Skip to content

Convergence and Divergence of Sequences: Practice Problems

Type
Practice problems
Problems
15
Levels
Easy · Medium · Hard

Understanding the convergence and divergence of sequences is a cornerstone of real analysis and calculus. A sequence {an}\{a_n\} converges to a finite limit LL if lim⁡n→∞an=L\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.

Contents7 sections

Convergence by Direct Limit Computation

These problems develop the fundamental skill of determining whether a sequence converges by computing lim⁡n→∞an\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≥1n \geq 1:

  1. Determine whether an=3n2−5n+12n2+7a_n = \dfrac{3n^2 - 5n + 1}{2n^2 + 7} converges or diverges. If it converges, find its limit.
  2. Determine whether bn=4n3+nn2+1b_n = \dfrac{4n^3 + n}{n^2 + 1} converges or diverges. Justify your answer clearly.
Hint
Divide both numerator and denominator by the highest power of nn present in the expression. Identify which terms vanish as n→∞n \to \infty and which dominate.
View Solution
Solution to question 1: Divide numerator and denominator by n2n^2: an=3−5n+1n22+7n2→n→∞3−0+02+0=32. 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 32\dfrac{3}{2}.

Solution to question 2: Divide numerator and denominator by n2n^2: bn=4n+1n1+1n2. b_n = \frac{4n + \tfrac{1}{n}}{1 + \tfrac{1}{n^2}}. As n→∞n \to \infty, the numerator 4n→∞4n \to \infty while the denominator →1\to 1, so bn→+∞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:

  1. Determine whether cn=2nn!c_n = \dfrac{2^n}{n!} converges or diverges. If it converges, state the limit.
  2. Determine whether dn=n101.01nd_n = \dfrac{n^{10}}{1.01^n} converges or diverges.
Hint
For (1), write out several terms or bound the sequence using the fact that n!≥2n−1n! \geq 2^{n-1} for large nn. For (2), recall that exponential growth always dominates polynomial growth; consider using L’Hôpital’s rule or the ratio dn+1/dnd_{n+1}/d_n.
View Solution
Solution to question 1: Write cn=2nn!=21⋅22⋅23⋯2n. c_n = \frac{2^n}{n!} = \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdots \frac{2}{n}. For n≥3n \geq 3, each factor 2k≤23<1\tfrac{2}{k} \leq \tfrac{2}{3} < 1, so 0<cn≤21⋅22⋅(23)n−2=2⋅(23)n−2→0. 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, cn→0c_n \to 0. The sequence converges to 0.

Solution to question 2: Consider the ratio dn+1dn=(n+1)101.01n+1⋅1.01nn10=11.01(1+1n)10→11.01<1\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 dn→0d_n \to 0. The sequence converges to 0.

Problem 3: Sequences with Logarithms and Roots

Easy

Investigate the convergence of the following sequences:

  1. Does en=ln⁡nne_n = \dfrac{\ln n}{n} converge or diverge? Find the limit if it exists.
  2. Does fn=n1/nf_n = n^{1/n} converge or diverge? Find the limit if it exists.
Hint
Both sequences can be analyzed using the continuous function f(x)f(x) corresponding to ana_n. Apply L’Hôpital’s rule to handle the indeterminate forms. For fn=n1/nf_n = n^{1/n}, take the natural logarithm first.
View Solution
Solution to question 1: The function f(x)=ln⁡xxf(x) = \tfrac{\ln x}{x} satisfies f(n)=enf(n) = e_n. Applying L’Hôpital’s rule: lim⁡x→∞ln⁡xx=lim⁡x→∞1/x1=0. \lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0. Therefore en→0e_n \to 0. The sequence converges to 0.

Solution to question 2: Let L=lim⁡n→∞n1/nL = \lim_{n \to \infty} n^{1/n}. Then ln⁡L=lim⁡n→∞ln⁡nn=0(from part 1). \ln L = \lim_{n \to \infty} \frac{\ln n}{n} = 0 \quad (\text{from part 1}). Since ln⁡L=0\ln L = 0, we have L=e0=1L = e^0 = 1. The sequence converges to 1.

Geometric Sequences and the Role of the Common Ratio

Geometric sequences of the form {rn}\{r^n\} exhibit a sharp dichotomy: they converge when ∣r∣<1|r| < 1 or r=1r = 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.

  1. an=(−34)na_n = \left(-\dfrac{3}{4}\right)^n
  2. an=(−1)na_n = (-1)^n
  3. an=1.001na_n = 1.001^n
Hint
Recall the key theorem: {rn}\{r^n\} converges if and only if −1<r≤1-1 < r \leq 1. When r=−1r = -1, the terms oscillate between −1-1 and 11 without settling, which indicates divergence.
View Solution
Solution to question 1: Here r=−34r = -\tfrac{3}{4} and ∣r∣=34<1|r| = \tfrac{3}{4} < 1, so (−34)n→0\left(-\tfrac{3}{4}\right)^n \to 0. The sequence converges to 0.

Solution to question 2: Here r=−1r = -1. The terms alternate: −1,1,−1,1,…-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>1r = 1.001 > 1, so 1.001n→+∞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:

  1. Determine whether an=n⋅2n3na_n = \dfrac{n \cdot 2^n}{3^n} converges or diverges.
  2. Determine whether bn=(−1)n⋅nn+1b_n = \dfrac{(-1)^n \cdot n}{n + 1} converges or diverges. Justify your answer rigorously.
Hint
For (1), rewrite as n⋅(23)nn \cdot \left(\tfrac{2}{3}\right)^n and recall that exponential decay beats polynomial growth. For (2), consider the subsequences of even and odd indices separately.
View Solution
Solution to question 1: an=n⋅(23)n. a_n = n \cdot \left(\frac{2}{3}\right)^n. Since (23)n→0\left(\tfrac{2}{3}\right)^n \to 0 exponentially and nn grows polynomially, exponential decay dominates. More formally, for any ε>0\varepsilon > 0, one can find NN such that n<(1+ε)n/Cn < (1+\varepsilon)^n / C for a suitable constant CC. Therefore an→0a_n \to 0. The sequence converges to 0.

Solution to question 2: Consider two subsequences: when n=2kn = 2k (even), b2k=2k2k+1→1b_{2k} = \tfrac{2k}{2k+1} \to 1. When n=2k−1n = 2k-1 (odd), b2k−1=−2k−12k→−1b_{2k-1} = -\tfrac{2k-1}{2k} \to -1. Since these two subsequences converge to different values (1≠−11 \neq -1), the sequence {bn}\{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 an≤sn≤bna_n \leq s_n \leq b_n for all sufficiently large nn and both ana_n and bnb_n converge to the same limit LL, then sn→Ls_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:

  1. Prove that an=sin⁡(n2)na_n = \dfrac{\sin(n^2)}{n} converges and find its limit.
  2. Determine whether bn=n+cos⁡nnb_n = \dfrac{n + \cos n}{n} converges. If so, state the limit.
Hint
For both problems, use the universal bound −1≤sin⁡(⋅)≤1-1 \leq \sin(\cdot) \leq 1 and −1≤cos⁡(⋅)≤1-1 \leq \cos(\cdot) \leq 1. Construct a simple lower bound and upper bound for each sequence, then let n→∞n \to \infty.
View Solution
Solution to question 1: Since −1≤sin⁡(n2)≤1-1 \leq \sin(n^2) \leq 1 for all nn, we have −1n≤sin⁡(n2)n≤1n. -\frac{1}{n} \leq \frac{\sin(n^2)}{n} \leq \frac{1}{n}. Both bounds converge to 0 as n→∞n \to \infty. By the Squeeze Theorem, an→0a_n \to 0. The sequence converges to 0.

Solution to question 2: Using −1≤cos⁡n≤1-1 \leq \cos n \leq 1: 1−1n=n−1n≤n+cos⁡nn≤n+1n=1+1n. 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, bn→1b_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:

  1. Show that an=(nn+1)na_n = \left(\dfrac{n}{n+1}\right)^n converges and identify the limit.
  2. Show that bn=(1+3n)nb_n = \left(1 + \dfrac{3}{n}\right)^n converges and find its limit.
Hint
Both expressions are related to the definition of ee. Recall that lim⁡n→∞(1+xn)n=ex\lim_{n \to \infty} \left(1 + \tfrac{x}{n}\right)^n = e^x. For (1), rewrite nn+1=11+1/n\tfrac{n}{n+1} = \tfrac{1}{1 + 1/n} and use this standard limit.
View Solution
Solution to question 1: an=(nn+1)n=(11+1/n)n=1(1+1n)n. 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 (1+1n)n→e\left(1 + \tfrac{1}{n}\right)^n \to e, we get an→1e=e−1a_n \to \dfrac{1}{e} = e^{-1}. The sequence converges to e−1e^{-1}.

Solution to question 2: Using the standard limit with x=3x = 3: bn=(1+3n)n→e3. b_n = \left(1 + \frac{3}{n}\right)^n \to e^3. The sequence converges to e3e^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 an+1/ana_{n+1}/a_n or the difference an+1−ana_{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.

  1. Let an=2nn+1a_n = \dfrac{2n}{n+1}. Show the sequence is increasing and bounded above, and find its limit.
  2. Let bn=(13)n+1b_n = \left(\dfrac{1}{3}\right)^n + 1. Show the sequence is decreasing and bounded below, and find its limit.
Hint
For monotonicity, compute an+1−ana_{n+1} - a_n (or the ratio) and determine its sign. For bounds, find explicit constants m≤an≤Mm \leq a_n \leq M that hold for all n≥1n \geq 1.
View Solution
Solution to question 1: Monotone: an+1−an=2(n+1)n+2−2nn+1=2(n+1)2−2n(n+2)(n+2)(n+1)=2(n+2)(n+1)>0. 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 {an}\{a_n\} is strictly increasing. Bounded: an=2nn+1<2a_n = \tfrac{2n}{n+1} < 2 for all n≥1n \geq 1, and an≥1a_n \geq 1 for n≥1n \geq 1. By MCT, the limit exists. Direct computation: lim⁡n→∞2nn+1=2\lim_{n\to\infty}\tfrac{2n}{n+1} = 2. The sequence converges to 2.

Solution to question 2: Monotone: bn+1−bn=(13)n+1−(13)n=(13)n(13−1)=−23(13)n<0b_{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 {bn}\{b_n\} is strictly decreasing. Bounded: bn=(13)n+1>1b_n = \left(\tfrac{1}{3}\right)^n + 1 > 1 for all nn, and b1=43b_1 = \tfrac{4}{3}. By MCT, bn→1b_n \to 1. The sequence converges to 1.

Problem 9: A Recursive Sequence and the Monotone Convergence Theorem

Hard

Define a sequence recursively by a1=1a_1 = 1 and an+1=2+ana_{n+1} = \sqrt{2 + a_n} for all n≥1n \geq 1.

  1. Prove by induction that an≤2a_n \leq 2 for all n≥1n \geq 1.
  2. Prove that the sequence {an}\{a_n\} is strictly increasing.
  3. Conclude that {an}\{a_n\} converges and determine its limit.
Hint
For (1), assume an≤2a_n \leq 2 and show an+1=2+an≤2+2=2a_{n+1} = \sqrt{2 + a_n} \leq \sqrt{2 + 2} = 2. For (3), if L=lim⁡anL = \lim a_n exists, then taking limits on both sides of the recurrence gives L=2+LL = \sqrt{2 + L}. Solve this equation for LL, keeping only the positive root.
View Solution
Solution to question 1 (Boundedness by induction): Base case: a1=1≤2a_1 = 1 \leq 2. ✓ Inductive step: Assume an≤2a_n \leq 2. Then an+1=2+an≤2+2=4=2. a_{n+1} = \sqrt{2 + a_n} \leq \sqrt{2 + 2} = \sqrt{4} = 2. By induction, an≤2a_n \leq 2 for all n≥1n \geq 1. Since each an>0a_n > 0, the sequence is also bounded below by 0.

Solution to question 2 (Monotonicity): We need an+1>ana_{n+1} > a_n, i.e., 2+an>an\sqrt{2 + a_n} > a_n. Squaring (both sides positive): 2+an>an22 + a_n > a_n^2, i.e., an2−an−2<0a_n^2 - a_n - 2 < 0, i.e., (an−2)(an+1)<0(a_n - 2)(a_n + 1) < 0. Since 0<an≤20 < a_n \leq 2, we have an−2≤0a_n - 2 \leq 0 and an+1>0a_n + 1 > 0, so the product ≤0\leq 0. Strict inequality holds for an<2a_n < 2. Since a1=1<2a_1 = 1 < 2, all terms satisfy an<2a_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, {an}\{a_n\} is bounded and increasing, so L=lim⁡n→∞anL = \lim_{n \to \infty} a_n exists. Taking limits in the recurrence: L=2+L  ⟹  L2=2+L  ⟹  L2−L−2=0  ⟹  (L−2)(L+1)=0. L = \sqrt{2 + L} \implies L^2 = 2 + L \implies L^2 - L - 2 = 0 \implies (L-2)(L+1) = 0. Since L≥a1=1>0L \geq a_1 = 1 > 0, we discard L=−1L = -1. Therefore L=2L = 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.

  1. an=(−1)n⋅na_n = (-1)^n \cdot n
  2. bn=ln⁡nb_n = \ln n
  3. cn=(−2)nc_n = (-2)^n
Hint
For (1), examine the subsequences along even and odd indices. For (2), recall that ln⁡n→+∞\ln n \to +\infty. For (3), consider the magnitude ∣(−2)n∣=2n|(-2)^n| = 2^n alongside the alternating sign.
View Solution
Solution to question 1: Even terms: a2k=2k→+∞a_{2k} = 2k \to +\infty. Odd terms: a2k−1=−(2k−1)→−∞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\ln n is strictly increasing and unbounded above, bn→+∞b_n \to +\infty. This is divergence to +∞+\infty.

Solution to question 3: ∣cn∣=2n→+∞|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 {an}\{a_n\} converges to LL if and only if every subsequence also converges to LL.

  1. Use the subsequence criterion to prove that an=sin⁡ ⁣(nπ2)a_n = \sin\!\left(\dfrac{n\pi}{2}\right) diverges.
  2. Prove that bn=cos⁡(nπ)b_n = \cos(n\pi) diverges by identifying two convergent subsequences with different limits.
Hint
For (1), compute ana_n for n=1,2,3,4,5,…n = 1, 2, 3, 4, 5, \ldots and identify recurring values. Then pick two subsequences with different limits. For (2), recall that cos⁡(nπ)=(−1)n\cos(n\pi) = (-1)^n.
View Solution
Solution to question 1: The sequence cycles: a1=1,  a2=0,  a3=−1,  a4=0,  a5=1,…a_1 = 1,\; a_2 = 0,\; a_3 = -1,\; a_4 = 0,\; a_5 = 1, \ldots Consider the subsequence a4k+1=sin⁡ ⁣((4k+1)π2)=1→1a_{4k+1} = \sin\!\left(\tfrac{(4k+1)\pi}{2}\right) = 1 \to 1, and the subsequence a4k+3=sin⁡ ⁣((4k+3)π2)=−1→−1a_{4k+3} = \sin\!\left(\tfrac{(4k+3)\pi}{2}\right) = -1 \to -1. Since two subsequences converge to different limits (1≠−11 \neq -1), the sequence {an}\{a_n\} diverges.

Solution to question 2: bn=cos⁡(nπ)=(−1)nb_n = \cos(n\pi) = (-1)^n. Even subsequence: b2k=1→1b_{2k} = 1 \to 1. Odd subsequence: b2k−1=−1→−1b_{2k-1} = -1 \to -1. Since 1≠−11 \neq -1, by the subsequence criterion, {bn}\{b_n\} diverges.

Formal Epsilon-N Proofs of Convergence

The epsilon-N (ε-N\varepsilon\text{-}N) definition of convergence provides the rigorous foundation of all limit results. A sequence {an}\{a_n\} converges to LL if: for every ε>0\varepsilon > 0, there exists N∈NN \in \mathbb{N} such that n>N  ⟹  ∣an−L∣<εn > 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:

  1. Prove that lim⁡n→∞1n=0\lim_{n \to \infty} \dfrac{1}{n} = 0.
  2. Prove that lim⁡n→∞2n+1n+3=2\lim_{n \to \infty} \dfrac{2n+1}{n+3} = 2.
Hint
Start by writing out ∣an−L∣|a_n - L| and simplify. Then solve the inequality ∣an−L∣<ε|a_n - L| < \varepsilon for nn to find an appropriate NN in terms of ε\varepsilon. It is acceptable to choose any NN that works; you do not need the smallest such NN.
View Solution
Solution to question 1: Let ε>0\varepsilon > 0. We need ∣1n−0∣=1n<ε\left|\tfrac{1}{n} - 0\right| = \tfrac{1}{n} < \varepsilon, which requires n>1εn > \tfrac{1}{\varepsilon}. Choose N=⌈1ε⌉N = \left\lceil \tfrac{1}{\varepsilon} \right\rceil. Then for all n>Nn > N: ∣1n−0∣=1n<1N≤ε. \left|\frac{1}{n} - 0\right| = \frac{1}{n} < \frac{1}{N} \leq \varepsilon.
Solution to question 2: ∣2n+1n+3−2∣=∣2n+1−2(n+3)n+3∣=∣−5n+3∣=5n+3. \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 5n+3<ε\dfrac{5}{n+3} < \varepsilon, i.e., n>5ε−3n > \dfrac{5}{\varepsilon} - 3. Choose N=⌈5ε⌉N = \left\lceil \dfrac{5}{\varepsilon} \right\rceil. Then for all n>Nn > N: ∣2n+1n+3−2∣=5n+3<5n≤5N≤ε. \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:

  1. Prove that lim⁡n→∞n2n2+1=1\lim_{n \to \infty} \dfrac{n^2}{n^2 + 1} = 1.
  2. Prove that lim⁡n→∞n+1−n=0\lim_{n \to \infty} \sqrt{n+1} - \sqrt{n} = 0.
Hint
For (1), simplify ∣an−1∣|a_n - 1| and bound it above by a simpler expression involving only nn. For (2), rationalize the expression n+1−n\sqrt{n+1} - \sqrt{n} by multiplying and dividing by n+1+n\sqrt{n+1} + \sqrt{n}.
View Solution
Solution to question 1: ∣n2n2+1−1∣=∣−1n2+1∣=1n2+1<1n2. \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 ε>0\varepsilon > 0, choose N=⌈1ε⌉N = \left\lceil \tfrac{1}{\sqrt{\varepsilon}} \right\rceil. For n>Nn > N: ∣n2n2+1−1∣<1n2<1N2≤ε. \left|\frac{n^2}{n^2+1} - 1\right| < \frac{1}{n^2} < \frac{1}{N^2} \leq \varepsilon.
Solution to question 2: Rationalize: n+1−n=(n+1)−nn+1+n=1n+1+n<12n. \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 ε>0\varepsilon > 0, we need 12n<ε\dfrac{1}{2\sqrt{n}} < \varepsilon, i.e., n>14ε2n > \dfrac{1}{4\varepsilon^2}. Choose N=⌈14ε2⌉N = \left\lceil \dfrac{1}{4\varepsilon^2} \right\rceil. For n>Nn > N: ∣n+1−n−0∣<12n<12N≤ε. \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 (lim sup⁡\limsup) and limit inferior (lim inf⁡\liminf) extend the notion of a limit to all sequences, even divergent ones. A sequence converges if and only if lim sup⁡n→∞an=lim inf⁡n→∞an\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 lim sup⁡n→∞an\limsup_{n \to \infty} a_n and lim inf⁡n→∞an\liminf_{n \to \infty} a_n. State whether the sequence converges.

  1. an=(−1)n+1na_n = (-1)^n + \dfrac{1}{n}
  2. bn=2+(−1)n⋅nnb_n = \dfrac{2 + (-1)^n \cdot n}{n}
Hint
Recall that lim sup⁡an=lim⁡N→∞sup⁡n≥Nan\limsup a_n = \lim_{N \to \infty} \sup_{n \geq N} a_n and lim inf⁡an=lim⁡N→∞inf⁡n≥Nan\liminf a_n = \lim_{N \to \infty} \inf_{n \geq N} a_n. Separate even and odd index terms and find the supremum and infimum of the tail of each subsequence.
View Solution
Solution to question 1: For even n=2kn = 2k: a2k=1+12k→1a_{2k} = 1 + \tfrac{1}{2k} \to 1 (from above, approaching 1). For odd n=2k−1n = 2k-1: a2k−1=−1+12k−1→−1a_{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: lim sup⁡n→∞an=1,lim inf⁡n→∞an=−1. \limsup_{n \to \infty} a_n = 1, \qquad \liminf_{n \to \infty} a_n = -1. Since lim sup⁡≠lim inf⁡\limsup \neq \liminf, the sequence diverges.

Solution to question 2: Rewrite: bn=2n+(−1)nb_n = \tfrac{2}{n} + (-1)^n. For even nn: b2k=1k+1→1b_{2k} = \tfrac{1}{k} + 1 \to 1. For odd nn: b2k−1=22k−1−1→−1b_{2k-1} = \tfrac{2}{2k-1} - 1 \to -1. Therefore: lim sup⁡n→∞bn=1,lim inf⁡n→∞bn=−1. \limsup_{n \to \infty} b_n = 1, \qquad \liminf_{n \to \infty} b_n = -1. Since lim sup⁡≠lim inf⁡\limsup \neq \liminf, the sequence diverges.

Problem 15: Cauchy Sequences and the Completeness of the Reals

Hard

A sequence {an}\{a_n\} is called a Cauchy sequence if for every ε>0\varepsilon > 0 there exists N∈NN \in \mathbb{N} such that m,n>N  ⟹  ∣am−an∣<εm, n > N \implies |a_m - a_n| < \varepsilon.

  1. Prove that every convergent sequence is a Cauchy sequence.
  2. Show directly (using the Cauchy criterion) that the sequence an=1na_n = \dfrac{1}{n} is a Cauchy sequence.
Hint
For (1), use the triangle inequality: ∣am−an∣≤∣am−L∣+∣an−L∣|a_m - a_n| \leq |a_m - L| + |a_n - L|. Choose NN so that both terms are less than ε2\tfrac{\varepsilon}{2}. For (2), estimate ∣am−an∣=∣1m−1n∣|a_m - a_n| = \left|\tfrac{1}{m} - \tfrac{1}{n}\right| using the triangle inequality.
View Solution
Solution to question 1: Suppose an→La_n \to L. Let ε>0\varepsilon > 0. Since an→La_n \to L, there exists NN such that n>N  ⟹  ∣an−L∣<ε2n > N \implies |a_n - L| < \tfrac{\varepsilon}{2}. For m,n>Nm, n > N, by the triangle inequality: ∣am−an∣≤∣am−L∣+∣an−L∣<ε2+ε2=ε. |a_m - a_n| \leq |a_m - L| + |a_n - L| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. Therefore {an}\{a_n\} is a Cauchy sequence.

Solution to question 2: Let ε>0\varepsilon > 0. For m,n>Nm, n > N: ∣1m−1n∣≤1m+1n<1N+1N=2N. \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=⌈2ε⌉N = \left\lceil \tfrac{2}{\varepsilon} \right\rceil. Then 2N≤ε\tfrac{2}{N} \leq \varepsilon, so ∣1m−1n∣<ε\left|\tfrac{1}{m} - \tfrac{1}{n}\right| < \varepsilon for all m,n>Nm, n > N. Therefore {1n}\left\{\tfrac{1}{n}\right\} is a Cauchy sequence.