
\[ \lim_{n\to+\infty}\left(2-\frac3n\right)=2. \]
Solution
Consider
\[ a_n=2-\frac3n. \]
The next term is
\[ a_{n+1}=2-\frac{3}{n+1}. \]
Since
\[ \frac{3}{n+1}<\frac3n, \]
changing sign we obtain
\[ -\frac{3}{n+1}>-\frac3n. \]
Adding \(2\):
\[ 2-\frac{3}{n+1}>2-\frac3n. \]
Hence
\[ a_{n+1}>a_n. \]
The sequence is strictly increasing.
Moreover, for every \(n\geq1\),
\[ 2-\frac3n<2. \]
Thus \(2\) is an upper bound.
By the monotone convergence theorem, the sequence converges to its supremum.
Since \(\displaystyle \frac3n\to0\), the terms approach \(2\) from below. Therefore
\[ \sup\left\{2-\frac3n:n\in\mathbb N,\ n\geq1\right\}=2. \]
We conclude that
\[ \lim_{n\to+\infty}\left(2-\frac3n\right)=2. \]
Exercise 8 — level ★★☆☆☆
Study the sequence
\[ a_n=5+\frac{2}{n}. \]
Determine whether it is monotone and bounded, and find its limit.
Answer
The sequence is decreasing, bounded below, and
\[ \lim_{n\to+\infty}\left(5+\frac2n\right)=5. \]
Solution
Consider
\[ a_n=5+\frac2n. \]
The next term is
\[ a_{n+1}=5+\frac{2}{n+1}. \]
Since
\[ \frac{2}{n+1}<\frac2n, \]
adding \(5\) to both sides we obtain
\[ 5+\frac{2}{n+1}<5+\frac2n. \]
Hence
\[ a_{n+1}<a_n. \]
The sequence is strictly decreasing.
Moreover, for every \(n\geq1\),
\[ 5+\frac2n>5. \]
Thus \(5\) is a lower bound.
A decreasing sequence that is bounded below converges to its infimum.
Since \(\displaystyle \frac2n\to0\), the terms approach \(5\) from above. Hence
\[ \inf\left\{5+\frac2n:n\in\mathbb N,\ n\geq1\right\}=5. \]
Therefore
\[ \lim_{n\to+\infty}\left(5+\frac2n\right)=5. \]
Exercise 9 — level ★★★☆☆
Study the sequence
\[ a_n=\frac{2n+1}{n+1}. \]
Determine whether it is monotone and bounded, and find its limit.
Answer
The sequence is increasing, bounded above, and
\[ \lim_{n\to+\infty}\frac{2n+1}{n+1}=2. \]
Solution
Let us rewrite the general term:
\[ \frac{2n+1}{n+1} = \frac{2(n+1)-1}{n+1} = 2-\frac{1}{n+1}. \]
Hence
\[ a_n=2-\frac{1}{n+1}. \]
Since \(\displaystyle \frac{1}{n+1}\) is decreasing, the term
\[ -\frac{1}{n+1} \]
is increasing. Hence \(a_n\) is increasing.
Let us also check with the next term:
\[ a_{n+1}=2-\frac{1}{n+2}. \]
Since
\[ \frac{1}{n+2}<\frac{1}{n+1}, \]
we have
\[ -\frac{1}{n+2}>-\frac{1}{n+1}. \]
Adding \(2\):
\[ 2-\frac{1}{n+2}>2-\frac{1}{n+1}. \]
Therefore
\[ a_{n+1}>a_n. \]
The sequence is strictly increasing.
Moreover
\[ a_n=2-\frac{1}{n+1}<2, \]
so \(2\) is an upper bound.
The sequence is increasing and bounded above, hence it converges to its supremum.
Since \(\displaystyle \frac1{n+1}\to0\), the supremum is \(2\). Hence
\[ \lim_{n\to+\infty}\frac{2n+1}{n+1}=2. \]
Exercise 10 — level ★★★☆☆
Study the sequence
\[ a_n=\frac{3n+4}{n}. \]
Determine whether it is monotone and bounded, and find its limit.
Answer
The sequence is decreasing, bounded below, and
\[ \lim_{n\to+\infty}\frac{3n+4}{n}=3. \]
Solution
Let us rewrite the sequence:
\[ a_n=\frac{3n+4}{n}=3+\frac4n. \]
Since \(\displaystyle \frac4n\) is decreasing, the expression
\[ 3+\frac4n \]
is also decreasing.
Indeed
\[ a_{n+1}=3+\frac{4}{n+1}. \]
Since
\[ \frac{4}{n+1}<\frac4n, \]
we have
\[ 3+\frac{4}{n+1}<3+\frac4n. \]
Hence
\[ a_{n+1}<a_n. \]
The sequence is strictly decreasing.
Moreover
\[ 3+\frac4n>3 \]
for every \(n\geq1\), so \(3\) is a lower bound.
By the monotone convergence theorem, the sequence converges to its infimum.
Since \(\displaystyle \frac4n\to0\), the terms approach \(3\) from above. Therefore
\[ \lim_{n\to+\infty}\frac{3n+4}{n}=3. \]
Exercise 11 — level ★★★☆☆
Study the sequence
\[ a_n=\frac{n^2}{n^2+1}. \]
Determine whether it is monotone and bounded, and find its limit.
Answer
The sequence is increasing, bounded above, and
\[ \lim_{n\to+\infty}\frac{n^2}{n^2+1}=1. \]
Solution
Let us rewrite the sequence:
\[ a_n=\frac{n^2}{n^2+1}=1-\frac{1}{n^2+1}. \]
Since \(n^2+1\) increases as \(n\) increases, the quantity
\[ \frac{1}{n^2+1} \]
decreases.
Consequently
\[ 1-\frac{1}{n^2+1} \]
increases.
Hence the sequence is increasing.
Moreover, for every \(n\geq1\),
\[ \frac{n^2}{n^2+1}<1. \]
Thus \(1\) is an upper bound.
The sequence is increasing and bounded above, hence convergent.
By the monotone convergence theorem, its limit is the supremum of the set of its values.
Since
\[ \frac{1}{n^2+1}\to0, \]
we obtain
\[ a_n=1-\frac{1}{n^2+1}\to1. \]
Hence
\[ \lim_{n\to+\infty}\frac{n^2}{n^2+1}=1. \]
Exercise 12 — level ★★★☆☆
Study the sequence
\[ a_n=\frac{n^2+1}{n^2}. \]
Determine whether it is monotone and bounded, and find its limit.
Answer
The sequence is decreasing, bounded below, and
\[ \lim_{n\to+\infty}\frac{n^2+1}{n^2}=1. \]
Solution
Let us rewrite the sequence:
\[ a_n=\frac{n^2+1}{n^2}=1+\frac1{n^2}. \]
The sequence \(\displaystyle \frac1{n^2}\) is decreasing, because \(n^2\) increases as \(n\) increases.
Hence the expression
\[ 1+\frac1{n^2} \]
is also decreasing.
Moreover, for every \(n\geq1\),
\[ 1+\frac1{n^2}>1. \]
Thus \(1\) is a lower bound.
The sequence is decreasing and bounded below, hence it converges to its infimum.
Since
\[ \frac1{n^2}\to0, \]
we have
\[ 1+\frac1{n^2}\to1. \]
Therefore
\[ \lim_{n\to+\infty}\frac{n^2+1}{n^2}=1. \]
Exercise 13 — level ★★★☆☆
Study the sequence
\[ a_n=\frac{2^n}{2^n+1}. \]
Determine whether it is monotone and bounded, and find its limit.
Answer
The sequence is increasing, bounded above, and
\[ \lim_{n\to+\infty}\frac{2^n}{2^n+1}=1. \]
Solution
Let us rewrite the sequence:
\[ a_n=\frac{2^n}{2^n+1}=1-\frac{1}{2^n+1}. \]
Since \(2^n\) increases as \(n\) increases, so does \(2^n+1\). Hence
\[ \frac{1}{2^n+1} \]
decreases.
Consequently
\[ 1-\frac{1}{2^n+1} \]
increases.
The sequence is therefore increasing.
Moreover, for every \(n\geq1\),
\[ \frac{2^n}{2^n+1}<1. \]
Thus \(1\) is an upper bound.
The sequence is increasing and bounded above; therefore it converges to its supremum.
Since
\[ \frac{1}{2^n+1}\to0, \]
it follows that
\[ a_n=1-\frac{1}{2^n+1}\to1. \]
Therefore
\[ \lim_{n\to+\infty}\frac{2^n}{2^n+1}=1. \]
Exercise 14 — level ★★★☆☆
Study the sequence
\[ a_n=\frac{3^n+1}{3^n}. \]
Determine whether it is monotone and bounded, and find its limit.
Answer
The sequence is decreasing, bounded below, and
\[ \lim_{n\to+\infty}\frac{3^n+1}{3^n}=1. \]
Solution
Let us rewrite the general term:
\[ a_n=\frac{3^n+1}{3^n}=1+\frac{1}{3^n}. \]
Since \(3^n\) increases as \(n\) increases, the sequence
\[ \frac{1}{3^n} \]
is decreasing.
Hence
\[ a_n=1+\frac{1}{3^n} \]
is decreasing.
Moreover, for every \(n\geq1\),
\[ 1+\frac{1}{3^n}>1. \]
Thus \(1\) is a lower bound.
The sequence is decreasing and bounded below, hence it converges to its infimum.
Since
\[ \frac1{3^n}\to0, \]
it follows that
\[ 1+\frac1{3^n}\to1. \]
Hence
\[ \lim_{n\to+\infty}\frac{3^n+1}{3^n}=1. \]
Exercise 15 — level ★★★★☆
Study the sequence
\[ a_n=\frac{n^2+n}{n^2+n+1}. \]
Determine whether it is monotone and bounded, and find its limit.
Answer
The sequence is increasing, bounded above, and
\[ \lim_{n\to+\infty}\frac{n^2+n}{n^2+n+1}=1. \]
Solution
Let us rewrite the sequence:
\[ a_n=\frac{n^2+n}{n^2+n+1}=1-\frac{1}{n^2+n+1}. \]
The denominator
\[ n^2+n+1 \]
increases as \(n\) increases. Indeed, passing from \(n\) to \(n+1\) we obtain
\[ (n+1)^2+(n+1)+1=n^2+3n+3, \]
which is greater than
\[ n^2+n+1. \]
Hence
\[ \frac{1}{n^2+n+1} \]
is decreasing.
Consequently
\[ 1-\frac{1}{n^2+n+1} \]
is increasing.
Moreover, for every \(n\geq1\),
\[ a_n=\frac{n^2+n}{n^2+n+1}<1. \]
Thus \(1\) is an upper bound.
The sequence is increasing and bounded above, hence convergent.
Since
\[ \frac{1}{n^2+n+1}\to0, \]
we have
\[ a_n=1-\frac{1}{n^2+n+1}\to1. \]
Therefore
\[ \lim_{n\to+\infty}\frac{n^2+n}{n^2+n+1}=1. \]
Exercise 16 — level ★★★★☆
Study the sequence
\[ a_n=\frac{n^2+2}{n^2+1}. \]
Determine whether it is monotone and bounded, and find its limit.
Answer
The sequence is decreasing, bounded below, and
\[ \lim_{n\to+\infty}\frac{n^2+2}{n^2+1}=1. \]
Solution
Let us rewrite:
\[ a_n=\frac{n^2+2}{n^2+1} = \frac{n^2+1+1}{n^2+1} = 1+\frac{1}{n^2+1}. \]
Since \(n^2+1\) increases as \(n\) increases, the quantity
\[ \frac{1}{n^2+1} \]
decreases.
Hence
\[ a_n=1+\frac{1}{n^2+1} \]
is decreasing.
Moreover, for every \(n\geq1\),
\[ a_n>1. \]
Thus \(1\) is a lower bound.
By the monotone convergence theorem, the sequence converges to its infimum.
Since
\[ \frac{1}{n^2+1}\to0, \]
we obtain
\[ a_n\to1. \]
Hence
\[ \lim_{n\to+\infty}\frac{n^2+2}{n^2+1}=1. \]
Exercise 17 — level ★★★★☆
Study the sequence
\[ a_n=\sqrt{n}. \]
Determine whether it is monotone and bounded, and find its limit.
Answer
The sequence is increasing, not bounded above, and
\[ \lim_{n\to+\infty}\sqrt n=+\infty. \]
Solution
The sequence is
\[ a_n=\sqrt n. \]
Since \(n+1>n\) and the square root is order-preserving on the non-negative numbers, we have
\[ \sqrt{n+1}>\sqrt n. \]
Hence
\[ a_{n+1}>a_n. \]
The sequence is strictly increasing.
We now show that it is not bounded above. Given \(M>0\), we seek \(n\) such that
\[ \sqrt n>M. \]
This inequality is equivalent to
\[ n>M^2. \]
It is always possible to choose a natural number \(n\) greater than \(M^2\). Hence the sequence is not bounded above.
Being increasing and not bounded above, by the monotone convergence theorem it diverges to \(+\infty\).
Therefore
\[ \lim_{n\to+\infty}\sqrt n=+\infty. \]
Exercise 18 — level ★★★★☆
Study the sequence
\[ a_n=-\sqrt n. \]
Determine whether it is monotone and bounded, and find its limit.
Answer
The sequence is decreasing, not bounded below, and
\[ \lim_{n\to+\infty}(-\sqrt n)=-\infty. \]
Solution
The sequence is
\[ a_n=-\sqrt n. \]
Since
\[ \sqrt{n+1}>\sqrt n, \]
multiplying by \(-1\) reverses the inequality:
\[ -\sqrt{n+1}<-\sqrt n. \]
Hence
\[ a_{n+1}<a_n. \]
The sequence is strictly decreasing.
It is not bounded below. Indeed, given \(M>0\), we seek \(n\) such that
\[ -\sqrt n<-M. \]
Multiplying by \(-1\), the inequality reverses:
\[ \sqrt n>M. \]
This inequality holds when
\[ n>M^2. \]
Hence the terms fall below any negative threshold.
Being decreasing and not bounded below, the sequence diverges to \(-\infty\).
Therefore
\[ \lim_{n\to+\infty}(-\sqrt n)=-\infty. \]
Exercise 19 — level ★★★★★
Let \((a_n)\) be an increasing sequence such that
\[ a_n<4 \]
for every \(n\geq1\). Determine whether the sequence has a limit and state what can be said about its value.
Answer
The sequence is convergent. Its limit \(L\) exists and equals
\[ L=\sup\{a_n:n\in\mathbb N,\ n\geq1\}. \]
Moreover \(L\leq4\).
Solution
The sequence \((a_n)\) is increasing by hypothesis. Moreover, for every \(n\geq1\),
\[ a_n<4. \]
Hence \(4\) is an upper bound of the set of values of the sequence:
\[ \{a_n:n\in\mathbb N,\ n\geq1\}. \]
The sequence is therefore increasing and bounded above.
By the monotone convergence theorem, an increasing sequence that is bounded above converges to its supremum.
Hence the finite limit
\[ L=\lim_{n\to+\infty}a_n \]
exists.
Moreover
\[ L=\sup\{a_n:n\in\mathbb N,\ n\geq1\}. \]
Since \(4\) is an upper bound, the supremum cannot exceed \(4\). Thus
\[ L\leq4. \]
We cannot, however, necessarily conclude that \(L=4\). For instance, an increasing sequence that always stays below \(4\) might converge to \(4\), but it might equally converge to some smaller number.
Hence the only definite information is:
\[ \text{the sequence converges and its limit satisfies } L\leq4. \]
Exercise 20 — level ★★★★★
Let \((a_n)\) be a decreasing sequence such that
\[ a_n>-2 \]
for every \(n\geq1\). Determine whether the sequence has a limit and state what can be said about its value.
Answer
The sequence is convergent. Its limit \(L\) exists and equals
\[ L=\inf\{a_n:n\in\mathbb N,\ n\geq1\}. \]
Moreover \(L\geq -2\).
Solution
The sequence \((a_n)\) is decreasing by hypothesis. Moreover, for every \(n\geq1\),
\[ a_n>-2. \]
Hence \(-2\) is a lower bound of the set of values of the sequence:
\[ \{a_n:n\in\mathbb N,\ n\geq1\}. \]
The sequence is therefore decreasing and bounded below.
By the monotone convergence theorem, a decreasing sequence that is bounded below converges to its infimum.
Hence the finite limit
\[ L=\lim_{n\to+\infty}a_n \]
exists.
Moreover
\[ L=\inf\{a_n:n\in\mathbb N,\ n\geq1\}. \]
Since \(-2\) is a lower bound, the infimum cannot be less than \(-2\). Thus
\[ L\geq -2. \]
We cannot, however, necessarily conclude that \(L=-2\). The sequence might tend to \(-2\), but it might equally tend to a larger number.
Hence the only definite information is:
\[ \text{the sequence converges and its limit satisfies } L\geq -2. \]