Let's solve each part step by step: --- ### **4. The series \(\sum_{n=1}^\infty \frac{1}{(n^2+1)^{3/2}}\) is convergent and satisfies the hypotheses of the integral test.** #### **a. Find \(\int_{4}^{\infty} \frac{1}{(x^2+1)^{3/2}} dx\).** Let’s compute the improper integral: \[ \int_{4}^{\infty} \frac{1}{(x^2+1)^{3/2}} dx \] Let’s use the substitution \(x = \tan \theta\), so \(dx = \sec^2\theta d\theta\), and \(x^2+1 = \tan^2\theta + 1 = \sec^2\theta\): \[ (x^2+1)^{3/2} = (\sec^2\theta)^{3/2} = \sec^3\theta \] Therefore, \[ \int \frac{1}{(x^2+1)^{3/2}} dx = \int \frac{1}{\sec^3\theta} \cdot \sec^2\theta d\theta = \int \cos\theta d\theta = \sin\theta + C \] Now, \(x = \tan\theta \implies \theta = \arctan x\): So, \[ \int_{4}^{\infty} \frac{1}{(x^2+1)^{3/2}} dx = \left. \sin(\arctan x) \right|_{x=4}^{x \to \infty} \] Recall that \(\sin(\arctan x) = \frac{x}{\sqrt{x^2+1}}\). So, \[ \int_{4}^{\infty} \frac{1}{(x^2+1)^{3/2}} dx = \left. \frac{x}{\sqrt{x^2+1}} \right|_{4}^{\infty} \] As \(x \to \infty\), \(\frac{x}{\sqrt{x^2+1}} \to 1\), and at \(x = 4\), \(\frac{4}{\sqrt{17}}\): \[ \int_{4}^{\infty} \frac{1}{(x^2+1)^{3/2}} dx = 1 - \frac{4}{\sqrt{17}} \] Now, \(\sqrt{17} \approx 4.12311\): \[ \frac{4}{\sqrt{17}} \approx \frac{4}{4.12311} \approx 0.97014 \] So, \[ 1 - 0.97014 = 0.02986 \] **Final answer for (a):** \[ \boxed{0.02986} \] --- #### **b. Find \(s_4\) for the series, and use the result from part (a) to estimate the sum of the series. Round your answer to 5 decimal places.** First, calculate \(s_4\): \[ s_4 = \sum_{n=1}^4 \frac{1}{(n^2+1)^{3/2}} \] Let's compute each term: - \(n=1\): \(\frac{1}{(1^2+1)^{3/2}} = \frac{1}{2^{3/2}} = \frac{1}{2\sqrt{2}} \approx 0.35355\) - \(n=2\): \(\frac{1}{(4+1)^{3/2}} = \frac{1}{5^{3/2}} = \frac{1}{5\sqrt{5}} \approx 0.08944\) - \(n=3\): \(\frac{1}{(9+1)^{3/2}} = \frac{1}{10^{3/2}} = \frac{1}{10\sqrt{10}} \approx 0.03162\) - \(n=4\): \(\frac{1}{(16+1)^{3/2}} = \frac{1}{17^{3/2}} = \frac{1}{17\sqrt{17}} \approx 0.01419\) Sum: \[ s_4 \approx 0.35355 + 0.08944 + 0.03162 + 0.01419 \approx 0.48880 \] Now, estimate the sum of the series using the integral test estimate: \[ \text{Sum} \approx s_4 + \int_{4}^{\infty} \frac{1}{(x^2+1)^{3/2}} dx \] \[ \approx 0.48880 + 0.02986 = 0.51866 \] **Final answer for (b):** \[ \boxed{0.51866} \] --- ### **5. Determine whether the series converges or diverges.** #### **a. \(\sum_{n=1}^\infty \frac{2n+1}{\sqrt{9n^4 - n^2}}\)** For large \(n\): - Numerator: \(2n+1 \sim 2n\) - Denominator: \(\sqrt{9n^4 - n^2} \approx \sqrt{9n^4} = 3n^2\) So, term behaves like: \[ \frac{2n}{3n^2} = \frac{2}{3n} \] \(\sum \frac{1}{n}\) diverges (harmonic series), so this series **diverges**. --- #### **b. \(\sum_{n=1}^\infty \frac{3^n \sin^2 n}{n^4 + 4n}\)** For large \(n\): - Numerator: \(3^n \sin^2 n\), which is bounded between \(0\) and \(3^n\) - Denominator: \(n^4 + 4n \sim n^4\) But \(3^n\) grows much faster than \(n^4\). The terms do **not** go to zero; in fact, they grow exponentially. So, this series **diverges**. --- #### **c. \(\sum_{n=1}^\infty \frac{\sqrt{n^2+9}}{n^3+8} \tan^{-1}(e^n)\)** For large \(n\): - \(\sqrt{n^2+9} \sim n\) - \(n^3+8 \sim n^3\) - \(\tan^{-1}(e^n) \to \frac{\pi}{2}\) as \(n \to \infty\) So, for large \(n\), the term behaves like: \[ \frac{n}{n^3} \cdot \frac{\pi}{2} = \frac{\pi}{2n^2} \] \(\sum \frac{1}{n^2}\) converges, so this series **converges**. --- ## **Summary of Answers:** ### **4a.** \(\boxed{0.02986}\) ### **4b.** \(\boxed{0.51866}\) ### **5a.** Diverges ### **5b.** Diverges ### **5c.** Converges Let me know if you want detailed working for any step!