CALC-2.3

4-5 Hours

Fundamental Convergence Theorems

Core Theorems

6 Sections

Essential for Analysis

Learning Objectives

By the end of this course, you will be able to:

State and apply the Squeeze Theorem (Sandwich Theorem)
Understand and prove the Monotone Bounded Theorem
Derive the important limit e = lim(1 + 1/n)^n
Apply the Stolz Theorem for indeterminate forms
Understand infinitesimals and their order comparison
Master techniques for proving convergence without knowing the limit

Section 1: The Squeeze Theorem

Theorem 3.1: Squeeze Theorem (Sandwich Theorem)

Let $\{a_n\}$ , $\{b_n\}$ , $\{c_n\}$ be sequences satisfying:

$\exists N_0 \in \mathbb{N}$ : $a_n \leq b_n \leq c_n$ for all $n \geq N_0$
$\lim_{n \to \infty} a_n = L = \lim_{n \to \infty} c_n$

Then $\lim_{n \to \infty} b_n = L$ .

Proof of Theorem 3.1:

Given $\epsilon > 0$ :

$\exists N_1$ : $n > N_1 \Rightarrow |a_n - L| < \epsilon \Rightarrow L - \epsilon < a_n$
$\exists N_2$ : $n > N_2 \Rightarrow |c_n - L| < \epsilon \Rightarrow c_n < L + \epsilon$

For $n > \max\{N_0, N_1, N_2\}$ :

L - \epsilon < a_n \leq b_n \leq c_n < L + \epsilon

Therefore $|b_n - L| < \epsilon$ .

∎

Corollary 3.1: Zero Limit Equivalence

\lim_{n \to \infty} a_n = 0 \Leftrightarrow \lim_{n \to \infty} |a_n| = 0

Corollary 3.2: Domination Principle

If $|a_n| \leq b_n$ and $\lim_{n \to \infty} b_n = 0$ , then $\lim_{n \to \infty} a_n = 0$ .

Example 3.1: Proving √[n]{n} → 1

Problem: Prove $\lim_{n \to \infty} \sqrt[n]{n} = 1$ .

Solution:

Let $\sqrt[n]{n} = 1 + h_n$ where $h_n \geq 0$ for $n \geq 1$ .

Then $n = (1 + h_n)^n \geq 1 + \binom{n}{2}h_n^2 = 1 + \frac{n(n-1)}{2}h_n^2$ (Binomial theorem)

So $h_n^2 \leq \frac{2(n-1)}{n(n-1)} = \frac{2}{n}$ , giving $0 \leq h_n \leq \sqrt{\frac{2}{n}}$ .

Since $\sqrt{\frac{2}{n}} \to 0$ , by Squeeze theorem $h_n \to 0$ , so $\sqrt[n]{n} \to 1$ .

Example 3.2: Maximum of Powers

Problem: Prove $\lim_{n \to \infty} \sqrt[n]{a^n + b^n} = \max\{a, b\}$ for $a, b > 0$ .

Solution:

WLOG assume $a \geq b > 0$ . Then:

a = \sqrt[n]{a^n} \leq \sqrt[n]{a^n + b^n} \leq \sqrt[n]{2a^n} = a \cdot \sqrt[n]{2}

Since $\sqrt[n]{2} \to 1$ , by Squeeze theorem the limit is $a = \max\{a, b\}$ .

Section 2: Monotone Bounded Theorem

Definition 3.1: Monotone Sequence

A sequence $\{a_n\}$ is:

Monotonically increasing if $a_n \leq a_{n+1}$ for all $n$
Strictly increasing if $a_n < a_{n+1}$ for all $n$
Monotonically decreasing if $a_n \geq a_{n+1}$ for all $n$
Strictly decreasing if $a_n > a_{n+1}$ for all $n$

Theorem 3.2: Monotone Bounded Theorem

A bounded monotone sequence converges.

Proof of Theorem 3.2 (Increasing Case):

Let $E = \{a_n : n \in \mathbb{N}\}$ . Since $E$ is non-empty and bounded above, by the Completeness Axiom, $\alpha = \sup E$ exists.

We claim $\lim_{n \to \infty} a_n = \alpha$ .

Given $\epsilon > 0$ :

By supremum property: $\exists N$ such that $a_N > \alpha - \epsilon$
By monotonicity: $n > N \Rightarrow a_n \geq a_N > \alpha - \epsilon$
By upper bound: $a_n \leq \alpha < \alpha + \epsilon$

Therefore $|a_n - \alpha| < \epsilon$ for $n > N$ .

∎

Methods to Check Monotonicity

1. Difference Method

Compute $a_{n+1} - a_n$ and check its sign.

2. Ratio Method

For positive sequences, compute $\frac{a_{n+1}}{a_n}$ and compare to 1.

Section 3: The Important Limit e

Definition 3.2: Sequences Defining e

Define sequences:

x_n = \left(1 + \frac{1}{n}\right)^n, \quad y_n = \left(1 + \frac{1}{n}\right)^{n+1}

Theorem 3.3: Properties of e

$\{x_n\}$ is strictly increasing and bounded above by 3
$\{y_n\}$ is strictly decreasing and bounded below by 2
Both converge to the same limit, denoted $e$

Key Result: Euler's Number

e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \approx 2.71828...

Important Inequalities:

From $x_n < e < y_n$ :

\frac{1}{n+1} < \ln\left(1 + \frac{1}{n}\right) < \frac{1}{n}

Euler-Mascheroni Constant:

The sequence $\gamma_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n} - \ln n$ converges to $\gamma \approx 0.5772$ .

Section 4: Infinitesimals and Order Comparison

Definition 3.3: Infinitesimal

A sequence $\{x_n\}$ is an infinitesimal if $\lim_{n \to \infty} x_n = 0$ , written $x_n = o(1)$ .

Order Comparison Notation

Notation	Definition	Meaning
$x_n \sim y_n$	$\lim \frac{x_n}{y_n} = 1$	Equivalent (asymptotically equal)
$x_n = o(y_n)$	$\lim \frac{x_n}{y_n} = 0$	Higher order infinitesimal (xₙ is much smaller)
$x_n = O(y_n)$	$\|x_n\| \leq C\|y_n\| \text{ for large } n$	Same order or smaller (big-O notation)

Section 5: Stolz Theorem

Theorem 3.4: Stolz Theorem

Let $\{x_n\}$ , $\{y_n\}$ satisfy:

$\{y_n\}$ is strictly increasing and $y_n \to +\infty$
$\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = L$ (finite or infinite)

Then $\lim_{n \to \infty} \frac{x_n}{y_n} = L$ .

Remark 3.1: Discrete L'Hôpital's Rule

Stolz theorem is the discrete analog of L'Hôpital's rule for $\frac{\infty}{\infty}$ forms. It allows us to simplify limits by looking at differences rather than the sequences themselves.

Corollary 3.3: Cesàro Mean

If $\lim_{n \to \infty} x_n = A$ , then:

\lim_{n \to \infty} \frac{x_1 + x_2 + \cdots + x_n}{n} = A

Proof of Cesàro Mean:

Apply Stolz with $y_n = n$ . Then:

\frac{x_{n+1} - x_n}{(n+1) - n} = x_{n+1} - x_n \to A - A = 0?

Wait, let's reconsider. Let $S_n = x_1 + \cdots + x_n$ . Then:

\frac{S_{n+1} - S_n}{(n+1) - n} = x_{n+1} \to A

By Stolz, $\frac{S_n}{n} \to A$ .

∎

Remark 3.2: Converse is False

The converse is false. Cesàro mean can converge even when the original sequence diverges.

Example: $x_n = (-1)^n$ diverges, but $\frac{1}{n}\sum_{k=1}^{n}(-1)^k \to 0$ .

Example 3.3: Applying Stolz Theorem

Problem: Find $\lim_{n \to \infty} \frac{1^2 + 2^2 + \cdots + n^2}{n^3}$ .

Solution using Stolz:

Let $x_n = 1^2 + 2^2 + \cdots + n^2$ and $y_n = n^3$ .

\frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \frac{(n+1)^2}{(n+1)^3 - n^3} = \frac{(n+1)^2}{3n^2 + 3n + 1}

Taking the limit:

\lim_{n \to \infty} \frac{(n+1)^2}{3n^2 + 3n + 1} = \lim_{n \to \infty} \frac{n^2 + 2n + 1}{3n^2 + 3n + 1} = \frac{1}{3}

Example 3.4: Recursive Sequence with Monotone Bounded

Problem: Let $x_1 = 1$ , $x_{n+1} = \sqrt{2 + x_n}$ . Find $\lim_{n \to \infty} x_n$ .

Solution:

Step 1: Assume limit $L$ exists. Then $L = \sqrt{2 + L}$ .

L^2 = 2 + L \Rightarrow L^2 - L - 2 = 0 \Rightarrow (L-2)(L+1) = 0

So $L = 2$ (since $x_n > 0$ ).

Step 2: Prove $x_n < 2$ by induction:

$x_1 = 1 < 2$ ✓
If $x_n < 2$ , then $x_{n+1} = \sqrt{2 + x_n} < \sqrt{2 + 2} = 2$ ✓

Step 3: Prove increasing:

x_{n+1}^2 - x_n^2 = 2 + x_n - x_n^2 = -(x_n - 2)(x_n + 1) > 0

since $x_n < 2$ and $x_n > 0$ . So $x_{n+1} > x_n$ .

Conclusion: By Monotone Bounded Theorem, $x_n \to 2$ .

Example 3.5: Proving √[n]{n!} / n → 1/e

Problem: Prove $\lim_{n \to \infty} \frac{\sqrt[n]{n!}}{n} = \frac{1}{e}$ .

Solution using Cesàro Mean:

Let $a_n = \ln\frac{\sqrt[n]{n!}}{n} = \frac{1}{n}\ln(n!) - \ln n = \frac{1}{n}\sum_{k=1}^{n}\ln k - \ln n$

This equals $\frac{1}{n}\sum_{k=1}^{n}\ln\frac{k}{n}$ .

As $n \to \infty$ , this is a Riemann sum for:

\int_0^1 \ln x \, dx = [x\ln x - x]_0^1 = -1

Therefore $\lim_{n \to \infty} \frac{\sqrt[n]{n!}}{n} = e^{-1} = \frac{1}{e}$ .

Summary: Convergence Theorem Applications

Theorem	When to Use	Key Requirement
Squeeze	Can bound sequence between two with same limit	$a_n \leq x_n \leq b_n$
Monotone Bounded	Recursive sequences, proving existence	Monotonic + Bounded
Stolz	Indeterminate $\frac{\infty}{\infty}$ forms	$y_n \to \infty$ strictly increasing
Cesàro Mean	Average of sequence terms	Original sequence converges

Theorem 3.5: Ratio Test for Sequences

Let $\{a_n\}$ be a sequence with $a_n > 0$ . If $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$ , then:

If $L < 1$ , then $\lim_{n \to \infty} a_n = 0$
If $L > 1$ , then $\lim_{n \to \infty} a_n = +\infty$
If $L = 1$ , the test is inconclusive

Example 3.6: Using Ratio Test

Problem: Find $\lim_{n \to \infty} \frac{n^{10}}{2^n}$ .

Solution:

Let $a_n = \frac{n^{10}}{2^n}$ . Compute the ratio:

\frac{a_{n+1}}{a_n} = \frac{(n+1)^{10}}{2^{n+1}} \cdot \frac{2^n}{n^{10}} = \frac{1}{2}\left(\frac{n+1}{n}\right)^{10} = \frac{1}{2}\left(1 + \frac{1}{n}\right)^{10}

As $n \to \infty$ : $\frac{a_{n+1}}{a_n} \to \frac{1}{2} < 1$

By the Ratio Test, $\lim_{n \to \infty} \frac{n^{10}}{2^n} = 0$ .

This shows exponential growth dominates polynomial growth.

Section 6: Important Limit Computations

Theorem 3.6: Fundamental Limits

The following limits are fundamental and should be memorized:

\lim_{n \to \infty} \sqrt[n]{n} = 1

\lim_{n \to \infty} \sqrt[n]{a} = 1 \quad (a > 0)

\lim_{n \to \infty} \frac{a^n}{n!} = 0 \quad (\forall a)

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

Example 3.7: Proving √[n]{n} → 1

Problem: Prove $\lim_{n \to \infty} \sqrt[n]{n} = 1$ .

Solution:

Let $\sqrt[n]{n} = 1 + h_n$ where $h_n > 0$ for $n \geq 2$ .

Then $n = (1 + h_n)^n$ . By binomial theorem:

n = (1 + h_n)^n \geq 1 + nh_n + \frac{n(n-1)}{2}h_n^2 > \frac{n(n-1)}{2}h_n^2

So $h_n^2 < \frac{2}{n-1}$ , giving $0 < h_n < \sqrt{\frac{2}{n-1}} \to 0$ .

Therefore $\sqrt[n]{n} = 1 + h_n \to 1$ .

Example 3.8: Nested Radical

Problem: Find $\lim_{n \to \infty} \sqrt[n]{1 + 2^n + 3^n}$ .

Solution:

Note that $3^n \leq 1 + 2^n + 3^n \leq 3 \cdot 3^n$ .

Taking $n$ -th roots:

3 \leq \sqrt[n]{1 + 2^n + 3^n} \leq 3 \cdot \sqrt[n]{3}

Since $\sqrt[n]{3} \to 1$ , by Squeeze Theorem:

\lim_{n \to \infty} \sqrt[n]{1 + 2^n + 3^n} = 3

Example 3.9: The Number e Revisited

Problem: Prove $\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n+1} = e$ .

Solution:

\left(1 + \frac{1}{n}\right)^{n+1} = \left(1 + \frac{1}{n}\right)^n \cdot \left(1 + \frac{1}{n}\right)

As $n \to \infty$ :

$\left(1 + \frac{1}{n}\right)^n \to e$
$\left(1 + \frac{1}{n}\right) \to 1$

By product rule: $\left(1 + \frac{1}{n}\right)^{n+1} \to e \cdot 1 = e$ .

Remark 3.3: Variations of e

All of the following equal $e$ :

\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e

\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^{-n} = e

\lim_{n \to \infty} \left(1 + \frac{k}{n}\right)^n = e^k

\lim_{n \to \infty} \left(\frac{n+1}{n}\right)^n = e

Problem-Solving Strategy

Identify the dominant term - Which term grows fastest as $n \to \infty$ ?
Factor out the dominant term - Express others relative to it.
Apply standard limits - Use memorized results for $\sqrt[n]{n}$ , $e$ , etc.
Use Squeeze when possible - Bound between simpler expressions.

Example 3.10: Double Application of Stolz

Problem: Find $\lim_{n \to \infty} \frac{1^3 + 2^3 + \cdots + n^3}{n^4}$ .

Solution:

Let $x_n = \sum_{k=1}^n k^3$ , $y_n = n^4$ . Apply Stolz:

\frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \frac{(n+1)^3}{(n+1)^4 - n^4}

We have $(n+1)^4 - n^4 = 4n^3 + 6n^2 + 4n + 1$ , so:

\frac{(n+1)^3}{4n^3 + 6n^2 + 4n + 1} = \frac{n^3 + 3n^2 + 3n + 1}{4n^3 + 6n^2 + 4n + 1} \to \frac{1}{4}

Remark 3.4: Alternative Formula

The sum $\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$ gives directly:

\frac{\sum_{k=1}^n k^3}{n^4} = \frac{n^2(n+1)^2}{4n^4} = \frac{(n+1)^2}{4n^2} \to \frac{1}{4}

Example 3.11: Applying Squeeze to Oscillating Sequences

Problem: Find $\lim_{n \to \infty} \frac{\sin(n^2)}{\sqrt{n}}$ .

Solution:

Since $-1 \leq \sin(n^2) \leq 1$ , we have:

\frac{-1}{\sqrt{n}} \leq \frac{\sin(n^2)}{\sqrt{n}} \leq \frac{1}{\sqrt{n}}

Both bounds tend to 0 as $n \to \infty$ .

By Squeeze Theorem: $\lim_{n \to \infty} \frac{\sin(n^2)}{\sqrt{n}} = 0$ .

Example 3.12: Nested Sequence with e

Problem: Find $\lim_{n \to \infty} \left(\frac{n+2}{n}\right)^n$ .

Solution:

\left(\frac{n+2}{n}\right)^n = \left(1 + \frac{2}{n}\right)^n

Using the limit $\lim_{n \to \infty} \left(1 + \frac{k}{n}\right)^n = e^k$ with $k = 2$ :

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

Quick Reference: Key Formulas

Power Sums

\sum_{k=1}^n k = \frac{n(n+1)}{2}

\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}

Growth Comparison

\ln n \ll n^\alpha \ll a^n \ll n!

Section 7: Advanced Techniques

Theorem 3.7: Root Test for Sequences

Let $\{a_n\}$ be a sequence with $a_n \geq 0$ . If $\lim_{n \to \infty} \sqrt[n]{a_n} = L$ , then:

If $L < 1$ , then $\lim a_n = 0$
If $L > 1$ , then $\{a_n\}$ diverges to $+\infty$

Example 3.13: Root Test Application

Problem: Find $\lim_{n \to \infty} \left(\frac{n}{n+1}\right)^{n^2}$ .

Solution:

Let $a_n = \left(\frac{n}{n+1}\right)^{n^2}$ . Compute:

\sqrt[n]{a_n} = \left(\frac{n}{n+1}\right)^n = \left(1 - \frac{1}{n+1}\right)^n

As $n \to \infty$ : $\left(1 - \frac{1}{n+1}\right)^n \to e^{-1} = \frac{1}{e} < 1$

By Root Test: $\lim_{n \to \infty} \left(\frac{n}{n+1}\right)^{n^2} = 0$ .

Theorem 3.8: Toeplitz Lemma

If $\lim_{n \to \infty} x_n = L$ and $\{w_{n,k}\}$ is a triangular array with:

$w_{n,k} \geq 0$ , $\sum_{k=1}^{n} w_{n,k} = 1$
$\lim_{n \to \infty} w_{n,k} = 0$ for each fixed $k$

Then $\lim_{n \to \infty} \sum_{k=1}^{n} w_{n,k} x_k = L$ .

Example 3.14: Power Mean Convergence

Problem: If $x_n \to L$ , prove $\sqrt[n]{x_1 x_2 \cdots x_n} \to L$ (for $x_n > 0$ ).

Solution:

Taking logarithms:

\ln\sqrt[n]{x_1 \cdots x_n} = \frac{1}{n}\sum_{k=1}^{n}\ln x_k

Since $\ln x_n \to \ln L$ , by Cesàro mean:

\frac{1}{n}\sum_{k=1}^{n}\ln x_k \to \ln L

Therefore $\sqrt[n]{x_1 \cdots x_n} \to e^{\ln L} = L$ .

Remark 3.5: Comparison with Series

Many sequence techniques parallel series convergence tests:

Sequences

Ratio Test: $\frac{a_{n+1}}{a_n} \to L$

Series

d'Alembert: $\frac{a_{n+1}}{a_n} \to L < 1$ ⟹ convergent

Example 3.15: Factorial Growth

Problem: Find $\lim_{n \to \infty} \frac{n^n}{n!}$ .

Solution:

Using Ratio Test:

\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{(n+1)!} \cdot \frac{n!}{n^n} = \frac{(n+1)^{n+1}}{(n+1) \cdot n^n} = \left(\frac{n+1}{n}\right)^n

As $n \to \infty$ : $\left(1 + \frac{1}{n}\right)^n \to e > 1$

Therefore $\frac{n^n}{n!} \to +\infty$ .

Chapter Summary

Squeeze: Bound between two sequences with same limit
Monotone Bounded: Monotonic + bounded ⟹ convergent
Stolz: Discrete L'Hôpital for ∞/∞ forms
e definition: $\lim\left(1 + \frac{1}{n}\right)^n = e$

Example 3.16: Limit of Factorial Ratio

Problem: Find $\lim_{n \to \infty} \frac{(2n)!}{(n!)^2 \cdot 4^n}$ .

Solution using Stirling's approximation:

$n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$

\frac{(2n)!}{(n!)^2 \cdot 4^n} \approx \frac{\sqrt{4\pi n}(2n/e)^{2n}}{2\pi n(n/e)^{2n} \cdot 4^n} = \frac{1}{\sqrt{\pi n}} \to 0

Theorem 3.9: Bernoulli Inequality

For $h > -1$ and $n \in \mathbb{N}$ :

(1 + h)^n \geq 1 + nh

This is essential for proving exponential limits.

Example 3.17: Using Bernoulli

Problem: Prove $\lim_{n \to \infty} \frac{n}{2^n} = 0$ .

Solution:

Write $2^n = (1 + 1)^n \geq 1 + n$ by Bernoulli.

Actually, for $n \geq 2$ : $2^n \geq \frac{n^2}{4}$ (by binomial expansion).

0 < \frac{n}{2^n} \leq \frac{n}{n^2/4} = \frac{4}{n} \to 0

Remark 3.6: Exponential Dominance

Key principle: Exponential growth dominates polynomial growth.

\lim_{n \to \infty} \frac{n^k}{a^n} = 0 \quad \text{for any } k \text{ and } a > 1

Similarly, factorial dominates exponential: $\lim \frac{a^n}{n!} = 0$ .

Example 3.18: Comparing Growth Rates

Problem: Arrange in order of growth: $n^{100}, 1.01^n, n!, n^n$ .

Solution:

From slowest to fastest growth:

n^{100} \ll 1.01^n \ll n! \ll n^n

Here $f \ll g$ means $\lim f/g = 0$ .

Standard Limits Reference

\lim_{n \to \infty} \sqrt[n]{n} = 1

\lim_{n \to \infty} \frac{\ln n}{n} = 0

\lim_{n \to \infty} n \sin\frac{1}{n} = 1

\lim_{n \to \infty} \frac{n!}{n^n} = 0

Example 3.19: Recursive Sequence

Problem: Let $x_1 = 1$ , $x_{n+1} = \sqrt{2 + x_n}$ . Find $\lim x_n$ .

Solution:

Step 1: Show bounded: $x_n < 2$ by induction.

Step 2: Show increasing: $x_{n+1}^2 - x_n^2 = 2 + x_n - x_n^2 = (2-x_n)(1+x_n) > 0$ .

Step 3: By Monotone Bounded Theorem, limit exists. If $L = \lim x_n$ :

L = \sqrt{2 + L} \Rightarrow L^2 = 2 + L \Rightarrow L = 2

Theorem 3.10: Weighted Cesàro Mean

If $x_n \to L$ and $w_n > 0$ with $\sum w_k \to \infty$ , then:

\frac{\sum_{k=1}^{n} w_k x_k}{\sum_{k=1}^{n} w_k} \to L

Remark 3.7: Squeeze Theorem Variants

Extensions of the Squeeze Theorem:

One-sided: If $0 \leq x_n \leq y_n$ and $y_n \to 0$ , then $x_n \to 0$
Ratio form: If $\frac{x_n}{y_n} \to 1$ and $y_n \to L$ , then $x_n \to L$
Absolute value: $|x_n| \to 0 \Leftrightarrow x_n \to 0$

Example 3.20: Squeeze with Oscillation

Problem: Find $\lim_{n \to \infty} \frac{\cos(n^2)}{n}$ .

Solution:

-\frac{1}{n} \leq \frac{\cos(n^2)}{n} \leq \frac{1}{n}

Both bounds → 0, so by Squeeze: $\frac{\cos(n^2)}{n} \to 0$ .

Theorem Selection Flowchart

1. Know the limit? → Use ε-N definition

2. Can bound both sides? → Squeeze Theorem

3. Monotonic sequence? → Monotone Bounded Theorem

4. Ratio of differences? → Stolz-Cesàro

5. Form $(1 + \frac{a}{n})^{bn}$ ? → Use $e$ limit

Example 3.21: Monotone Bounded Application

Problem: Let $x_1 = 2$ , $x_{n+1} = \frac{1}{2}(x_n + \frac{2}{x_n})$ . Find $\lim x_n$ .

Solution (Newton's method for √2):

Bounded below: AM-GM gives $x_{n+1} \geq \sqrt{2}$ .

Decreasing: $x_{n+1} - x_n = \frac{2-x_n^2}{2x_n} < 0$ for $x_n > \sqrt{2}$ .

Limit satisfies $L = \frac{1}{2}(L + \frac{2}{L})$ , giving $L = \sqrt{2}$ .

Theorem 3.11: Ratio-Root Comparison

If $\lim \frac{a_{n+1}}{a_n} = L$ , then $\lim \sqrt[n]{a_n} = L$ .

The converse is false.

Remark 3.8: Exponential Limit Forms

\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = e

\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n = \frac{1}{e}

\lim_{n\to\infty}\left(1+\frac{a}{n}\right)^n = e^a

\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n+k} = e

Chapter 2.3 Mastery Checklist

Squeeze Theorem application
Monotone Bounded Theorem
Stolz-Cesàro Theorem
Definition of $e$
Root and Ratio Tests

Example 3.22: Squeeze with Exponential

Problem: Find $\lim_{n \to \infty} \frac{2^n}{n!}$ .

Solution:

For $n \geq 4$ : $\frac{2^n}{n!} = \frac{2 \cdot 2 \cdot 2 \cdot 2}{1 \cdot 2 \cdot 3 \cdot 4} \cdot \frac{2^{n-4}}{5 \cdot 6 \cdots n} \leq \frac{16}{24} \cdot \left(\frac{2}{5}\right)^{n-4}$

Since $\left(\frac{2}{5}\right)^{n-4} \to 0$ , by Squeeze: $\frac{2^n}{n!} \to 0$ .

Remark 3.9: Key Takeaways

Squeeze Theorem is most useful when the sequence oscillates
Monotone Bounded is the go-to for recursive sequences
Stolz-Cesàro handles ∞/∞ forms elegantly
Always check if standard $e$ limit forms apply

Practice Quiz: Convergence Theorems

Questions

Correct

Accuracy

State the Squeeze Theorem: If

a_n \leq b_n \leq c_n

and

\lim a_n = \lim c_n = L

, then...

Easy

Not attempted

Is the sequence

\{1.001^n\}

bounded?

Easy

Not attempted

Find

\lim_{n \to \infty} \sqrt[n]{3^n + 4^n}

Medium

Not attempted

a_n \leq a_{n+1} \leq 5

for all

n

, what can we conclude?

Medium

Not attempted

Find

\lim_{n \to \infty} \left(1 + \frac{2}{n}\right)^n

Medium

Not attempted

Using Stolz theorem, find

\lim_{n \to \infty} \frac{1 + 2 + \cdots + n}{n^2}

Hard

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Evaluate

\lim_{n \to \infty} \frac{1^2 + 2^2 + \cdots + n^2}{n^3}

Hard

Not attempted

The harmonic series partial sums

H_n = \sum_{k=1}^n \frac{1}{k}

satisfy:

Hard

Not attempted

Frequently Asked Questions

When should I use the Squeeze Theorem vs. Monotone Bounded Theorem?

Use Squeeze when you can bound your sequence between two sequences with the same limit. Use Monotone Bounded when your sequence is monotonic (always increasing or decreasing) and bounded. Squeeze works even for oscillating sequences; Monotone Bounded requires monotonicity.

Why is e defined as a limit rather than just a decimal number?

The limit definition captures e's fundamental property: it's the natural base for exponential growth. Defining e ≈ 2.71828... is circular—we'd need limits to compute those decimals anyway. The limit (1 + 1/n)^n shows e arises naturally from compound interest and continuous growth.

What's the relationship between Stolz theorem and L'Hôpital's rule?

Stolz theorem is the discrete analog of L'Hôpital's rule. L'Hôpital handles continuous functions (derivatives), while Stolz handles sequences (differences). Both resolve indeterminate forms by looking at 'rates of change.'

Can the Cesàro mean converge when the original sequence diverges?

Yes! Example: xₙ = (-1)^n diverges, but (1/n)Σxₖ → 0. The converse is also important: if xₙ → L, then Cesàro mean → L too. But Cesàro convergence doesn't imply original convergence.

Why does the Monotone Bounded Theorem require both conditions?

Monotone alone isn't enough: n is increasing but unbounded, so diverges. Bounded alone isn't enough: (-1)^n is bounded but oscillates. Together, they force convergence: the sequence approaches its supremum (or infimum).

How do I find bounds for recursive sequences?

A useful trick: assume the limit L exists and solve L = f(L). The solutions are candidates for bounds. For example, if xₙ₊₁ = √(2 + xₙ), solving L = √(2 + L) gives L = 2. Then prove xₙ < 2 by induction and monotonicity.