Problem 32 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The sequence converges to 1.

Step 1: Analyze the Original Sequence

The sequence is given by \( a_n = \frac{n + (-1)^n}{n} \). Let's break down this expression. We have the numerator \( n + (-1)^n \), which means that when \( n \) is even, \( (-1)^n = 1 \) and when \( n \) is odd, \( (-1)^n = -1 \). Thus the sequence takes values \( \frac{n+1}{n} \) for even \( n \) and \( \frac{n-1}{n} \) for odd \( n \).

Step 2: Separate the Cases: Even and Odd

Let's evaluate two subsequences: one for even \( n \) and one for odd \( n \). For even \( n \), \( a_n = \frac{n+1}{n} = 1 + \frac{1}{n} \). For odd \( n \), \( a_n = \frac{n-1}{n} = 1 - \frac{1}{n} \).

Step 3: Determine the Limit for Even Terms

For even \( n \), the expression \( 1 + \frac{1}{n} \) approaches \( 1 \) as \( n \) goes to infinity. This is because \( \frac{1}{n} \) approaches zero.

Step 4: Determine the Limit for Odd Terms

For odd \( n \), the expression \( 1 - \frac{1}{n} \) also approaches \( 1 \) as \( n \) goes to infinity since \( \frac{1}{n} \) approaches zero.

Step 5: Conclude with the Overall Limit

Since both subsequences (odd and even terms) converge to \( 1 \), the whole sequence \( a_n \) converges to \( 1 \).

Key Concepts

Convergence of Sequences

The concept of convergence is crucial when analyzing sequences in mathematics. When a sequence converges, it means that as we move further along the sequence—keeping the index of the sequence increasing indefinitely—the terms of the sequence approach a specific value. This specific value is known as the "limit" of the sequence. The process of convergence is analyzed by considering the behavior of the sequence's terms when the index tends to infinity.

• If the sequence gets closer and closer to a certain number, we say it converges to that number.
• If the sequence doesn't approach any limit and stays erratic as you move along the terms or goes off to infinity, it's said to diverge.

In our example, the sequence \( a_n = \frac{n + (-1)^n}{n} \) is studied to determine convergence. By examining the separate behaviors of even and odd indices, we discover that both converge to the same limit. Therefore, the whole sequence converges to 1.

Limits of Sequences

In mathematics, the limit of a sequence refers to the value that the terms of a sequence get closer to as the index increases. Formally, for a sequence \( a_n \), we say that \( L \) is the limit if for any tiny positive number \( \varepsilon \), there is a position in the sequence beyond which all terms are within \( \varepsilon \) of \( L \). This exciting notion helps us understand the long-term behavior of sequences.
In the provided exercise, we find:

• For even n: The formula \( \frac{n+1}{n} = 1 + \frac{1}{n} \) shows that as \( n \) becomes large, \( \frac{1}{n} \) becomes very small, causing the sequence to approach 1.
• Similarly, for odd n: \( \frac{n-1}{n} = 1 - \frac{1}{n} \), the same logic applies, showing its limit is also 1.

Therefore, since both subsequences converge to the same limit 1, the entire sequence \( a_n \) also tends to the limit of 1.

Even and Odd Sequences

Understanding even and odd sequences often relies on recognizing differences in behavior for terms depending on whether their indices are even or odd. This separation deals with the principle that a sequence might behave differently if you only consider even indexed terms or odd indexed terms.
For the sequence \( a_n = \frac{n + (-1)^n}{n} \) in our exercise, there is a clear distinction based on the index:

• When \( n \) is even, \( (-1)^{n} = 1 \), transforming the sequence into \( \frac{n+1}{n} \).
• When \( n \) is odd, \( (-1)^{n} = -1 \), giving us \( \frac{n-1}{n} \).

This breakdown allows us to assess each subsequence independently regarding their limits. Since both subsequences reach the same limit, it simplifies the problem—confirming that despite the sequence's alternating behavior, convergence to a single limit, 1, is seen in the entire sequence. This analysis is key in handling more complex problems where sequences demonstrate varied behavior based on the parities of their terms.