91Ó°ÊÓ

Note that by definition of \(A_{n}\), we have \[A_{n+1}-A_{n} = \left(1+\frac{1}{2}+\cdots+\frac{1}{n+1}\right) - \left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right).\] Notice that when subtracting the terms, all of them will cancel out except for the last one, which is \(\frac{1}{n+1}\). Thus, \[A_{n+1}-A_{n} = \frac{1}{n+1}.\]

Now we need to show that the difference \(\frac{1}{n+1}\) tends to 0 as \(n\) tends to infinity. Since the numerator is a constant 1, and the denominator increases without bound, we have that \[\frac{1}{n+1} \rightarrow 0 \text{ as } n \rightarrow \infty.\]

In order to show that the sequence \((A_{n})\) is not Cauchy, let's first recall the definition of a Cauchy sequence. A sequence \((A_{n})\) is said to be Cauchy if for all \(\epsilon > 0\), there exists an \(N \in \mathbb{N}\) such that for all \(m, n > N\), we have \[|A_{m} - A_{n}| < \epsilon.\]

To show that \((A_{n})\) is not a Cauchy sequence, we will find an \(\epsilon > 0\) for which we cannot find an integer \(N\) satisfying the Cauchy property. Let's take \(\epsilon = 1\). We need to show that for any \(N \in \mathbb{N}\), we can find \(m, n > N\) such that \[|A_{m} - A_{n}| \geq 1.\] As a strategy, we try to find a suitable \(n\) for each choice of \(m\). We can express the difference \(|A_m-A_n|\) as a difference of sums, \[|A_m - A_n| = \left|\sum_{k=n+1}^m \frac{1}{k}\right|.\] By the definition of \(A_{n}\), each partial sum is positive and increasing. That is, \(A_m - A_n \geq 0\) for all \(m > n\). Thus, it is sufficient to consider the sum without the absolute value, \[\sum_{k=n+1}^m \frac{1}{k}.\] Now, we know that the harmonic series - i.e., the sum \(1 + \frac{1}{2} + \cdots + \frac{1}{k}\) (which are precisely the terms in the sequence \(A_{n}\)) - diverges. This means that we can always find terms with a difference greater than 1 if we sum up long enough. So, let \(m = N + t\), where \(t\) is a positive integer. Then the sum becomes \[\sum_{k=N+1}^{N+t} \frac{1}{k}.\] Since the harmonic series diverges, by choosing a large enough \(t\) (depending on \(N\)), we can make the sum above greater than or equal to 1. Therefore, we have found \(m, n > N\) with \(|A_m - A_n| \geq 1\). This means that the sequence \((A_{n})\) is not a Cauchy sequence, even though \((A_{n+1}-A_{n}) \rightarrow 0\) as \(n \rightarrow \infty\).