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Real Analysis

Convergent sequences

A sequence of real numbers is said to be convergent if it approaches a finite real limit as the index tends to infinity, forming the bedrock of modern analysis. This topic covers convergence tests, Cauchy's criterion, and monotone convergence theorems.

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Definition and statement

Definition of Convergence & Cauchy's General Principle of Convergence
A sequence $\langle a_n \rangle$ is said to converge to a real number $l$ if for every $\epsilon > 0$, there exists a positive integer $m = m(\epsilon)$ such that $|a_n - l| < \epsilon$ for all $n \ge m$. Symbolically, $\lim_{n \to \infty} a_n = l$. Furthermore, $\langle a_n \rangle$ converges if and only if for every $\epsilon > 0$, there exists $m \in \mathbb{N}$ such that $|a_m - a_n| < \epsilon$ for all $m, n \ge \nu(\epsilon)$.

📐 Geometric Interpretation

Geometrically, for any chosen arbitrarily small open interval $(l-\epsilon, l+\epsilon)$ around the limit $l$, all terms of the sequence $\langle a_n \rangle$ from some index onward lie permanently inside this neighborhood, with at most a finite number of exceptions lying outside.

⚙️ Algebraic & Structural Meaning

Algebraically, convergence transforms infinite limiting processes into manageable algebraic limit laws (sum, product, quotient rules), enabling the rigorous treatment of infinite series, power series, and continuity.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Boundedness for Monotone Sequences Domain of index set $\mathbb{N}$ Strictly Mandatory / Necessary A monotonic sequence that is not bounded diverges to $\pm \infty$ (e.g., $a_n = n$).
Cauchy Completeness of $\mathbb{R}$ Metric space $\mathbb{R}$ Strictly Mandatory Incomplete spaces (like $\mathbb{Q}$) contain Cauchy sequences that fail to converge within the space (e.g., sequence of rational approximations to $\sqrt{2}$).
Squeeze (Sandwich) Inequality Index inequality $a_n \le b_n \le c_n$ Essential for conclusion If bounding sequences do not converge to the exact same limit $l$, the squeeze theorem fails to apply.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Cesaro-Stolz Theorem (L'Hopital for Sequences)

$$ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} $$

Rule: If $\langle b_n \rangle$ is a strictly monotonic divergent sequence, replace indeterminate sequence ratios with consecutive differences.

Application: Use this result in the stated type of problem.
02

Ratio Test for Sequences ($n$-th root / D'Alembert)

$$ \text{If } \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = l < 1, \text{ then } \lim_{n \to \infty} a_n = 0. $$

Rule: Use when factorial or exponential terms are present in the numerator and denominator.

03

Standard Algebraic Limits

$$ \lim_{n \to \infty} n^{1/n} = 1, \quad \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = e^x $$

Rule: Direct substitution of standard asymptotic growth rates. Exponential beats polynomial; factorial beats exponential.

Application: Directly appears as 1-mark objective questions in UP TGT/PGT.
04

Monotone Bounded Convergence

$$ \text{If } a_{n+1} \ge a_n \text{ and } a_n \le M \implies \lim a_n = \sup(a_n) $$

Rule: For recursively defined sequences $x_{n+1} = f(x_n)$, equate $x_{n+1} = x_n = L$ and solve the algebraic equation $f(L) = L$.

05

Cauchy's First Theorem on Limits (Arithmetic Mean)

$$ \text{If } \lim_{n \to \infty} a_n = l, \text{ then } \lim_{n \to \infty} \frac{a_1 + a_2 + \dots + a_n}{n} = l. $$

Rule: The limit of the arithmetic mean equals the limit of the sequence itself.

06

Rapid Elimination via Boundedness

$$ \text{Unbounded } \implies \text{ Divergent} $$

Rule: If any subsequence tends to infinity or oscillates unboundedly, eliminate 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: Assuming every bounded sequence is convergent without checking monotonicity.

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

Why it fails: The sequence is bounded ($|a_n| \le 1$), but it oscillates between $-1$ and $1$, failing to converge.

🛡️ Exam Defense: Boundedness is necessary for convergence, but only sufficient when combined with monotonicity.
⚠️ COMMON ERROR #02

Algebraic Limit Law Violation on Infinity

Common Pitfall: Applying limit laws like $\lim(a_n + b_n) = \lim a_n + \lim b_n$ when individual limits do not exist or yield $\infty - \infty$.

Counterexample: $a_n = n$, $b_n = -n + \frac{1}{n}$

Why it fails: Individual sequences diverge or lead to indeterminate forms if combined improperly.

🛡️ Exam Defense: Always ensure individual sequences are convergent before applying sum/product/quotient limit theorems.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Monotone Convergence TheoremMonotone and BoundedLimit equals Supremum or InfimumA curve moving strictly in one direction hitting a ceiling/floorRecursive sequences $x_{n+1} = f(x_n)$
Cauchy's Convergence CriterionMetric space completeness$|a_m - a_n| < \epsilon$ for large $m,n$Terms eventually cluster arbitrarily close togetherProving convergence without knowing the limit
Sandwich (Squeeze) Theorem$a_n \le b_n \le c_n$ and $\lim a_n = \lim c_n = l$$\lim b_n = l$A function trapped between two converging envelopesTrigonometric ratios divided by $n$ or powers
Cesaro-Stolz TheoremMonotonic divergent denominatorRatio of differences equals ratio of termsRatio of accumulated sums matches instantaneous ratioSummations divided by $n^k$

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Evaluate the limit of the recursive sequence given 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) \infty
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: Prove boundedness $x_n < 2$ by induction. Step 2: Prove monotonicity $x_{n+1} > x_n$. Step 3: Let $\lim x_n = L$. Step 4: Solve $L = \sqrt{2 + L} \implies L^2 - L - 2 = 0 \implies L = 2$ (rejecting negative root).
⚡ Method B: 25-Second Exam Shortcut
Directly put $x_{n+1} = x_n = L$ in $L = \sqrt{2 + L}$ and solve quadratic equation in 5 seconds.
💡 Tip: Classic UP PGT repeat question. Always check positive branch.
CANONICAL PROBLEM 02 Dual-Method Breakdown

Determine the limit $\lim_{n \to \infty} \left( \frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + \dots + \frac{1}{\sqrt{n^2+n}} \right).$

(A) 0(B) 1(C) \frac{1}{2}(D) Does not exist
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Use sandwich theorem. Each term is bounded between $\frac{n}{\sqrt{n^2+n}}$ and $\frac{n}{\sqrt{n^2+1}}$. Taking limits as $n \to \infty$ yields 1 for both bounds.
⚡ Method B: 25-Second Exam Shortcut
Count number of terms ($n$), multiply by minimum/maximum term, and evaluate limit.
💡 Tip: Look out for summation of $n$ terms where each term depends on $n$.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: The sequence $a_n = \frac{n!}{n^n}$ is:
A Convergent to 0
B Convergent to 1
C Divergent to $\infty$
D Oscillatory
Derivation & Analysis:

Using the ratio test: $\frac{a_{n+1}}{a_n} = \left(\frac{n}{n+1}\right)^n = \frac{1}{(1 + 1/n)^n} \to \frac{1}{e} < 1$. Thus, $\lim a_n = 0$.

Q2
MCQ 2: The sequence $x_n = 1 + \frac{1}{1!} + \frac{1}{2!} + \dots + \frac{1}{n!}$ converges to:
A 1
B e
C e-1
D $\pi$
Derivation & Analysis:

This is the standard definition of Euler's number $e$ as the limit of the partial sums of the exponential series.

Q3
MCQ 3: Which of the following sequences is convergent?
A $a_n = (-1)^n n$
B $b_n = \sin\left(\frac{n\pi}{2}\right)$
C $c_n = \frac{2n + 1}{n + 3}$
D $d_n = \ln(n)$
Derivation & Analysis:

For $c_n = \frac{2n+1}{n+3} = \frac{2 + 1/n}{1 + 3/n} \to 2$. All others are unbounded or oscillate.

Q4
MCQ 4: If $\lim_{n \to \infty} a_n = L$, then $\lim_{n \to \infty} a_{2n}$ is:
A L/2
B 2L
C L
D Does not exist
Derivation & Analysis:

Every subsequence of a convergent sequence converges to the same limit $L$.

Q5
MCQ 5: The sequence $a_n = \frac{sin(n)}{n}$ satisfies:
A $\lim a_n = 1$
B $\lim a_n = \infty$
C $\lim a_n = 0$
D Does not exist
Derivation & Analysis:

Since $-1 \le \sin(n) \le 1$, we have $-\frac{1}{n} \le \frac{\sin(n)}{n} \le \frac{1}{n}$. By Squeeze Theorem, limit is 0.

Q6
MCQ 6: Every bounded monotonic sequence is:
A Divergent
B Convergent
C Periodic
D Oscillatory
Derivation & Analysis:

This is the Monotone Convergence Theorem, a fundamental axiom equivalent to the completeness of $\mathbb{R}$.

Q7
MCQ 7: The value of $\lim_{n \to \infty} n^{1/n}$ is:
A 0
B 1
C e
D $\infty$
Derivation & Analysis:

Standard limit result proven using logarithmic forms or the binomial theorem: $\lim n^{1/n} = 1$.

Q8
MCQ 8: Assertion (A): Every Cauchy sequence in $\mathbb{R}$ is bounded. Reason (R): Convergent sequences are bounded.
A Both (A) and (R) are true and (R) is the correct explanation of (A)
B Both (A) and (R) are true but (R) is not the correct explanation
C (A) is true but (R) is false
D (A) is false but (R) is true
Derivation & Analysis:

Every Cauchy sequence is convergent in $\mathbb{R}$ (completeness), and every convergent sequence is bounded. Hence both are true and linked via convergence.

Q9
MCQ 9: The limit $\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{2n}$ equals:
A e
B $e^2$
C $\sqrt{e}$
D 1
Derivation & Analysis:

$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{2n} = \left[ \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \right]^2 = e^2$.

Q10
MCQ 10: If $0 < y_1 < x_1$, $x_{n+1} = \frac{x_n + y_n}{2}$ and $y_{n+1} = \sqrt{x_n y_n}$, then sequences $\langle x_n \rangle$ and $\langle y_n \rangle$:
A Diverge to infinity
B Converge to different limits
C Converge to the same limit (AGM)
D Oscillate finitely
Derivation & Analysis:

These are arithmetic-geometric mean sequences. Both converge to the same common limit known as the Arithmetic-Geometric Mean.

Common questions

No. By the uniqueness of limits in Hausdorff spaces (and specifically in metric spaces like $\mathbb{R}$), a convergent sequence can have only one unique limit.

Failing to check whether a sequence is both bounded AND monotonic when applying the Monotone Convergence Theorem, or incorrectly assuming boundedness guarantees convergence.

Memorize standard limits (like $n^{1/n}$, $(1+1/n)^n$, ratio test results for factorials) and utilize the algebraic substitution trick $x_{n+1} = x_n = L$ for recursive sequence limits.

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Divergent sequences