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Real Analysis

Bounded sequences

A sequence \langle x_n \rangle in \mathbb{R} is bounded if its range set is bounded above and below, meaning both supremum and infimum exist finitely.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade
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Definition and statement

Definition of Bounded Sequence & Boundedness Criterion
A sequence \langle x_n \rangle is said to be bounded if \exists M \in \mathbb{R}, M > 0 such that |x_n| \le M \ \forall n \in \mathbb{N}. Equivalently, \langle x_n \rangle is bounded above if \exists K \in \mathbb{R} such that x_n \le K \ \forall n, and bounded below if \exists k \in \mathbb{R} such that x_n \ge k \ \forall n.

📐 Geometric Interpretation

Geometrically, all terms of a bounded sequence lie entirely within a finite closed interval $[-M, M]$ on the real number line, preventing terms from escaping to infinity or minus infinity.

⚙️ Algebraic & Structural Meaning

Boundedness is a necessary (though not sufficient) condition for convergence, and forms the structural bedrock for the Bolzano-Weierstrass theorem, Cauchy's general principle of convergence, and Supremum/Infimum properties of \mathbb{R}.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Range Set Boundedness Subset of \mathbb{R} Strictly Mandatory / Necessary If the range set fails to be bounded above or below, the sequence diverges to \pm\infty or oscillates infinitely without bounds (e.g., \langle n \rangle or \langle (-1)^n n \rangle).
Domain of Sequence \mathbb{N} (Natural Numbers) Strictly Mandatory Sequences must have domain \mathbb{N}; failing to index over discrete natural steps invalidates standard limit definitions as n \to \infty.
Field Properties of \mathbb{R} Archimedean Ordered Field Essential for conclusion In non-complete ordered fields (like \mathbb{Q}), bounded sets may fail to have suprema or infima within the field, breaking completeness arguments.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Monotone Bounded Test

$$ \text{If } \langle x_n \rangle \text{ is monotonic and bounded} \implies \lim_{n \to \infty} x_n \text{ exists (converges).} $$

Rule: Never compute limits explicitly for recursive sequences like $x_{n+1} = \sqrt{2 + x_n}$; just prove monotonicity and boundedness to guarantee convergence, then solve $L = \sqrt{2 + L}$.

Application: undefined
02

Sandwich (Squeeze) Theorem for Bounding

$$ \text{If } a_n \le x_n \le b_n \text{ and } \lim a_n = \lim b_n = L \implies \lim x_n = L $$

Rule: For sequences of the form \sum_{i=1}^n \frac{1}{n^2 + i}, bound terms by replacing index i with 1 and n to establish limits instantly.

Application: Frequent in DSSSB and KVS PGT for series and sequence bounds.
03

Algebraic Growth Order Rule

$$ \lim_{n \to \infty} \frac{n^k}{a^n} = 0 \quad (a > 1), \quad \lim_{n \to \infty} \frac{\log n}{n^k} = 0 \quad (k > 0) $$

Rule: Exponential growth strictly dominates polynomial growth, which strictly dominates logarithmic growth when determining sequence bounds and limits.

Application: Instant elimination tool for options in limit-evaluation questions.
04

Cauchy's First Theorem on Limits

$$ \text{If } \lim x_n = L \implies \lim \frac{x_1 + x_2 + \dots + x_n}{n} = L $$

Rule: Use arithmetic mean limits when direct evaluation of $x_n$ is impossible or indeterminate.

Application: Standard recurring question type in UP PGT Mathematics.
05

Boundedness of Trigonometric Sequences

$$ -1 \le \sin(n), \cos(n) \le 1 \quad \forall n \in \mathbb{N} $$

Rule: Any sequence multiplied by bounded periodic terms \sin(n\theta) or \cos(n\theta) shares the same limit behavior as the envelope if the base term goes to 0.

Application: Quick check for convergence of oscillating sequences like $\frac{\sin n}{n}$. '0'.
06

Rapid Elimination via Subsequences

$$ \text{If } \langle x_n \rangle \text{ has two convergent subsequences converging to different limits} \implies \text{Unbounded or Divergent} $$

Rule: Check alternating or modular subsequences to instantly reject convergence claims.

Application: Used in statement-type questions in PGT screening tests.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Boundedness Implies Convergence Fallacy

Common Pitfall: Candidates assume that because a sequence is bounded, it must converge.

Counterexample: x_n = (-1)^n

Why it fails: The sequence is bounded ($-1 \le (-1)^n \le 1$), but it oscillates and has no limit.

🛡️ Exam Defense: Boundedness is necessary, not sufficient for convergence. Monotonicity is required.
⚠️ COMMON ERROR #02

Cauchy Sequence vs. Bounded Sequence Confusion

Common Pitfall: Assuming bounded sequences are always Cauchy.

Counterexample: x_n = (-1)^n

Why it fails: While every Cauchy sequence is bounded, the converse is false; bounded sequences can fail the Cauchy criterion.

🛡️ Exam Defense: Always verify distance condition $|x_m - x_n| < \epsilon$ for Cauchy property.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Bounded SequenceRange set bounded above and below$|x_n| \le M$Contained in finite interval $[-M, M]$Terms do not blow up to infinity
Monotone ConvergenceMonotonic + BoundedLimit exists ($L = \sup$ or $\inf$)One-directional motion settling at a boundaryRecursive relations ($x_{n+1} = f(x_n)$)
Bolzano-WeierstrassEvery bounded sequenceHas at least one convergent subsequenceAccumulation points exist within boundsKeywords: 'Subsequence', 'Bounded'
Cauchy Sequence$|x_m - x_n| < \epsilon$ for $m, n \ge N_0$Sequence converges in complete spacesTerms get arbitrarily close to each otherCauchy criterion for convergence

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Determine the boundedness and convergence of the sequence defined recursively by $x_1 = \sqrt{2}$ and $x_{n+1} = \sqrt{2 + x_n}$ for $n \ge 1$.

(A) Unbounded and divergent(B) Bounded and converges to 2(C) Bounded and converges to \sqrt{2}(D) Oscillatory
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: Show $x_n < 2$ by induction. Step 2: Show $x_{n+1} > x_n$ (monotonic increasing). Step 3: By Monotone Convergence Theorem, limit $L$ exists. Step 4: Solve $L = \sqrt{2 + L} \implies L^2 - L - 2 = 0 \implies L = 2$.
⚡ Method B: 25-Second Exam Shortcut
Instantly set $x_n = L = x_{n+1}$, solve $L = \sqrt{2+L} \implies L=2$ (rejecting negative root since terms are positive).
💡 Tip: For recursive square-root roots of the form $\sqrt{a + \sqrt{a + \dots}}$, the limit is always the positive root of $L^2 - L - a = 0$.
CANONICAL PROBLEM 02 Dual-Method Breakdown

The sequence $x_n = 1 + \frac{1}{1!} + \frac{1}{2!} + \dots + \frac{1}{n!}$ is:

(A) Bounded and divergent(B) Unbounded(C) Bounded and convergent to e(D) Oscillatory
✓ Correct Answer: (C)
📚 Method A: Formal Textbook Derivation
Step 1: Note that $x_{n+1} > x_n$ (strictly increasing). Step 2: Bound terms using geometric series: $1/2^k$. Step 3: Sequence is bounded above by 3. Step 4: By MCT, converges to e.
⚡ Method B: 25-Second Exam Shortcut
Recognize standard definition of Euler's number $e = \sum_{n=0}^{\infty} \frac{1}{n!}$, which is famously bounded and converges to $e \approx 2.718$.
💡 Tip: Memorize standard expansions for $e$ and $e^x$ to spot bounded convergent sums instantly.

Practice questions

Choose an answer to check it and see the solution:

Q1
Every convergent sequence is:
A Bounded
B Unbounded
C Monotone
D Geometric
Derivation & Analysis:

Every convergent sequence has a finite limit, and terms beyond a certain point stay within an $\epsilon$-neighborhood, making the sequence bounded.

Q2
The sequence $x_n = (-1)^n n$ is:
A Bounded above
B Bounded below
C Bounded
D Unbounded
Derivation & Analysis:

As $n \to \infty$, the terms alternate in sign with increasing magnitude, making the sequence completely unbounded.

Q3
Which of the following sequences is monotonic and bounded?
A $x_n = (-1)^n$
B $x_n = n$
C $x_n = 1 - \frac{1}{n}$
D $x_n = \sin(n)$
Derivation & Analysis:

$x_n = 1 - \frac{1}{n}$ is strictly increasing ($x_{n+1} > x_n$) and bounded above by 1.

Q4
The limit of the sequence $x_n = \frac{n}{n+1}$ is:
A 0
B 1
C $\infty$
D Does not exist
Derivation & Analysis:

Dividing numerator and denominator by $n$ gives $\frac{1}{1 + 1/n} \to 1$ as $n \to \infty$.

Q5
Bolzano-Weierstrass theorem for sequences states that every bounded sequence has:
A A unique limit
B At least one convergent subsequence
C No accumulation points
D Only finite terms
Derivation & Analysis:

This is the classic statement of the Bolzano-Weierstrass theorem in real analysis.

Q6
If $x_n > 0$ and $\lim_{n \to \infty} \frac{x_{n+1}}{x_n} = l < 1$, then $\lim x_n$ is:
A 1
B $\infty$
C 0
D Does not exist
Derivation & Analysis:

By D'Alembert's ratio test for sequences, if the limit of the ratio is less than 1, the sequence converges to 0.

Q7
The sequence $x_n = \sin\left(\frac{n\pi}{2}\right)$ is:
A Convergent
B Bounded but not convergent
C Unbounded
D Strictly increasing
Derivation & Analysis:

Terms cycle through $1, 0, -1, 0$, remaining bounded between $-1$ and $1$, but oscillating without settling at a limit.

Q8
A Cauchy sequence in the set of real numbers $\mathbb{R}$ is always:
A Divergent
B Unbounded
C Convergent
D Oscillatory
Derivation & Analysis:

Because $\mathbb{R}$ is complete, every Cauchy sequence of real numbers converges in $\mathbb{R}$.

Q9
If $a_n \le b_n \le c_n$ and $\lim a_n = 2, \lim c_n = 2$, then $\lim b_n$ is:
A 0
B 1
C 2
D Undefined
Derivation & Analysis:

By the Squeeze Theorem (Sandwich Theorem), $\lim b_n$ must equal 2.

Q10
Which of the following is a bounded sequence?
A $x_n = \log n$
B $x_n = 2^n$
C $x_n = \arctan(n)$
D $x_n = n^2$
Derivation & Analysis:

The range of $\arctan(n)$ is strictly contained within $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, making it bounded.

Common questions

No. Boundedness is necessary but not sufficient. For example, $x_n = (-1)^n$ is bounded between -1 and 1, but it oscillates and does not converge.

Use Mathematical Induction. Assume $x_n < K$ for some upper bound $K$, and prove that $x_{n+1} < K$ holds true using the recursive relation.

Yes! For instance, $x_n = (-1)^n$ is bounded and has two subsequential limits: 1 and -1.

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