Problem 119 - Thomas Calculus

Question

Which of the sequences in Exercises \(115-124\) converge, and which diverge? Give reasons for your answers. $$ a_{n}=\left((-1)^{n}+1\right)\left(\frac{n+1}{n}\right) $$

Step-by-Step Solution

Answer

The sequence diverges because it does not approach a single limit.

Step 1: Understand the Sequence

The given sequence is \( a_n = ((-1)^n + 1)\left(\frac{n+1}{n}\right) \). We need to analyze its behavior as \( n \to \infty \). Notice that \((-1)^n\) results alternately in \(-1\) and \(1\), making the sequence different for even and odd \(n\).

Step 2: Analyze for Even n

For even \(n\), \((-1)^n = 1\). The expression simplifies to:\[a_n = (1 + 1) \left(\frac{n+1}{n}\right) = 2 \left(\frac{n+1}{n}\right)\]Simplifying further:\[= 2 \left(1 + \frac{1}{n}\right) = 2 + \frac{2}{n}\]As \( n \to \infty \), \( \frac{2}{n} \to 0 \). Therefore, \( a_n \to 2 \).

Step 3: Analyze for Odd n

For odd \(n\), \((-1)^n = -1\). The expression becomes:\[a_n = ( -1 + 1) \left(\frac{n+1}{n}\right) = 0\]So, for odd \(n\), the sequence \(a_n = 0\).

Step 4: Conclude Convergence or Divergence

The sequence \(a_n\) approaches 2 for even values of \(n\) and 0 for odd values of \(n\). Because \(a_n\) doesn't approach a single value as \(n\) becomes large, the sequence does not converge.

Key Concepts

Alternating Sequence

An alternating sequence is a special type of sequence where the signs of the terms alternate. This happens due to a component like \((-1)^n\) which either equals \(-1\) or \(1\) based on whether \(n\) is odd or even. This alternating factor results in the sequence behavior significantly differing between consecutive terms. In the case of the sequence \(a_n = ((-1)^n + 1)\big(\frac{n+1}{n}\big)\), this is evident as even \(n\) results in a positive term, while odd \(n\) leads to zero.

• For even \(n\), the term simplifies by considering \((-1)^n = 1\), resulting in \(a_n = 2\big(1 + \frac{1}{n}\big)\)
• For odd \(n\), \((-1)^n = -1\) yields \(a_n = 0\)

Because the sequence changes behavior depending on whether \(n\) is odd or even, it alternates between these states, making it an alternating sequence. Understanding this helps in analyzing its convergence properties.

Limit of a Sequence

The limit of a sequence is an essential concept in understanding whether a sequence converges to a specific value. As \(n\) tends toward infinity, the terms of the sequence ideally approach a single finite limit. Let's examine what happens to both parts of our sequence \(a_n\).
For our sequence, depending on whether \(n\) is odd or even, the term changes:

• For even \(n\): \(a_n = 2 + \frac{2}{n}\). As \(n \ o \infty\), the fraction \(\frac{2}{n} \ o 0\), and \(a_n\) approaches 2.
• For odd \(n\): \(a_n = 0\). This stays constant as \(n\) increases.

A proper limit for the sequence must be the same regardless of whether \(n\) is odd or even. Here it is not the case, since the limiting behavior changes. Hence, the overall sequence does not have a single limit.

Convergent and Divergent Sequences

A convergent sequence is one where the terms approach a finite value as \(n\) gets very large. Conversely, a divergent sequence does not settle on a specific value. Let’s assess whether our sequence converges or diverges.
A sequence converges if all its terms stabilize to one single limit. Our sequence \(a_n = ((-1)^n + 1)\big(\frac{n+1}{n}\big)\), does not meet this criterion.

• Even \(n\) values push the sequence towards 2.
• Odd \(n\) values result in constant 0.
• The sequence lacks a consistent limiting value as all terms do not unify around a single point.

Therefore, due to varying limits based on the parity of \(n\), \(a_n\) is a divergent sequence. Recognizing divergent sequences is crucial for problems where understanding limits is necessary.