Let's tackle each part **step by step**: --- ### **(a) Convergence and Divergence of \(\sum_{n=1}^{\infty} \frac{(rn)^n}{n!}\)** #### **Convergence (\(0 \leq r < \frac{1}{e}\))** Consider the general term \(a_n = \frac{(rn)^n}{n!}\). Let's use the **Ratio Test**: \[ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{(r(n+1))^{n+1}}{(n+1)!} \cdot \frac{n!}{(rn)^n} = \lim_{n \to \infty} \frac{(r(n+1))^{n+1}}{(n+1)(rn)^n} \] Let's simplify: \[ = \lim_{n \to \infty} \frac{r^{n+1} (n+1)^{n+1}}{(n+1) r^n n^n} = \lim_{n \to \infty} r \cdot \frac{(n+1)^{n+1}}{(n+1)n^n} = r \lim_{n \to \infty} \frac{(n+1)^n}{n^n} = r \lim_{n \to \infty} \left( \frac{n+1}{n} \right)^n = r \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = r \cdot e \] **By the ratio test:** - If \(re < 1\) (\(r < \frac{1}{e}\)), the series converges. - If \(re > 1\) (\(r > \frac{1}{e}\)), the series diverges. --- ### **(b) If \(0 \leq \rho < \frac{1}{e}\), then \(a_n < \frac{(rn)^n}{n!}\) for large \(n\)** Given \(\rho = \lim_{n \to \infty} \sqrt[n]{\frac{n! a_n}{n^n}}\) and \(\rho < \frac{1}{e}\): - Pick \(r\) with \(\rho < r < \frac{1}{e}\). - By the definition of limit, there exists \(N\) such that for all \(n > N\): \[ \sqrt[n]{\frac{n! a_n}{n^n}} < r \implies \frac{n! a_n}{n^n} < r^n \implies a_n < \frac{(r n)^n}{n!} \] - Therefore, for all \(n > N\), \(a_n < \frac{(r n)^n}{n!}\). --- ### **(c) If \(\rho > \frac{1}{e}\), then \(a_n > \frac{(rn)^n}{n!}\) for large \(n\)** Given \(\rho = \lim_{n \to \infty} \sqrt[n]{\frac{n! a_n}{n^n}}\) and \(\rho > \frac{1}{e}\): - Pick \(r\) with \(\frac{1}{e} < r < \rho\). - By the definition of limit, there exists \(N\) such that for all \(n > N\): \[ \sqrt[n]{\frac{n! a_n}{n^n}} > r \implies \frac{n! a_n}{n^n} > r^n \implies a_n > \frac{(r n)^n}{n!} \] - Therefore, for all \(n > N\), \(a_n > \frac{(r n)^n}{n!}\). --- ### **(d) Prove the Barnum Test** **Statement:** Suppose \(\{a_n\}_{n=1}^\infty\) is a sequence of positive real numbers and \(\rho = \lim_{n \to \infty} \sqrt[n]{\frac{n! a_n}{n^n}}\) exists. - If \(\rho < \frac{1}{e}\), then \(\sum a_n\) converges. - If \(\rho > \frac{1}{e}\), then \(\sum a_n\) diverges. **Proof:** - From (b): If \(\rho < \frac{1}{e}\), for some \(r < \frac{1}{e}\), \(a_n < \frac{(r n)^n}{n!}\) for large \(n\). The series \(\sum \frac{(r n)^n}{n!}\) converges (from part a), so by comparison, \(\sum a_n\) converges. - From (c): If \(\rho > \frac{1}{e}\), for some \(r > \frac{1}{e}\), \(a_n > \frac{(r n)^n}{n!}\) for large \(n\). The series \(\sum \frac{(r n)^n}{n!}\) diverges, so by comparison, \(\sum a_n\) diverges. --- ### **(e) Apply the Barnum Test to:** \[ \sum_{n=1}^\infty \frac{\left( \frac{n^2 + 2}{n + 1} \arctan(3n) \right)^n}{(n+1)!} \] Let \(a_n = \frac{\left( \frac{n^2 + 2}{n + 1} \arctan(3n) \right)^n}{(n+1)!}\). Compute \[ \rho = \lim_{n \to \infty} \sqrt[n]{\frac{n! a_n}{n^n}} \] First, write \(a_n\) in terms of \(n!:\) \[ a_n = \frac{\left( \frac{n^2 + 2}{n + 1} \arctan(3n) \right)^n}{(n+1)!} \] \[ \Rightarrow n! a_n = \frac{n!}{(n+1)!} \left( \frac{n^2 + 2}{n + 1} \arctan(3n) \right)^n = \frac{1}{n+1} \left( \frac{n^2 + 2}{n + 1} \arctan(3n) \right)^n \] So \[ \frac{n! a_n}{n^n} = \frac{1}{n+1} \left( \frac{n^2 + 2}{n + 1} \frac{\arctan(3n)}{n} \right)^n \] Now, - \(\arctan(3n) \to \frac{\pi}{2}\) as \(n \to \infty\), - \(\frac{n^2 + 2}{n + 1} \sim n\). Thus, \[ \frac{n^2 + 2}{n + 1} \sim n, \] \[ \frac{n^2 + 2}{n + 1} \cdot \frac{\arctan(3n)}{n} \sim \frac{n \cdot \frac{\pi}{2}}{n} = \frac{\pi}{2} \] So, \[ \frac{n! a_n}{n^n} \sim \frac{1}{n+1} \left( \frac{\pi}{2} \right)^n \] Now, \[ \sqrt[n]{\frac{n! a_n}{n^n}} \sim \left( \frac{\pi}{2} \right) \cdot (n+1)^{-1/n} \] As \(n \to \infty\), \((n+1)^{-1/n} \to 1\). Therefore, \[ \rho = \frac{\pi}{2} \] Since \(\frac{\pi}{2} > \frac{1}{e}\) (since \(\frac{1}{e} \approx 0.3679\) and \(\frac{\pi}{2} \approx 1.5708\)), **By the Barnum Test, the series diverges.** --- ## **Summary of Answers** **(a)** The series \(\sum_{n=1}^\infty \frac{(rn)^n}{n!}\) converges if \(0 \leq r < \frac{1}{e}\) and diverges if \(r > \frac{1}{e}\). **(b)** For \(0 \leq \rho < \frac{1}{e}\), there exists \(r < \frac{1}{e}\) and \(N\) so that for \(n > N\), \(a_n < \frac{(rn)^n}{n!}\). **(c)** For \(\rho > \frac{1}{e}\), there exists \(r > \frac{1}{e}\) and \(N\) so that for \(n > N\), \(a_n > \frac{(rn)^n}{n!}\). **(d)** If \(\rho < \frac{1}{e}\), \(\sum a_n\) converges; if \(\rho > \frac{1}{e}\), \(\sum a_n\) diverges. **(e) Final Answer:** \[ \sum_{n=1}^\infty \frac{\left(\frac{n^2+2}{n+1} \arctan(3n)\right)^n}{(n+1)!} \] **diverges** by the Barnum Test, since \(\rho = \frac{\pi}{2} > \frac{1}{e}\). --- Let me know if you want a more detailed explanation on any specific step!