Problem 56 - Calculus for Biology and Medicine

Question

Formal Definition of Limits: \lim _{n \rightarrow \infty} a_{n}=a .\( Find the limit \)a\(, and determine \)N\( so that \)\left|a_{n}-a\right|<\epsilon\( for all \)n>N\( for the given value of \)\epsilon$. $$ a_{n}=\frac{1}{n^{2}}, \epsilon=0.001 $$

Step-by-Step Solution

Answer

The limit is 0, and \(N = 32\) ensures \(\left|a_n - a\right| < \epsilon = 0.001\).

Step 1: Identify Limit

First, identify the limit of the sequence \(a_n = \frac{1}{n^2}\) as \(n\) approaches infinity. As \(n\) increases, \(\frac{1}{n^2}\) approaches zero. Thus, the limit \(a\) is 0.

Step 2: Define the Inequality

Using the definition of the limit, we set up the inequality \(|a_n - a| < \epsilon\). Here, \(a_n = \frac{1}{n^2}\) and \(a = 0\), so we have \(\left|\frac{1}{n^2} - 0\right| = \frac{1}{n^2} < \epsilon\). Substitute \(\epsilon = 0.001\) into the inequality to get \(\frac{1}{n^2} < 0.001\).

Step 3: Solve the Inequality

Solve the inequality \(\frac{1}{n^2} < 0.001\) to find \(N\). This can be rewritten as \(n^2 > \frac{1}{0.001} = 1000\). Taking the square root of both sides gives \(n > \sqrt{1000}\).

Step 4: Calculate N

Calculate \(\sqrt{1000}\) which is approximately \(31.62\). Therefore, \(N\) is 32 because \(n\) must be an integer greater than 31.62, ensuring \(n > 31.62\).

Key Concepts

Limit of a Sequence

The limit of a sequence is a fundamental concept in calculus and analysis. It represents the value that the terms of a sequence approach as the index number increases toward infinity. Imagine you have a sequence of numbers, like with beads on a string, each bead representing a term in the sequence. As you move further down the string, closer to infinity, the beads (or terms) start converging to a single value. This value is known as the limit.
For the sequence given here, \( a_n = \frac{1}{n^2} \), it’s easy to see that as \( n \to \infty \), each term gets closer to zero. Always keep in mind, when dealing with limits, you need to ensure that the term gets arbitrarily close to, but never actually equaling, the limiting value. Here, the limit \( a \) is zero. This understanding forms the basis for determining how well a sequence approximates its limit, much like estimating how a shadow outlines an object without ever touching it.

Epsilon-Delta Definition

The epsilon-delta definition is a rigorous way of defining what it means for a sequence to have a limit. It's quite precise, which is crucial for mathematical proofs. The central idea is that for every small positive number \( \epsilon \) (epsilon) you choose, you can find a number \( N \) such that all terms of the sequence beyond that index \( N \) have a value less than \( \epsilon \) away from the limit \( a \).
Think of \( \epsilon \) as a tiny margin of error. If you can make the sequence terms fall within this error margin by simply choosing a large enough index \( N \), then the sequence indeed approaches the limit, satisfying the definition.

• Given: \( \epsilon = 0.001 \)
• Sequence: \( a_n = \frac{1}{n^2} \)
• Limit: \( a = 0 \)

To apply the epsilon-delta definition here, you first set up \( |a_n - a| < \epsilon \), which simplifies to \( \frac{1}{n^2} < 0.001 \). Solving this inequality helps you find \( N \), the threshold beyond which all terms of \( a_n \) are quite close to zero, closer than \( 0.001 \). This process ensures a robust understanding of what it means for a sequence to zero in on its limit.

Convergence

Convergence is about how a sequence closes in on its limit as the sequence progresses. When we say a sequence converges, we are asserting that as you proceed down the sequence, the terms get "convergingly" close to potential limit \( a \). This is not an ordinary closeness, but one that can be controlled as precisely as needed by choosing a suitable position in the sequence.
For example, in the case of \( a_n = \frac{1}{n^2} \), the sequence converges to 0. By determining the smallest whole number \( N \) such that any term \( a_n \) with \( n > N \) lies within the tiny band of \( \epsilon \) around zero, the sequence is effectively converging. Here, we calculated \( N \approx 32 \), so beyond this point, all terms are remarkably close to zero with an error margin of just 0.001.
Understanding convergence requires grasping that while terms of the sequence are always getting nearer to the limit, they do not necessarily equal the limit within a finite number of steps.

• Beyond any point \( N \), the terms hover around \( 0 \) within any given small tolerance, here \( 0.001 \).
• This convergence demonstrates the infinite process but a definite trend towards a particular value.

Thus, convergence is the story of how sequences grow nearer and nearer a target value as you run through its course to infinity.