Let's solve each problem step by step **using the Limit Comparison Test** as instructed. --- ### **19.** \(\sum_{n=1}^{\infty} \frac{2n^2 - 1}{3n^5 + 2n + 1}\) #### **Step 1: Identify dominant terms** - Numerator: \(2n^2\) (as \(n \to \infty\)) - Denominator: \(3n^5\) (as \(n \to \infty\)) So, for large \(n\), \(\frac{2n^2 - 1}{3n^5 + 2n + 1} \sim \frac{2n^2}{3n^5} = \frac{2}{3n^3}\). #### **Step 2: Choose comparison series** Compare with \(b_n = \frac{1}{n^3}\), which is a **p-series** with \(p = 3 > 1\) (converges). #### **Step 3: Compute the limit** \[ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{2n^2 - 1}{3n^5 + 2n + 1}}{\frac{1}{n^3}} = \lim_{n \to \infty} \frac{(2n^2 - 1)n^3}{3n^5 + 2n + 1} \] \[ = \lim_{n \to \infty} \frac{2n^5 - n^3}{3n^5 + 2n + 1} \] Divide numerator and denominator by \(n^5\): \[ = \lim_{n \to \infty} \frac{2 - \frac{1}{n^2}}{3 + \frac{2}{n^4} + \frac{1}{n^5}} = \frac{2}{3} \] #### **Step 4: Conclude** Limit is a positive, finite number (\(\frac{2}{3}\)), so both series **converge or diverge together**. Since \(\sum \frac{1}{n^3}\) converges, **the given series converges**. --- ### **21.** \(\sum_{n=1}^{\infty} \frac{n + 3}{n(n+2)}\) #### **Step 1: Identify dominant terms** - Numerator: \(n\) - Denominator: \(n^2\) So for large \(n\), \(\frac{n + 3}{n(n+2)} \sim \frac{n}{n^2} = \frac{1}{n}\). #### **Step 2: Choose comparison series** Compare with \(b_n = \frac{1}{n}\) (**harmonic series**, diverges). #### **Step 3: Compute the limit** \[ \lim_{n \to \infty} \frac{\frac{n+3}{n(n+2)}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n+3}{n(n+2)} \cdot n = \lim_{n \to \infty} \frac{n(n+3)}{n(n+2)} = \lim_{n \to \infty} \frac{n+3}{n+2} = 1 \] #### **Step 4: Conclude** Limit is 1 (positive, finite). The series behaves like the harmonic series, which **diverges**. **Final Answer:** The given series **diverges**. --- ### **23.** \(\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n^2 + 1}}\) #### **Step 1: Identify dominant terms** For large \(n\), \(\sqrt{n^2+1} \sim n\), so denominator is \(n \cdot n = n^2\). So, \(\frac{1}{n\sqrt{n^2 + 1}} \sim \frac{1}{n^2}\). #### **Step 2: Choose comparison series** Compare with \(b_n = \frac{1}{n^2}\), a **p-series** with \(p = 2 > 1\) (converges). #### **Step 3: Compute the limit** \[ \lim_{n \to \infty} \frac{\frac{1}{n\sqrt{n^2+1}}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^2}{n\sqrt{n^2+1}} = \lim_{n \to \infty} \frac{n}{\sqrt{n^2+1}} \] \[ = \lim_{n \to \infty} \frac{1}{\sqrt{1 + \frac{1}{n^2}}} = 1 \] #### **Step 4: Conclude** Limit is 1. The comparison series converges, therefore, **Final Answer:** The given series **converges**. --- ## **Summary of Answers** - **19:** Converges - **21:** Diverges - **23:** Converges Let me know if you want further explanation or details on any step!