Understanding the convergence and divergence of sequences is a cornerstone of real analysis and calculus. A sequence \(\{a_n\}\) converges to a finite limit \(L\) if \(\lim_{n \to \infty} a_n = L\); otherwise it diverges. Mastering this topic requires fluency with limit laws, the Squeeze Theorem, geometric sequences, bounded and monotone sequences, and the formal epsilon-N definition of convergence. The problems below are organized from foundational limit computations to rigorous epsilon-N arguments, providing a complete pedagogical path for students at every level.
Convergence by Direct Limit Computation
These problems develop the fundamental skill of determining whether a sequence converges by computing \(\lim_{n \to \infty} a_n\) directly. You will apply algebraic simplification, L’Hôpital’s rule (when the sequence is derived from a continuous function), and standard limit laws. This section builds the computational fluency needed for all subsequent topics.
Problem 1: Rational Sequences and Dominant Terms
Easy
Consider the following two sequences defined for \(n \geq 1\):
- Determine whether \(a_n = \dfrac{3n^2 – 5n + 1}{2n^2 + 7}\) converges or diverges. If it converges, find its limit.
- Determine whether \(b_n = \dfrac{4n^3 + n}{n^2 + 1}\) converges or diverges. Justify your answer clearly.
Hint
View Solution
Divide numerator and denominator by \(n^2\):
\[
a_n = \frac{3 – \tfrac{5}{n} + \tfrac{1}{n^2}}{2 + \tfrac{7}{n^2}} \xrightarrow{n \to \infty} \frac{3 – 0 + 0}{2 + 0} = \frac{3}{2}.
\]
The sequence converges to \(\dfrac{3}{2}\).
Solution to question 2:
Divide numerator and denominator by \(n^2\):
\[
b_n = \frac{4n + \tfrac{1}{n}}{1 + \tfrac{1}{n^2}}.
\]
As \(n \to \infty\), the numerator \(4n \to \infty\) while the denominator \(\to 1\), so \(b_n \to +\infty\). The sequence diverges (to \(+\infty\)).
Problem 2: Sequences Involving Exponentials and Factorials
Easy
Analyze the long-term behavior of the following sequences:
- Determine whether \(c_n = \dfrac{2^n}{n!}\) converges or diverges. If it converges, state the limit.
- Determine whether \(d_n = \dfrac{n^{10}}{1.01^n}\) converges or diverges.
Hint
View Solution
Write
\[
c_n = \frac{2^n}{n!} = \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdots \frac{2}{n}.
\]
For \(n \geq 3\), each factor \(\tfrac{2}{k} \leq \tfrac{2}{3} < 1\), so \[ 0 < c_n \leq \frac{2}{1} \cdot \frac{2}{2} \cdot \left(\frac{2}{3}\right)^{n-2} = 2 \cdot \left(\frac{2}{3}\right)^{n-2} \to 0. \] By the Squeeze Theorem, \(c_n \to 0\). The sequence converges to 0.
Solution to question 2:
Consider the ratio \(\dfrac{d_{n+1}}{d_n} = \dfrac{(n+1)^{10}}{1.01^{n+1}} \cdot \dfrac{1.01^n}{n^{10}} = \dfrac{1}{1.01}\left(1+\tfrac{1}{n}\right)^{10} \to \dfrac{1}{1.01} < 1\).
Since the ratio eventually falls below 1, the terms decrease to 0, so \(d_n \to 0\). The sequence converges to 0.
Problem 3: Sequences with Logarithms and Roots
Easy
Investigate the convergence of the following sequences:
- Does \(e_n = \dfrac{\ln n}{n}\) converge or diverge? Find the limit if it exists.
- Does \(f_n = n^{1/n}\) converge or diverge? Find the limit if it exists.
Hint
View Solution
The function \(f(x) = \tfrac{\ln x}{x}\) satisfies \(f(n) = e_n\). Applying L’Hôpital’s rule:
\[
\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0.
\]
Therefore \(e_n \to 0\). The sequence converges to 0.
Solution to question 2:
Let \(L = \lim_{n \to \infty} n^{1/n}\). Then
\[
\ln L = \lim_{n \to \infty} \frac{\ln n}{n} = 0 \quad (\text{from part 1}).
\]
Since \(\ln L = 0\), we have \(L = e^0 = 1\). The sequence converges to 1.
Geometric Sequences and the Role of the Common Ratio
Geometric sequences of the form \(\{r^n\}\) exhibit a sharp dichotomy: they converge when \(|r| < 1\) or \(r = 1\), and diverge otherwise. This section explores that behavior and its application to more complex sequences built on exponential growth or decay. Understanding geometric sequences is essential for studying series convergence tests and recursive models.
Problem 4: Pure Geometric Sequences
Easy
For each of the following, determine whether the sequence converges or diverges. If it converges, state the limit.
- \(a_n = \left(-\dfrac{3}{4}\right)^n\)
- \(a_n = (-1)^n\)
- \(a_n = 1.001^n\)
Hint
View Solution
Here \(r = -\tfrac{3}{4}\) and \(|r| = \tfrac{3}{4} < 1\), so \(\left(-\tfrac{3}{4}\right)^n \to 0\). The sequence converges to 0.
Solution to question 2:
Here \(r = -1\). The terms alternate: \(-1, 1, -1, 1, \ldots\). Since there is no single finite value they approach, the sequence diverges (it oscillates).
Solution to question 3:
Here \(r = 1.001 > 1\), so \(1.001^n \to +\infty\). The sequence diverges.
Problem 5: Geometric-Type Sequences with Algebraic Factors
Medium
Analyze the convergence of the following sequences, which combine polynomial and geometric components:
- Determine whether \(a_n = \dfrac{n \cdot 2^n}{3^n}\) converges or diverges.
- Determine whether \(b_n = \dfrac{(-1)^n \cdot n}{n + 1}\) converges or diverges. Justify your answer rigorously.
Hint
View Solution
\[
a_n = n \cdot \left(\frac{2}{3}\right)^n.
\]
Since \(\left(\tfrac{2}{3}\right)^n \to 0\) exponentially and \(n\) grows polynomially, exponential decay dominates. More formally, for any \(\varepsilon > 0\), one can find \(N\) such that \(n < (1+\varepsilon)^n / C\) for a suitable constant \(C\). Therefore \(a_n \to 0\). The sequence converges to 0.
Solution to question 2:
Consider two subsequences: when \(n = 2k\) (even), \(b_{2k} = \tfrac{2k}{2k+1} \to 1\). When \(n = 2k-1\) (odd), \(b_{2k-1} = -\tfrac{2k-1}{2k} \to -1\). Since these two subsequences converge to different values (\(1 \neq -1\)), the sequence \(\{b_n\}\) diverges by oscillation.
The Squeeze Theorem for Sequences
The Squeeze Theorem (also called the Sandwich Theorem) is an indispensable tool when direct limit computation is difficult. If \(a_n \leq s_n \leq b_n\) for all sufficiently large \(n\) and both \(a_n\) and \(b_n\) converge to the same limit \(L\), then \(s_n \to L\) as well. This section trains students to construct appropriate bounding sequences, a skill that also underlies many formal convergence proofs.
Problem 6: Applying the Squeeze Theorem with Trigonometric Bounds
Medium
Use the Squeeze Theorem to analyze the following sequences:
- Prove that \(a_n = \dfrac{\sin(n^2)}{n}\) converges and find its limit.
- Determine whether \(b_n = \dfrac{n + \cos n}{n}\) converges. If so, state the limit.
Hint
View Solution
Since \(-1 \leq \sin(n^2) \leq 1\) for all \(n\), we have
\[
-\frac{1}{n} \leq \frac{\sin(n^2)}{n} \leq \frac{1}{n}.
\]
Both bounds converge to 0 as \(n \to \infty\). By the Squeeze Theorem, \(a_n \to 0\). The sequence converges to 0.
Solution to question 2:
Using \(-1 \leq \cos n \leq 1\):
\[
1 – \frac{1}{n} = \frac{n – 1}{n} \leq \frac{n + \cos n}{n} \leq \frac{n + 1}{n} = 1 + \frac{1}{n}.
\]
Both bounds converge to 1, so by the Squeeze Theorem, \(b_n \to 1\). The sequence converges to 1.
Problem 7: Squeeze Theorem with Exponential and Power Bounds
Medium
Apply the Squeeze Theorem to the following sequences involving non-obvious bounds:
- Show that \(a_n = \left(\dfrac{n}{n+1}\right)^n\) converges and identify the limit.
- Show that \(b_n = \left(1 + \dfrac{3}{n}\right)^n\) converges and find its limit.
Hint
View Solution
\[
a_n = \left(\frac{n}{n+1}\right)^n = \left(\frac{1}{1 + 1/n}\right)^n = \frac{1}{\left(1 + \frac{1}{n}\right)^n}.
\]
Since \(\left(1 + \tfrac{1}{n}\right)^n \to e\), we get \(a_n \to \dfrac{1}{e} = e^{-1}\). The sequence converges to \(e^{-1}\).
Solution to question 2:
Using the standard limit with \(x = 3\):
\[
b_n = \left(1 + \frac{3}{n}\right)^n \to e^3.
\]
The sequence converges to \(e^3\).
Bounded and Monotone Sequences
The Monotone Convergence Theorem (MCT) guarantees that every sequence which is both bounded and monotone must converge, even when an explicit limit formula is unavailable. This section develops the skill of proving monotonicity (via the ratio \(a_{n+1}/a_n\) or the difference \(a_{n+1} – a_n\)) and establishing bounds, which are critical techniques in analysis and the study of recursive sequences.
Problem 8: Proving Convergence via the Monotone Convergence Theorem
Medium
For each sequence, prove it is both bounded and monotone, then conclude whether it converges.
- Let \(a_n = \dfrac{2n}{n+1}\). Show the sequence is increasing and bounded above, and find its limit.
- Let \(b_n = \left(\dfrac{1}{3}\right)^n + 1\). Show the sequence is decreasing and bounded below, and find its limit.
Hint
View Solution
Monotone:
\[
a_{n+1} – a_n = \frac{2(n+1)}{n+2} – \frac{2n}{n+1} = \frac{2(n+1)^2 – 2n(n+2)}{(n+2)(n+1)} = \frac{2}{(n+2)(n+1)} > 0.
\]
So \(\{a_n\}\) is strictly increasing.
Bounded: \(a_n = \tfrac{2n}{n+1} < 2\) for all \(n \geq 1\), and \(a_n \geq 1\) for \(n \geq 1\). By MCT, the limit exists. Direct computation: \(\lim_{n\to\infty}\tfrac{2n}{n+1} = 2\). The sequence converges to 2.
Solution to question 2:
Monotone: \(b_{n+1} – b_n = \left(\tfrac{1}{3}\right)^{n+1} – \left(\tfrac{1}{3}\right)^n = \left(\tfrac{1}{3}\right)^n\left(\tfrac{1}{3}-1\right) = -\tfrac{2}{3}\left(\tfrac{1}{3}\right)^n < 0\). So \(\{b_n\}\) is strictly decreasing.
Bounded: \(b_n = \left(\tfrac{1}{3}\right)^n + 1 > 1\) for all \(n\), and \(b_1 = \tfrac{4}{3}\). By MCT, \(b_n \to 1\). The sequence converges to 1.
Problem 9: A Recursive Sequence and the Monotone Convergence Theorem
Hard
Define a sequence recursively by \(a_1 = 1\) and \(a_{n+1} = \sqrt{2 + a_n}\) for all \(n \geq 1\).
- Prove by induction that \(a_n \leq 2\) for all \(n \geq 1\).
- Prove that the sequence \(\{a_n\}\) is strictly increasing.
- Conclude that \(\{a_n\}\) converges and determine its limit.
Hint
View Solution
Base case: \(a_1 = 1 \leq 2\). ✓
Inductive step: Assume \(a_n \leq 2\). Then
\[
a_{n+1} = \sqrt{2 + a_n} \leq \sqrt{2 + 2} = \sqrt{4} = 2.
\]
By induction, \(a_n \leq 2\) for all \(n \geq 1\). Since each \(a_n > 0\), the sequence is also bounded below by 0.
Solution to question 2 (Monotonicity):
We need \(a_{n+1} > a_n\), i.e., \(\sqrt{2 + a_n} > a_n\). Squaring (both sides positive): \(2 + a_n > a_n^2\), i.e., \(a_n^2 – a_n – 2 < 0\), i.e., \((a_n - 2)(a_n + 1) < 0\). Since \(0 < a_n \leq 2\), we have \(a_n - 2 \leq 0\) and \(a_n + 1 > 0\), so the product \(\leq 0\). Strict inequality holds for \(a_n < 2\). Since \(a_1 = 1 < 2\), all terms satisfy \(a_n < 2\) strictly (the bound is never attained), so the sequence is strictly increasing.
Solution to question 3 (Finding the limit):
By the Monotone Convergence Theorem, \(\{a_n\}\) is bounded and increasing, so \(L = \lim_{n \to \infty} a_n\) exists. Taking limits in the recurrence:
\[
L = \sqrt{2 + L} \implies L^2 = 2 + L \implies L^2 – L – 2 = 0 \implies (L-2)(L+1) = 0.
\]
Since \(L \geq a_1 = 1 > 0\), we discard \(L = -1\). Therefore \(L = 2\), and the sequence converges to 2.
Divergent Sequences: Types and Identification
A sequence diverges when it does not converge to any finite value. There are three qualitatively distinct modes of divergence: the sequence grows to \(+\infty\), decreases to \(-\infty\), or oscillates without settling near any fixed value. Identifying the correct mode of divergence is as important as proving convergence, especially when applying divergence tests to series. This section covers all three modes through carefully graded examples.
Problem 10: Identifying the Mode of Divergence
Easy
For each sequence below, determine whether it diverges to \(+\infty\), to \(-\infty\), or by oscillation. Justify each conclusion.
- \(a_n = (-1)^n \cdot n\)
- \(b_n = \ln n\)
- \(c_n = (-2)^n\)
Hint
View Solution
Even terms: \(a_{2k} = 2k \to +\infty\). Odd terms: \(a_{2k-1} = -(2k-1) \to -\infty\). The sequence oscillates with terms growing in magnitude in opposite directions. This is divergence by oscillation (with unbounded amplitude).
Solution to question 2:
Since \(\ln n\) is strictly increasing and unbounded above, \(b_n \to +\infty\). This is divergence to \(+\infty\).
Solution to question 3:
\(|c_n| = 2^n \to +\infty\), but the sign alternates. Even terms \(\to +\infty\) and odd terms \(\to -\infty\). This is again divergence by oscillation with unbounded amplitude (similar to case 1 but with exponential growth).
Problem 11: Subsequence Criterion for Divergence
Medium
The following problem uses the subsequence criterion: a sequence \(\{a_n\}\) converges to \(L\) if and only if every subsequence also converges to \(L\).
- Use the subsequence criterion to prove that \(a_n = \sin\!\left(\dfrac{n\pi}{2}\right)\) diverges.
- Prove that \(b_n = \cos(n\pi)\) diverges by identifying two convergent subsequences with different limits.
Hint
View Solution
The sequence cycles: \(a_1 = 1,\; a_2 = 0,\; a_3 = -1,\; a_4 = 0,\; a_5 = 1, \ldots\)
Consider the subsequence \(a_{4k+1} = \sin\!\left(\tfrac{(4k+1)\pi}{2}\right) = 1 \to 1\), and the subsequence \(a_{4k+3} = \sin\!\left(\tfrac{(4k+3)\pi}{2}\right) = -1 \to -1\). Since two subsequences converge to different limits (\(1 \neq -1\)), the sequence \(\{a_n\}\) diverges.
Solution to question 2:
\(b_n = \cos(n\pi) = (-1)^n\). Even subsequence: \(b_{2k} = 1 \to 1\). Odd subsequence: \(b_{2k-1} = -1 \to -1\). Since \(1 \neq -1\), by the subsequence criterion, \(\{b_n\}\) diverges.
Formal Epsilon-N Proofs of Convergence
The epsilon-N (\(\varepsilon\text{-}N\)) definition of convergence provides the rigorous foundation of all limit results. A sequence \(\{a_n\}\) converges to \(L\) if: for every \(\varepsilon > 0\), there exists \(N \in \mathbb{N}\) such that \(n > N \implies |a_n – L| < \varepsilon\). The problems in this section develop the craft of constructing such proofs, an essential skill for analysis courses and for rigorously verifying convergence claims that cannot rely solely on limit laws.
Problem 12: Epsilon-N Proof for a Simple Rational Sequence
Medium
Use the formal epsilon-N definition of convergence to prove the following:
- Prove that \(\lim_{n \to \infty} \dfrac{1}{n} = 0\).
- Prove that \(\lim_{n \to \infty} \dfrac{2n+1}{n+3} = 2\).
Hint
View Solution
Let \(\varepsilon > 0\). We need \(\left|\tfrac{1}{n} – 0\right| = \tfrac{1}{n} < \varepsilon\), which requires \(n > \tfrac{1}{\varepsilon}\). Choose \(N = \left\lceil \tfrac{1}{\varepsilon} \right\rceil\). Then for all \(n > N\):
\[
\left|\frac{1}{n} – 0\right| = \frac{1}{n} < \frac{1}{N} \leq \varepsilon. \]
Solution to question 2:
\[
\left|\frac{2n+1}{n+3} – 2\right| = \left|\frac{2n+1 – 2(n+3)}{n+3}\right| = \left|\frac{-5}{n+3}\right| = \frac{5}{n+3}.
\]
We need \(\dfrac{5}{n+3} < \varepsilon\), i.e., \(n > \dfrac{5}{\varepsilon} – 3\). Choose \(N = \left\lceil \dfrac{5}{\varepsilon} \right\rceil\). Then for all \(n > N\):
\[
\left|\frac{2n+1}{n+3} – 2\right| = \frac{5}{n+3} < \frac{5}{n} \leq \frac{5}{N} \leq \varepsilon. \]
Problem 13: Epsilon-N Proof Involving an Absolute Value and a Square Root
Hard
Write rigorous epsilon-N proofs for the following convergence statements:
- Prove that \(\lim_{n \to \infty} \dfrac{n^2}{n^2 + 1} = 1\).
- Prove that \(\lim_{n \to \infty} \sqrt{n+1} – \sqrt{n} = 0\).
Hint
View Solution
\[
\left|\frac{n^2}{n^2+1} – 1\right| = \left|\frac{-1}{n^2+1}\right| = \frac{1}{n^2+1} < \frac{1}{n^2}. \] Given \(\varepsilon > 0\), choose \(N = \left\lceil \tfrac{1}{\sqrt{\varepsilon}} \right\rceil\). For \(n > N\):
\[
\left|\frac{n^2}{n^2+1} – 1\right| < \frac{1}{n^2} < \frac{1}{N^2} \leq \varepsilon. \]
Solution to question 2:
Rationalize:
\[
\sqrt{n+1} – \sqrt{n} = \frac{(n+1) – n}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{2\sqrt{n}}. \] Given \(\varepsilon > 0\), we need \(\dfrac{1}{2\sqrt{n}} < \varepsilon\), i.e., \(n > \dfrac{1}{4\varepsilon^2}\). Choose \(N = \left\lceil \dfrac{1}{4\varepsilon^2} \right\rceil\). For \(n > N\):
\[
\left|\sqrt{n+1} – \sqrt{n} – 0\right| < \frac{1}{2\sqrt{n}} < \frac{1}{2\sqrt{N}} \leq \varepsilon. \]
Limit Superior, Limit Inferior, and Advanced Convergence
The limit superior (\(\limsup\)) and limit inferior (\(\liminf\)) extend the notion of a limit to all sequences, even divergent ones. A sequence converges if and only if \(\limsup_{n\to\infty} a_n = \liminf_{n\to\infty} a_n\). These tools are essential for handling oscillatory sequences, proving convergence in advanced contexts, and applying the Root Test for series. This section also introduces Cauchy sequences, which provide an intrinsic convergence criterion that does not require prior knowledge of the limit.
Problem 14: Computing Limit Superior and Limit Inferior
Hard
For each of the following sequences, compute \(\limsup_{n \to \infty} a_n\) and \(\liminf_{n \to \infty} a_n\). State whether the sequence converges.
- \(a_n = (-1)^n + \dfrac{1}{n}\)
- \(b_n = \dfrac{2 + (-1)^n \cdot n}{n}\)
Hint
View Solution
For even \(n = 2k\): \(a_{2k} = 1 + \tfrac{1}{2k} \to 1\) (from above, approaching 1).
For odd \(n = 2k-1\): \(a_{2k-1} = -1 + \tfrac{1}{2k-1} \to -1\) (from above, approaching -1).
Since the even terms approach 1 from above and the odd terms approach -1 from above:
\[
\limsup_{n \to \infty} a_n = 1, \qquad \liminf_{n \to \infty} a_n = -1.
\]
Since \(\limsup \neq \liminf\), the sequence diverges.
Solution to question 2:
Rewrite: \(b_n = \tfrac{2}{n} + (-1)^n\). For even \(n\): \(b_{2k} = \tfrac{1}{k} + 1 \to 1\). For odd \(n\): \(b_{2k-1} = \tfrac{2}{2k-1} – 1 \to -1\). Therefore:
\[
\limsup_{n \to \infty} b_n = 1, \qquad \liminf_{n \to \infty} b_n = -1.
\]
Since \(\limsup \neq \liminf\), the sequence diverges.
Problem 15: Cauchy Sequences and the Completeness of the Reals
Hard
A sequence \(\{a_n\}\) is called a Cauchy sequence if for every \(\varepsilon > 0\) there exists \(N \in \mathbb{N}\) such that \(m, n > N \implies |a_m – a_n| < \varepsilon\).
- Prove that every convergent sequence is a Cauchy sequence.
- Show directly (using the Cauchy criterion) that the sequence \(a_n = \dfrac{1}{n}\) is a Cauchy sequence.
Hint
View Solution
Suppose \(a_n \to L\). Let \(\varepsilon > 0\). Since \(a_n \to L\), there exists \(N\) such that \(n > N \implies |a_n – L| < \tfrac{\varepsilon}{2}\). For \(m, n > N\), by the triangle inequality:
\[
|a_m – a_n| \leq |a_m – L| + |a_n – L| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \] Therefore \(\{a_n\}\) is a Cauchy sequence.Solution to question 2:
Let \(\varepsilon > 0\). For \(m, n > N\):
\[
\left|\frac{1}{m} – \frac{1}{n}\right| \leq \frac{1}{m} + \frac{1}{n} < \frac{1}{N} + \frac{1}{N} = \frac{2}{N}. \] Choose \(N = \left\lceil \tfrac{2}{\varepsilon} \right\rceil\). Then \(\tfrac{2}{N} \leq \varepsilon\), so \(\left|\tfrac{1}{m} - \tfrac{1}{n}\right| < \varepsilon\) for all \(m, n > N\). Therefore \(\left\{\tfrac{1}{n}\right\}\) is a Cauchy sequence.