Let's solve each part step by step: --- ### (a) **Integral Test for Convergence** We are given the series: \[ \sum_{n=0}^\infty \frac{1}{(n+2)^5} \] **Integral Test:** If \( f(n) = \frac{1}{(n+2)^5} \) is positive, continuous, and decreasing for \( n \geq 0 \), then the convergence of the series is determined by the convergence of the improper integral: \[ \int_0^\infty \frac{1}{(x+2)^5} dx \] **Evaluate the Integral:** \[ \int_0^\infty \frac{1}{(x+2)^5} dx \] Let \( u = x+2 \implies du = dx \) When \( x = 0, u = 2 \) When \( x \to \infty, u \to \infty \) So, \[ \int_2^\infty u^{-5} du = \left[ \frac{u^{-4}}{-4} \right]_2^\infty = \left( 0 - \frac{2^{-4}}{-4} \right) = \frac{1}{4 \cdot 16} = \frac{1}{64} \] **Conclusion:** The integral converges, hence **the series converges** by the Integral Test. --- ### (b) **Evaluate the Integral** \[ \int_1^\infty \frac{2}{(x+1)^5} dx \] Let \( u = x+1 \implies du = dx \) When \( x = 1, u = 2 \) When \( x \to \infty, u \to \infty \) So, \[ \int_2^\infty 2u^{-5} du = 2 \int_2^\infty u^{-5} du = 2 \left[ \frac{u^{-4}}{-4} \right]_2^\infty = 2 \left( 0 - \frac{2^{-4}}{-4} \right) = 2 \cdot \frac{1}{4 \cdot 16} = \frac{1}{32} \] **Final Answer:** \[ \boxed{\frac{1}{32}} \] --- ### (c) **Convergence of Power Series** \[ \sum_{k=0}^\infty \frac{2^k (x^2+1)^k}{k!} \] Rewrite: \[ \sum_{k=0}^\infty \frac{[2(x^2+1)]^k}{k!} \] This is the Taylor expansion of \( e^{2(x^2+1)} \). The exponential series converges for all real \( x \): **Conclusion:** The series **converges for every \( x \in \mathbb{R} \)**. --- ### (d) **Convergence of the Integral** \[ \int_1^\infty \frac{1}{x^6+5} dx \] **Comparison Test:** For \( x \geq 1 \), \( x^6+5 > x^6 \), so \( \frac{1}{x^6+5} < \frac{1}{x^6} \). \[ \int_1^\infty \frac{1}{x^6} dx = \left[ \frac{x^{-5}}{-5} \right]_1^\infty = 0 - \left(-\frac{1}{5}\right) = \frac{1}{5} \] which converges. By comparison, **the given integral also converges**. --- ## **Summary of Answers** --- **(a):** The series converges (by the Integral Test). **(b):** The value is \( \boxed{\frac{1}{32}} \). **(c):** The series converges for all \( x \in \mathbb{R} \). **(d):** The integral converges (by comparison with \( \frac{1}{x^6} \)).