Calculus II: Sequences and Their Limits

In engineering, you don't just model static quantities; you analyze processes that evolve step-by-step. The trajectory of a robotic arm, the damping of a spring-mass system over discrete intervals, or the algorithmic steps in a numerical simulation—these are fundamentally described by sequences. Understanding how sequences behave as their index grows indefinitely, specifically whether they approach a stable value (converge) or not (diverge), is critical for predicting long-term system behavior, ensuring computational stability, and solving advanced problems in differential equations and signal processing.

Defining Sequence Convergence

A sequence is an ordered list of numbers $a_1,a_2,a_3,...,a_n,...$, typically denoted as $\{a_n{\}}_{n=1}^{\infty }$ or simply $\{a_n\}$. The core question is: As $n$ becomes arbitrarily large, does $a_n$ approach a specific, finite number?

Formally, we say the sequence $\{a_n\}$ converges to the limit $L$ if, for every positive number $ϵ$ (no matter how small), there exists a corresponding integer $N$ such that for all $n>N$, the distance $|a_n-L|<ϵ$. In simpler terms, you can force the terms of the sequence to stay within any arbitrarily tiny "error band" around $L$ by going far enough out in the sequence.

If no such finite limit $L$ exists, the sequence diverges. Divergence can be to positive or negative infinity, or it can be oscillatory with no settling point (e.g., $\{(-1{)}^n\}$).

Example: Consider $a_n=\frac{n+1}{2n}$. Intuitively, as $n$ grows large, the "+1" and factor of "2" become less significant, suggesting a limit of $\frac{1}{2}$. Using the formal definition, for a given $ϵ>0$, we find an $N$ such that $|\frac{n+1}{2n}-\frac{1}{2}|=|\frac{1}{2n}|<ϵ$. This holds if $n>\frac{1}{2ϵ}$. So, choosing $N$ to be any integer greater than $\frac{1}{2ϵ}$ satisfies the condition, proving convergence to $L=\frac{1}{2}$.

Limit Laws and the Squeeze Theorem

Just like limits of functions, limits of sequences obey algebraic laws that make computation straightforward. If ${\lim }_{n\to \infty }{a}_n=A$ and ${\lim }_{n\to \infty }{b}_n=B$, and $c$ is a constant, then:

• ${\lim }_{n\to \infty }(a_n\pm b_n)=A\pm B$
• ${\lim }_{n\to \infty }(c\cdot a_n)=cA$
• ${\lim }_{n\to \infty }(a_n\cdot b_n)=A\cdot B$
• ${\lim }_{n\to \infty }(a_n/b_n)=A/B$, provided $B\ne 0$
• ${\lim }_{n\to \infty }(a_n{)}^p=A^p$, for $p>0$ and $A>0$ if $p$ is not an integer.

These laws allow you to break down complex sequences into simpler components.

When direct application of limit laws is impossible—often because a sequence is difficult to handle algebraically—the Squeeze Theorem is a powerful tool. It states: If $a_n\le b_n\le c_n$ for all $n$ beyond some point, and ${\lim }_{n\to \infty }{a}_n={\lim }_{n\to \infty }{c}_n=L$, then ${\lim }_{n\to \infty }{b}_n=L$ as well. You "squeeze" the unknown sequence between two sequences with known, identical limits.

Example: Find ${\lim }_{n\to \infty }\frac{\sin (n)}{n}$. We cannot use the quotient law directly because $\lim \sin (n)$ does not exist. However, we know $-1\le \sin (n)\le 1$ for all $n$. Therefore: $$-\frac{1}{n}\le \frac{\sin (n)}{n}\le \frac{1}{n}$$ Since ${\lim }_{n\to \infty }(-\frac{1}{n})=0$ and ${\lim }_{n\to \infty }(\frac{1}{n})=0$, by the Squeeze Theorem, ${\lim }_{n\to \infty }\frac{\sin (n)}{n}=0$.

The Bounded Monotone Convergence Theorem

This theorem provides a powerful existence guarantee for convergence without requiring you to find the limit's exact value—a common scenario in iterative engineering algorithms. It has two key parts:

1. Monotonic: A sequence is monotonic if it is either entirely non-decreasing ($a_{n+1}\ge a_n$) or entirely non-increasing ($a_{n+1}\le a_n$).
2. Bounded: A sequence is bounded if there exists a number $M$ such that $|a_n|\le M$ for all $n$.

The Bounded Monotone Convergence Theorem states: Every bounded, monotonic sequence converges.

• If $\{a_n\}$ is bounded above and non-decreasing, it converges to its least upper bound.
• If $\{a_n\}$ is bounded below and non-increasing, it converges to its greatest lower bound.

Example: Consider the recursive sequence $a_1=2$, $a_{n+1}=\frac{1}{2}(a_n+\frac{6}{a_n})$, used in algorithms for calculating square roots. You can show it is non-increasing and bounded below (by $\sqrt{6}$). The theorem guarantees it converges to some limit $L$. You can then find $L$ by taking the limit of both sides of the recursive definition: $L=\frac{1}{2}(L+\frac{6}{L})$, which solves to $L=\sqrt{6}$.

Dominance Relationships and Growth Rates

When sequences involve different classes of functions, direct limit evaluation often results in indeterminate forms like $\frac{\infty }{\infty }$ or $\infty \cdot 0$. To resolve these, you must understand the dominance hierarchy of growth rates as $n\to \infty$. From slowest-growing to fastest-growing, the standard hierarchy is: $$\text{Logarithms}\ll \text{Polynomials}\ll \text{Exponentials}\ll \text{Factorials}$$ More precisely: For any constants $p>0$, $b>1$, and $r>0$, $$\underset{n\to \infty }{\lim }\frac{\ln (n)}{n^p}=0,\underset{n\to \infty }{\lim }\frac{n^p}{b^n}=0,\underset{n\to \infty }{\lim }\frac{b^n}{n!}=0.$$ This means factorials dominate exponentials, which dominate polynomials, which dominate logarithms.

Application Example: Determine ${\lim }_{n\to \infty }\frac{n^{100}+5^n}{e^n+n!}$. The dominant term in the numerator is $5^n$ (an exponential). The dominant term in the denominator is $n!$ (a factorial). Since factorials dominate exponentials, the terms $5^n$ and $e^n$ are essentially negligible compared to $n!$. Thus, the limit behaves like ${\lim }_{n\to \infty }\frac{5^n}{n!}=0$.

Applying Techniques to Determine Limits

The skill lies in selecting the right tool. Follow this decision framework:

1. Direct Evaluation/Substitution: Try plugging in a large $n$. If it yields a finite number, you likely have convergence. If it yields $\frac{\text{constant}}{\infty }$ or $\frac{\text{constant}}{0}$ (non-zero), the limit is 0 or infinity, respectively.
2. Algebraic Manipulation: For rational functions (polynomial over polynomial), factor out the highest power of $n$ from both numerator and denominator.

$$\underset{n\to \infty }{\lim }\frac{3n^2-n}{5n^2+7}=\underset{n\to \infty }{\lim }\frac{n^2(3-\frac{1}{n})}{n^2(5+\frac{7}{n^2})}=\frac{3}{5}$$

1. L'Hôpital's Rule: If $a_n=f(n)$ where $f(x)$ is a real differentiable function and the limit yields $\frac{0}{0}$ or $\frac{\infty }{\infty }$, you can often apply L'Hôpital's Rule to the function $f(x)$.
2. Dominance Comparison: For complex forms, identify the dominant term in the numerator and denominator using the growth hierarchy. This often resolves $\frac{\infty }{\infty }$ forms instantly.
3. Squeeze Theorem: Use this when the sequence involves oscillatory terms (like $\sin (n)$, $\cos (n)$) or when you can find natural bounding sequences.
4. Bounded Monotone Theorem: Use this for recursively defined sequences or when you need to prove convergence exists before finding the value.

Common Pitfalls

1. Misapplying L'Hôpital's Rule to Discrete $n$: L'Hôpital's Rule applies to continuous functions. You can only use it on a sequence $a_n=f(n)$ if $f(x)$ is a differentiable function defined for real numbers $x>0$. You cannot differentiate a formula that is only defined at integer values.

• Correction: Ensure $a_n$ is given by a formula that makes sense for all real $x$ (e.g., $n^2$, $\ln (n)$, $e^{-n}$). If it's purely discrete (like $n!$), you cannot use L'Hôpital's directly.

1. Assuming Boundedness Implies Convergence: A bounded sequence is not necessarily convergent. For example, the sequence $a_n=\sin (\frac{n\pi }{2})$ oscillates between -1 and 1 (it's bounded) but never settles on a limit.

• Correction: Remember convergence requires boundedness and monotonicity (or another specific behavior). Boundedness alone is not sufficient.

1. Incorrect Dominance Intuition: Thinking that a higher-degree polynomial always dominates any exponential for large $n$ is a critical error.

• Correction: Memorize the hierarchy: Any exponential $b^n$ ($b>1$) will eventually outgrow any polynomial $n^p$, no matter how large $p$ is. For ${\lim }_{n\to \infty }\frac{n^{1000}}{1.001^n}$, the exponential $1.001^n$ in the denominator wins, and the limit is 0.

1. Neglecting Absolute Values in the Formal Definition: When working with the $ϵ$-$N$ definition, the condition is $|a_n-L|<ϵ$. Forgetting the absolute value can lead to incorrect inequalities, especially if the sequence approaches the limit from below.

• Correction: Always start with $|a_n-L|$, simplify the expression inside the absolute value, and then find $N$ such that this simplified expression is less than $ϵ$.

Summary

• A sequence converges to a limit $L$ if its terms can be made arbitrarily close to $L$ by taking the index $n$ sufficiently large. The formal $ϵ$-$N$ definition makes this intuitive idea mathematically precise.
• Limit Laws allow for the algebraic manipulation of convergent sequences, while the Squeeze Theorem is invaluable for handling oscillatory or hard-to-simplify terms.
• The Bounded Monotone Convergence Theorem is a fundamental existence theorem: if a sequence is always increasing (or decreasing) and cannot exceed (or fall below) a certain value, it must converge.