Problem 70 - Calculus for Biology and Medicine

Question

\(\lim _{n \rightarrow \infty} a_{n}=a\). Find the limit \(a\), and determine \(N\) so that \(\left|a_{n}-a\right|<\epsilon\) for all \(n>N\) for the given value of \(\epsilon\) $$ a_{n}=\ln \left(1+\frac{1}{n}\right), \epsilon=0.1 $$

Step-by-Step Solution

Answer

The limit is \(a = 0\) and \(N = 10\) ensures \(|a_n - a| < 0.1\).

Step 1: Identify the Form of the Sequence

We are given the sequence \(a_{n} = \ln(1 + \frac{1}{n})\). We need to find the limit of this sequence as \(n\) approaches infinity.

Step 2: Determine the Limit

Evaluate the limit: \(\lim_{n \to \infty} \ln(1 + \frac{1}{n})\). As \(n\) becomes very large, \(\frac{1}{n}\) approaches zero, which means the expression \(1 + \frac{1}{n}\) approaches 1. Therefore, the limit of \(\ln(1 + \frac{1}{n})\) as \(n\) approaches infinity is \(\ln(1) = 0\). Thus, \(a = 0\).

Step 3: Understand the Condition \(|a_n - a| < \epsilon\)

We need to ensure that \(|a_n - a| < \epsilon\) for all \(n > N\). In simpler terms, we have \(|\ln(1 + \frac{1}{n}) - 0| < 0.1\). Since \(|\ln(1 + \frac{1}{n})| = \ln(1 + \frac{1}{n})\) for \(n > 0\), we need \(\ln(1 + \frac{1}{n}) < 0.1\).

Step 4: Solve for \(N\)

We solve the inequality \(\ln(1 + \frac{1}{n}) < 0.1\) by exponentiating both sides: \(1 + \frac{1}{n} < e^{0.1}\). Calculate \(e^{0.1} \approx 1.10517\). Thus, \(1 + \frac{1}{n} < 1.10517\) implies \(\frac{1}{n} < 0.10517\), which simplifies to \(n > \frac{1}{0.10517} \approx 9.51\). We choose \(N = 10\) to meet this condition for all integer \(n\).

Key Concepts

Convergence of Sequences

When we talk about the convergence of sequences in calculus, we are exploring whether a sequence of numbers "settles down" to a single number as it progresses to infinity. If it does, it is convergent. Otherwise, it's divergent.

- **What is a Sequence?**A sequence is a set of numbers arranged in a specific order. Typically, this order follows a rule or a formula, like our sequence here: \(a_n = \ln(1 + \frac{1}{n})\).

- **Convergence Defined**A sequence \(a_n\) converges to a limit \(a\) if for any small positive number \(\epsilon\), there's a point in the sequence after which all following terms are within \(\epsilon\) of \(a\).

In our exercise, as \(n\) approaches infinity, \(\ln(1 + \frac{1}{n})\) approaches 0. Thus, the limit \(a\) is 0, exemplifying convergence to this point.

Understanding convergence is crucial for analyzing behaviors in calculus and determining limits of functions and sequences under different conditions.

Natural Logarithm

The natural logarithm, denoted \(\ln\), is a unique and essential logarithm in mathematics with the base \(e\), the Euler's number, approximately equal to 2.71828.

- **Understanding \(\ln(x)\)**The natural logarithm \(\ln(x)\) represents the power to which \(e\) must be raised to obtain \(x\). For example, if \(\ln(x) = y\), then \(e^y = x\).

- **Natural Logarithm in Sequences**In our specific sequence \(a_n = \ln(1 + \frac{1}{n})\), as \(n\) increases, \(\frac{1}{n}\) shrinks towards zero. The logarithmic function lets us examine subtle changes as the values creep toward unity, influencing the limit result.

- **Properties of \(\ln(x)\)**

• \(\ln(1) = 0\)
• \(\ln(ab) = \ln(a) + \ln(b)\)
• \(\ln(\frac{a}{b}) = \ln(a) - \ln(b)\)
• \(\ln(a^b) = b \cdot \ln(a)\)

A strong grasp of these properties helps to solve limits and analyze exponential growth and decay behavior in various fields.

Epsilon-Delta Definition of Limit

The epsilon-delta definition is a foundation in calculus for formally defining limits. It rigorously explains what it means for a sequence or a function to approach a certain value, combining two principal ideas: the distance (\(\epsilon\)) and a threshold (\(\delta\)).

- **Core Idea**A sequence \(a_n\) approaches a limit \(a\) when, for any degree of closeness we want (\(\epsilon > 0\)), there's a certain point in the sequence from which onward all terms of the sequence are within \(\epsilon\) of \(a\). This point is denoted \(N\).

- **Relating to the Exercise**Here, \(|a_n - a| < \epsilon\) translates to \(\ln(1 + \frac{1}{n}) < 0.1\) for all \(n > N\) in our problem. Solving for \(N\) involved strategic inequality and understanding exponential functions, leading to \(N = 10\) for \(\epsilon = 0.1\).

This definition ensures predictability and precision in studies involving limits, crucial for both theoretical and practical applications in mathematics and beyond.