Real Analysis

Real sequences

A real sequence is a function mapping the set of natural numbers $\mathbb{N}$ into the set of real numbers $\mathbb{R}$, serving as the foundational stepping stone for limits, continuity, and convergence in Real Analysis.

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

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Definition and statement

Monotone Convergence Theorem (MCT) for Real Sequences

Every bounded monotone sequence of real numbers is convergent. Specifically, if $\langle x_n \rangle$ is monotonically increasing and bounded above, then \begin{equation*} \lim_{n \to \infty} x_n = \sup\{x_n : n \in \mathbb{N}\}. \end{equation*} If $\langle x_n \rangle$ is monotonically decreasing and bounded below, then \begin{equation*} \lim_{n \to \infty} x_n = \inf\{x_n : n \in \mathbb{N}\}. \end{equation*}

📐 Geometric Interpretation

Geometrically, points on the real number line move monotonically in one direction (either strictly increasing or decreasing) while being trapped by a finite upper or lower bound. Because the real number system $\mathbb{R}$ possesses the completeness property (supremum property), the sequence cannot wander indefinitely and must accumulate precisely at its least upper bound or greatest lower bound.

⚙️ Algebraic & Structural Meaning

Establishes the fundamental completeness axiom of the real number system in sequence form. It bridges order properties with topological limits, allowing the deduction of convergence without explicitly knowing or calculating the limiting value in advance.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Monotonicity (Order Property)	`Domain $\mathbb{N}$ (Discrete ordered set)`	Strictly Mandatory / Necessary	If monotonicity fails (e.g., alternating sequences like $\langle (-1)^n \rangle$), boundedness alone does not guarantee convergence; the sequence may oscillate between multiple limit points.
Boundedness (Magnitude Control)	`Range subset $[m, M] \subset \mathbb{R}$`	Strictly Mandatory	If boundedness fails (e.g., $x_n = n$), a monotonically increasing sequence diverges to infinity ($\to +\infty$) and possesses no finite limit in $\mathbb{R}$.
Completeness of the Codomain	`Codomain $\mathbb{R}$`	Essential for supremum/infimum existence	If evaluated over $\mathbb{Q}$, a bounded monotone sequence like decimal approximations of $\sqrt{2}$ would fail to converge within the space.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Algebraic Ratio Test for Sequences ($n$-th root or D'Alembert's ratio)

$$ \lim_{n \to \infty} \frac{x_{n+1}}{x_n} = L $$

Rule: If $L < 1$, then $\lim_{n \to \infty} x_n = 0$. If $L > 1$, the sequence diverges to infinity. If $L = 1$, the test fails and further analysis (like Raabe's Test) is required.

Application: Frequently asked in UP PGT and KVS to find limits of factorial-heavy or exponential terms like $x_n = \frac{n!}{n^n}$.

Cauchy's First Limit Theorem (Arithmetic Mean of Sequences)

$$ \text{If } \lim_{n \to \infty} x_n = L, \text{ then } \lim_{n \to \infty} \frac{x_1 + x_2 + \dots + x_n}{n} = L $$

Rule: The limit of the arithmetic mean of a convergent sequence is equal to the limit of the sequence itself. Useful when evaluating complex summation limits.

Application: Directly tested in DSSSB PGT and UP TGT to evaluate limits of averages such as $\frac{1}{n} \left(1 + \frac{1}{2} + \dots + \frac{1}{n}\right)$ or root-based combinations.

Cauchy's Second Limit Theorem (Geometric Mean)

$$ \text{If } x_n > 0 \text{ and } \lim_{n \to \infty} x_n = L, \text{ then } \lim_{n \to \infty} (x_1 x_2 \dots x_n)^{1/n} = L $$

Rule: The limit of the geometric mean of a positive convergent sequence matches the sequence limit.

Sandwich / Squeeze Theorem for Sequences

$$ \text{If } a_n \le x_n \le b_n \text{ and } \lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n = L, \text{ then } \lim_{n \to \infty} x_n = L $$

Rule: Squeeze complex fractional terms between two simpler geometric or power functions whose limits are easily evaluated as $0$ or a constant.

Application: Extensively used for evaluating limits involving $\frac{1}{n^p}$ sums or sine/cosine bounds.

Standard Growth Rate Hierarchy

$$ \ln n \ll n^p \ll a^n \ll n! \ll n^n \quad (a > 1, p > 0) $$

Rule: In any rational combination of these standard sequences as $n \to \infty$, the faster-growing term dominates. Numerator dominates $\implies \infty$; Denominator dominates $\implies 0$.

Application: Saves 45 seconds on polynomial-exponential limit problems in UP PGT.

Rapid Elimination via Subsequences

$$ \text{If } \langle x_{2n} \rangle \text{ and } \langle x_{2n-1} \rangle \text{ converge to distinct limits, } \langle x_n \rangle \text{ diverges.} $$

Rule: Instantly check even and odd terms. If they yield different limits, discard all options claiming convergence.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Boundedness Does Not Imply Convergence

Common Pitfall: Candidates assume that because a sequence remains within fixed bounds, it must converge to a specific single value.

Counterexample: $x_n = (-1)^n$

Why it fails: The sequence is bounded between $-1$ and $1$, but it oscillates indefinitely between $-1$ and $1$, failing to possess a unique limit.

🛡️ Exam Defense: Remember: Boundedness + Monotonicity = Convergence. Boundedness alone only guarantees the existence of at least one limit point (Bolzano-Weierstrass Theorem), not convergence.

⚠️ COMMON ERROR #02

Converse of Cauchy's First Theorem Failure

Common Pitfall: Assuming that if the arithmetic mean $\frac{1}{n}\sum x_n$ converges, the underlying sequence $\langle x_n \rangle$ must also converge.

Counterexample: $x_n = (-1)^n$

Why it fails: The arithmetic mean limit is $0$, but the sequence $\langle (-1)^n \rangle$ diverges.

🛡️ Exam Defense: The implication is strictly one-way: $x_n \to L \implies \text{Mean} \to L$. The converse is false.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Monotone Convergence Theorem	Sequence is monotone and bounded	$\lim x_n = \sup x_n$ or $\inf x_n$	One-directional flow trapped by a barrier	Proving convergence of recursive/iterative sequences
Cauchy's General Principle of Convergence	Metric space completeness of $\mathbb{R}$	$\|x_m - x_n\| < \varepsilon \quad \forall m, n \ge N$	Terms eventually cluster arbitrarily close to each other	Proving convergence without knowing the limit
Bolzano-Weierstrass Theorem	Sequence is bounded	Contains a convergent subsequence	Infinite points in a finite interval cluster somewhere	Identifying limit points and cluster sets
Cesaro's Theorem (Cauchy's 1st)	Sequence $\langle x_n \rangle$ converges to $L$	$\lim \frac{1}{n}\sum x_i = L$	Averaging smooths out fluctuations to the true limit	Evaluating limits of large sums divided by $n$

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

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

(A) $1$(B) $2$(C) $\sqrt{2}$(D) Does not exist

📚 Method A: Formal Textbook Derivation

Step 1: Show $x_n < 2$ by mathematical induction. Step 2: Show $x_{n+1} > x_n$ ($x_n$ is monotonically increasing). Step 3: By MCT, limit $L$ exists. Step 4: Solve $L = \sqrt{2 + L} \implies L^2 - L - 2 = 0 \implies L = 2$ (since $L > 0$).

⚡ Method B: 25-Second Exam Shortcut

Set $x_{n+1} = x_n = L$ directly in the recursive relation: $L = \sqrt{2 + L} \implies L^2 - L - 2 = 0 \implies L = 2$.

💡 Tip: For recursive relations of type $x_{n+1} = \sqrt{a + x_n}$, the positive root of $L^2 - L - a = 0$ is always the answer if $x_1 > 0$.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Determine the limit of the sequence $\langle x_n \rangle$ where $x_n = \left(1 + \frac{1}{n}\sup\right)^n$.

(A) $1$(B) $e$(C) $0$(D) $\infty$

📚 Method A: Formal Textbook Derivation

Expand using the Binomial Theorem for positive integers and apply the Monotone Convergence Theorem to prove boundedness and monotonicity, establishing the standard definition of Euler's constant $e$.

⚡ Method B: 25-Second Exam Shortcut

Direct standard definition memorization from real analysis: $\lim_{n \to \infty} (1 + \frac{1}{n})^n = e$.

💡 Tip: Direct standard result often repeated verbatim in UP PGT exams.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: Which of the following sequences is convergent?

A $x_n = (-1)^n n$

B $x_n = \frac{1}{n} \sin(n \frac{\pi}{2})$

C $x_n = n^{1/n}$

D $x_n = 1 + (-1)^n$

Derivation & Analysis:

By the Squeeze Theorem, $-\frac{1}{n} \le \frac{1}{n}\sin(\frac{n\pi}{2}) \le \frac{1}{n}$. Since limits of bounds are $0$, the sequence converges to $0$. Option A diverges to $\infty$ in magnitude, C converges to $1$, D oscillates between $0$ and $2$.

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

A Monotone increasing and divergent

B Monotone increasing and convergent

C Monotone decreasing and convergent

D Not monotonic

Derivation & Analysis:

Computing $x_{n+1} - x_n = \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} = \frac{1}{2n+1} - \frac{1}{2n+2} > 0$, so it is monotonically increasing. It is bounded above by $1$. By MCT, it converges.

MCQ 3: If $x_n = \frac{n!}{n^n}$, then $\lim_{n \to \infty} x_n$ is equal to:

A $1$

B $e$

C $0$

D $\infty$

Derivation & Analysis:

Using D'Alembert's ratio test for sequences: $\frac{x_{n+1}}{x_n} = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} = \left(\frac{n}{n+1}\right)^n = \frac{1}{(1 + \frac{1}{n})^n} \to \frac{1}{e} < 1$. Thus, $\lim x_n = 0$.

MCQ 4: Every bounded sequence has:

A A unique limit

B At least one limit point

C No limit points

D Only rational limit points

Derivation & Analysis:

This is the direct statement of the Bolzano-Weierstrass Theorem for sequences in $\mathbb{R}$.

MCQ 5: The limit $\lim_{n \to \infty} \left( \frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + \dots + \frac{1}{\sqrt{n^2+n}} \right)$ is:

A $0$

B $1$

C $1/2$

D Does not exist

Derivation & Analysis:

Bounding the sum: $n \cdot \frac{1}{\sqrt{n^2+n}} \le x_n \le n \cdot \frac{1}{\sqrt{n^2+1}}$. Taking limits as $n \to \infty$, both lower and upper bounds approach $1$. By Squeeze Theorem, the limit is $1$.

MCQ 6: If $x_1 = 1$ and $x_{n+1} = \frac{1}{3}(x_n + 1)$, then $\lim_{n \to \infty} x_n$ is:

A $1/2$

B $1/3$

C $1$

D $0$

Derivation & Analysis:

Set $L = \frac{1}{3}(L + 1) \implies 3L = L + 1 \implies 2L = 1 \implies L = 1/2$.

MCQ 7: Which of the following is a Cauchy sequence?

A $x_n = \sum_{k=1}^n \frac{1}{k}$

B $x_n = \sum_{k=1}^n \frac{(-1)^k}{k}$

C $x_n = \ln n$

D $x_n = (-1)^n n$

Derivation & Analysis:

Since $\mathbb{R}$ is complete, a sequence is Cauchy if and only if it converges. The alternating harmonic series partial sums form a convergent sequence, hence a Cauchy sequence. Harmonic series (A) diverges.

MCQ 8: If $\lim_{n \to \infty} x_n = L$, then $\lim_{n \to \infty} x_{n+k}$ (where $k$ is a fixed positive integer) is:

A $L + k$

B $L$

C $kL$

D Undefined

Derivation & Analysis:

Shifting finite indices does not alter the tail limit of a convergent sequence.

MCQ 9: The sequence $x_n = \frac{\sin n}{n}$ satisfies:

A Converges to $1$

B Converges to $0$

C Diverges

D Oscillates finitely

Derivation & Analysis:

Since $-1 \le \sin n \le 1$, dividing by $n > 0$ gives $-\frac{1}{n} \le \frac{\sin n}{n} \le \frac{1}{n}$. By Squeeze Theorem, limit is $0$.

Q10

MCQ 10: Let $x_n = (-1)^n \left(1 + \frac{1}{n}\right)$. The upper and lower limits (limsup and liminf) of $\langle x_n \rangle$ are respectively:

A $1$ and $-1$

B $-1$ and $1$

C $+\infty$ and $-\infty$

D $0$ and $0$

Derivation & Analysis:

The even terms approach $1$, and odd terms approach $-1$. Thus, $\limsup x_n = 1$ and $\liminf x_n = -1$.

Common questions

No. Boundedness alone only ensures that the sequence has at least one convergent subsequence (Bolzano-Weierstrass Theorem). For full convergence, the sequence must either be monotone or satisfy the Cauchy criterion.

Assuming that applying algebraic operations on divergent limits (like $\infty - \infty$ or $1^\infty$) works like ordinary arithmetic, leading to indeterminate forms without applying L'Hopital's rule or standard logarithmic transformations.

Instead of proving monotonicity and boundedness formally during the exam, assume convergence to $L$, substitute $L$ for $x_{n+1}$ and $x_n$ in the recursive equation, and solve the resulting algebraic equation.

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