Problem 70 - Thomas Calculus

Question

Which of the sequences \(\left\\{a_{n}\right\\}\) in Exercises \(27-90\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\left(\frac{n}{n+1}\right)^{n} $$

Step-by-Step Solution

Answer

The sequence converges, and its limit is \( \frac{1}{e} \).

Step 1: Understand the Problem

We are given the sequence \( a_n = \left( \frac{n}{n+1} \right)^n \) and need to determine if it converges or diverges, and find the limit if it converges.

Step 2: Analyze the Sequence

The sequence is \( a_n = \left( \frac{n}{n+1} \right)^n \). As \( n \to \infty \), the fraction \( \frac{n}{n+1} \to 1 \). Our task is to understand the behavior of the sequence as \( n \) becomes very large.

Step 3: Use Logarithms for Simplification

Consider \( \ln(a_n) = n \ln\left( \frac{n}{n+1} \right) \). Expanding this gives \( n(-\ln(1 + \frac{1}{n})) \). For large \( n \), we use the approximation \( \ln(1 + x) \approx x \) for small \( x \). Thus, \( \ln(1 + \frac{1}{n}) \approx \frac{1}{n} \).

Step 4: Simplify Using Approximation

Substitute the approximation: \( n(-\ln(1 + \frac{1}{n})) \approx n(-\frac{1}{n}) = -1 \). Thus, \( \ln(a_n) \approx -1 \) as \( n \rightarrow \infty \).

Step 5: Exponentiate to Find Limit

Exponentiating both sides gives \( a_n = e^{\ln(a_n)} \approx e^{-1} \). Thus, \( \lim_{n \to \infty} a_n = e^{-1} \), which means the sequence converges and its limit is \( \frac{1}{e} \).

Key Concepts

Sequence Divergence

Understanding sequence convergence and divergence is fundamental in math. A sequence is like a list of numbers that follow a specific order. It's called a sequence because numbers are usually indexed, which means ordered by natural numbers like first, second, third, etc. A sequence diverges when, as we list more and more numbers from the sequence, the numbers start to get further away from any fixed value. You can think of it as if you're trying to catch a moving train - no matter how hard you try to reach it, it just keeps speeding away.

It's important to differentiate between sequences that diverge and those that converge. For a sequence to diverge can mean a couple of things:

• The terms of the sequence grow indefinitely large.
• The terms do not settle around a particular number.

In contrast, if a sequence settles down to approach a single value, it is said to converge. Determining whether a sequence converges or diverges often involves considering the behavior of the terms as the sequence proceeds to infinity.

Limit of a Sequence

The limit of a sequence is the value that the terms of a sequence 'settle down' to as the sequence progresses towards infinity. It’s like zeroing in on a target - the closer you get, the nearer you are to the limit. Think of the limit as the eventual destination or the goal of the sequence.

Mathematically, if a sequence \(a_n\) has a limit \(L\), it means that for any tiny positive number you can think of, as \(n\) gets larger, the difference between \(a_n\) and \(L\) becomes smaller than that number.

• It's represented as \( \lim_{n \to \infty} a_n = L \).
• For a sequence to have a limit, its terms must get progressively closer to a single number.

In our given sequence \(a_n = \left( \frac{n}{n+1} \right)^n\), we found that it converges to \( \frac{1}{e} \), meaning the terms of the sequence tend towards this value as \(n\) increases.

Logarithmic Approximation

Logarithmic approximation is a handy mathematical tool often used to make complex expressions easier to work with, especially when dealing with sequences. It involves using the properties of logarithms to simplify expressions, making them easier to analyze.

In the context of sequences and limits, logarithmic approximation can simplify how we look at sequences as they head to infinity. One useful property is that \( \ln(1+x) \approx x \) when \(x\) is small. Here's how this helps:

• Assume \(x\) is a tiny number, making this formula approximate well.
• This transformation lets us break down the sequence into simpler pieces.

For the sequence \(a_n = \left( \frac{n}{n+1} \right)^n\), by taking the natural logarithm, we turned the expression into \(n \ln\left(\frac{n}{n+1}\right)\). With the approximation \( \ln(1 + \frac{1}{n}) \approx \frac{1}{n} \), it simplified to \(-1\), helping us see the sequence’s tendencies towards the limit \( \frac{1}{e} \). This is key for understanding the convergence of the given sequence.