Real Analysis

Cauchy's theorem on limits

Cauchy's theorem on limits establishes that if a sequence \{a_n\} converges to a limit $L$, then its arithmetic mean sequence also converges to $L$.

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

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Definition and statement

Cauchy's First Theorem on Limits

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

📐 Geometric Interpretation

Visually, as the terms of a convergent sequence cluster infinitely close to the limiting value $l$, their running arithmetic average is pulled toward the same limit $l$, smoothing out individual fluctuations.

⚙️ Algebraic & Structural Meaning

It provides a powerful analytical tool to evaluate limits of complex summation expressions involving sequences, converting difficult Cesàro sums into straightforward limit computations of the underlying sequence.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Existence of Sequence Limit	`N/A`	Strictly Mandatory / Necessary	If $\lim_{n \to \infty} a_n$ does not exist (e.g., $a_n = (-1)^n$), the Cesàro mean may still exist or oscillate, meaning the converse of Cauchy's theorem is false.
Finiteness of Limit $l$	`N/A`	Strictly Mandatory	If $l$ is infinite, specialized extensions apply, but standard formulation assumes a finite limit or requires careful adaptation for improper limits.
Summation Denominator Growth	`N/A`	Essential for conclusion	If the divisor does not grow unboundedly, the average limit ratio is undefined or indeterminate.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Cesàro Mean for $n^{-p} \sum k^{p-1}$

$$ \lim_{n \to \infty} \frac{1}{n^{p}} \sum_{k=1}^{n} k^{p-1} = \frac{1}{p} $$

Rule: For sums of powers divided by the next higher power, the limit is simply the reciprocal of the exponent $p$.

Geometric Sequence Average

$$ \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} r^k = 0 \quad (\text{for } |r| < 1) $$

Rule: The arithmetic mean of any geometric progression with common ratio strictly less than 1 vanishes in the limit.

Application: Frequently used in DSSSB PGT Mathematics to quickly evaluate infinite series limits.

Logarithmic Limit Conversion

$$ \lim_{n \to \infty} (n!)^{1/n} = \lim_{n \to \infty} \frac{n}{e} $$

Rule: Utilize Cauchy's second theorem on limits ($\lim \frac{a_{n+1}}{a_n}$) to evaluate factorial-based root limits instantly.

Bounded Sequence with Null Factor

$$ \lim_{n \to \infty} \frac{a_1 + a_2 + \dots + a_n}{n} = l \text{ if } a_n \to l $$

Rule: Do not attempt $\epsilon-N$ proofs in exam; substitute the asymptotic value directly into the general term if convergence is established.

Stolz-Cesàro Theorem Analogue

$$ \lim_{n \to \infty} \frac{x_n}{y_n} = \lim_{n \to \infty} \frac{x_n - x_{n-1}}{y_n - y_{n-1}} $$

Rule: For indeterminate forms $\frac{\infty}{\infty}$ with monotonic sequences, take differences instead of derivatives (discrete L'Hôpital's rule).

Application: Highly favored in premium PGT screening tests for ratio limit evaluations.

Rapid Elimination via Divergent Counterpart

$$ a_n = (-1)^n \implies \text{limit D.N.E., but mean } \to 0 $$

Rule: If the sequence oscillates, check if the mean converges; if the question assumes converse, eliminate immediately.

Application: Instant option elimination in assertion-reasoning items.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Assuming the Converse is True

Common Pitfall: Candidates mistakenly assume that if the arithmetic mean $\frac{1}{n}\sum a_n$ converges to $l$, then the sequence $a_n$ must also converge to $l$.

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

Why it fails: The sequence $a_n = (-1)^n$ oscillates between $-1$ and $1$ and has no limit, yet its arithmetic mean sequence converges to $0$.

🛡️ Exam Defense: Never apply Cauchy's theorem in reverse; the convergence of the mean does not imply the convergence of the sequence.

⚠️ COMMON ERROR #02

Misidentifying Cauchy's First and Second Theorems

Common Pitfall: Confusing Cauchy's First Theorem (arithmetic mean) with Cauchy's Second Theorem (geometric mean: if $a_n > 0$ and $\lim \frac{a_{n+1}}{a_n} = l$, then $\lim (a_n)^{1/n} = l$).

Counterexample: $a_n = n$

Why it fails: Applying arithmetic mean rules to geometric mean problems leads to false limit valuations.

🛡️ Exam Defense: Check whether the expression involves a sum (First Theorem) or a product/radical (Second Theorem).

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Cauchy's First Theorem	\lim a_n = l	\lim \frac{\sum a_k}{n} = l	Running average converges to sequence limit	Sum of sequence terms divided by n
Cauchy's Second Theorem	a_n > 0, \lim \frac{a_{n+1}}{a_n} = l	\lim (a_n)^{1/n} = l	Geometric mean tracks ratio limit	nth root of product or factorial
Stolz-Cesàro Theorem	y_n is strictly monotonic and \to \infty	\lim \frac{x_n}{y_n} = \lim \frac{x_n - x_{n-1}}{y_n - y_{n-1}}	Discrete L'Hôpital's rule for sequences	Ratio of two divergent summation series
Cesàro-Stolz Mean	\lim a_n = l	Weighted arithmetic average \to l	Weighted smoothing of limits	Weighted sequence sums

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

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

(A) $0$(B) $\log 2$(C) $1$(D) $\infty$

📚 Method A: Formal Textbook Derivation

Rewrite the expression as a Riemann sum: $\lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{1}{1 + r/n} = \int_{0}^{1} \frac{1}{1+x} \, dx = [\log(1+x)]_0^1 = \log 2$. Alternatively, apply Cauchy-Stolz theorem.

⚡ Method B: 25-Second Exam Shortcut

Recognize standard definite integral limit form $\frac{1}{n}\sum f(r/n) \to \int_0^1 f(x)dx$.

💡 Tip: A absolute classic in UP PGT exams. Memorize the result $\log 2$ for this exact series.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If $a_n = \frac{1}{\sqrt{n}} \left( 1 + \frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}} \right)$, find $\lim_{n \to \infty} a_n$.

(A) $0$(B) $1$(C) $2$(D) Does not exist

📚 Method A: Formal Textbook Derivation

Apply Stolz-Cesàro theorem. Let $x_n = \sum_{k=1}^n \frac{1}{\sqrt{k}}$ and $y_n = \sqrt{n}$. Then $\frac{x_n - x_{n-1}}{y_n - y_{n-1}} = \frac{1/\sqrt{n}}{\sqrt{n} - \sqrt{n-1}} = \frac{1/\sqrt{n}}{\frac{1}{\sqrt{n} + \sqrt{n-1}}} = \frac{\sqrt{n} + \sqrt{n-1}}{\sqrt{n}} = 1 + \sqrt{1 - 1/n} \to 2$.

⚡ Method B: 25-Second Exam Shortcut

Use the asymptotic expansion of harmonic/radical sums where $\sum_{k=1}^n \frac{1}{\sqrt{k}} \sim 2\sqrt{n}$, yielding $\frac{2\sqrt{n}}{\sqrt{n}} = 2$.

💡 Tip: Whenever denominator is $\sqrt{n}$ and numerator is radical sum, Stolz-Cesàro resolves it instantly.

Practice questions

Choose an answer to check it and see the solution:

If $\lim_{n \to \infty} a_n = l$, then $\lim_{n \to \infty} \frac{a_1 + 2a_2 + \dots + n a_n}{n^2}$ is equal to:

A l

B l/2

C 2l

D 0

Derivation & Analysis:

By application of Cauchy's theorem on limits or weighted Cesàro means, the limit evaluates to $l/2$ since the denominator weight sums to $n(n+1)/2 \sim n^2/2$.

The value of $\lim_{n \to \infty} \frac{1}{n} \left( 1 + 2^{1/2} + 3^{1/3} + \dots + n^{1/n} \right)$ is:

A 0

B 1/2

C 1

D e

Derivation & Analysis:

Since $\lim_{n \to \infty} n^{1/n} = 1$, by Cauchy's first theorem on limits, the arithmetic mean of the sequence also converges to 1.

Which of the following statements is true for Cauchy's First Theorem?

A It applies only to positive term sequences.

B The converse is always true.

C It states that the limit of the arithmetic mean equals the limit of the sequence.

D It requires the sequence to be strictly increasing.

Derivation & Analysis:

Cauchy's theorem states that if $a_n \to l$, then its arithmetic mean also converges to $l$. No strict monotonicity or positivity is required.

Evaluate $\lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^n \sin\left(\frac{r\pi}{n}\right)$:

A 0

B $2/\pi$

C $1/\pi$

D 1

Derivation & Analysis:

Recognize as Riemann integral $\int_0^1 \sin(\pi x) dx = \left[ -\frac{\cos(\pi x)}{\pi} \right]_0^1 = \frac{1 + 1}{\pi} = \frac{2}{\pi}$.

If $a_n = (-1)^n$, what is the behavior of the sequence $b_n = \frac{a_1 + \dots + a_n}{n}$?

A Diverges to $\infty$

B Oscillates finitely between -1 and 1

C Converges to 0

D Converges to 1

Derivation & Analysis:

For $a_n = (-1)^n$, the sum is either $0$ (for even $n$) or $-1$ (for odd $n$) divided by $n$, which limits to $0$ as $n \to \infty$. This demonstrates that the converse of Cauchy's theorem fails.

Cauchy's Second Theorem on limits states that if $a_n > 0$ and $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = l$, then $\lim_{n \to \infty} (a_n)^{1/n}$ is:

A 0

B l

C $l^2$

D $\infty$

Derivation & Analysis:

This is the exact statement of Cauchy's Second Theorem relating ratio limits to geometric mean/nth root limits.

The limit $\lim_{n \to \infty} \frac{1}{n} \left( 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \right)$ evaluates to:

A 1

B e

C 0

D $\infty$

Derivation & Analysis:

Since $\lim_{n \to \infty} \frac{1}{n} = 0$, by Cauchy's theorem, the arithmetic mean of the harmonic sequence terms converges to 0.

What is $\lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^n k$?

A 0

B 1/2

C 1

D $\infty$

Derivation & Analysis:

$\sum_{k=1}^n k = \frac{n(n+1)}{2}$. Dividing by $n^2$ gives $\frac{n^2 + n}{2n^2} \to \frac{1}{2}$ as $n \to \infty$.

If $\lim a_n = 3$, then $\lim \frac{a_1 + a_2 + \dots + a_n + 5n}{2n}$ is:

A 3

B 4

C 5

D 8

Derivation & Analysis:

Rewrite as $\frac{1}{2} \frac{\sum a_k}{n} + \frac{5}{2}$. By Cauchy's theorem, $\frac{\sum a_k}{n} \to 3$. Thus, $\frac{1}{2}(3) + \frac{5}{2} = \frac{8}{2} = 4$.

Q10

The Stolz-Cesàro theorem is primarily used for evaluating limits of the form:

A $0 \times \infty$

B $\frac{0}{0}$ or $\frac{\infty}{\infty}$ for sequences

C $1^\infty$

D $\infty - \infty$

Derivation & Analysis:

Stolz-Cesàro theorem is the discrete analogue of L'Hôpital's rule, dealing with indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$ for sequence ratios.

Common questions

Yes, extensions of Cauchy's theorem exist where if $a_n \to \infty$, then the arithmetic mean $\frac{1}{n}\sum a_n \to \infty$ as well. However, standard competitive exams primarily test finite limits.

Students frequently mistake the converse—assuming that if the Cesàro mean converges, the underlying sequence must also converge. Remember that oscillating sequences can have convergent means.

Memorize standard limits (such as $\lim n^{1/n} = 1$), master the conversion of series sums into definite Riemann integrals, and utilize the Stolz-Cesàro difference rule instead of cumbersome algebraic expansions.