91Ó°ÊÓ

: To understand the convergence of \(r^{\prime \prime}\), we have to notice that the second derivative indicates the curvature or concavity of the function. If the function's curvature changes infinitely as r increases, the sequence will diverge. Otherwise, it will converge. Since no additional information about the function that is being derived is given, we cannot determine the values of r for which \(r^{\prime \prime}\) converges. More context or information is needed.

: To analyze the convergence of \(n r^{n}\), we will use the root test. The root test states that if the nth root of the absolute value of the terms of a series converges, then the series converges. More formally, if we have a series \(\displaystyle\sum_{n=1}^{\infty} a_n\), and \( L = \lim_{n\to\infty} |a_n|^{\frac{1}{n}}, \) then, 1. If \(L < 1\), the series converges. 2. If \(L > 1\), the series diverges. 3. If \(L = 1\), the test is inconclusive. In our case, we have \(\displaystyle\sum_{n=1}^{\infty} n r^{n}\). Applying the root test, we get: \( L = \lim_{n\to\infty} \left| n r^{n} \right|^{\frac{1}{n}} = \lim_{n\to\infty} \left( n^{\frac{1}{n}} \cdot |r| \right). \) Now, we need to find the limit as n approaches infinity for \(n^{\frac{1}{n}}\): \( \lim_{n\to\infty} n^{\frac{1}{n}} = 1. \) Hence, \( L = \lim_{n\to\infty} \left( n^{\frac{1}{n}} \cdot |r| \right) = 1 \cdot |r| = |r|. \) Now, we compare L to 1: 1. If \(|r| < 1\), the series converges. 2. If \(|r| > 1\), the series diverges. 3. If \(|r| = 1\), the test is inconclusive. So, for our given series, the values of r for which the series converges are those that satisfy the inequality \(|r| < 1\). This means r is in the open interval \((-1, 1)\). The values of r for which the series diverges are those outside the interval, i.e., \(r \le -1\) and \(r \ge 1\). For \(r = -1\) and \(r = 1\), the root test is inconclusive, so we cannot make any assertion about the convergence of the series.