Let's analyze the convergence of each series: --- ### **(3) \(\sum_{n=0}^{+\infty} \frac{(-1)^n}{3n+7}\)** This is an **alternating series** (sign alternates due to \( (-1)^n \)). #### **Alternating Series Test (Leibniz Test):** A series \( \sum (-1)^n a_n \) converges if: 1. \( a_n \ge 0 \) 2. \( a_{n+1} \le a_n \) for all \( n \) (monotonically decreasing) 3. \( \lim_{n \to \infty} a_n = 0 \) Here, \( a_n = \frac{1}{3n+7} \): - \( a_n \ge 0 \): Yes. - \( a_{n+1} < a_n \): Yes, the denominator increases so terms decrease. - \( \lim_{n \to \infty} a_n = 0 \): Yes, as \( n \to \infty \), \( 3n+7 \to \infty \) so \( a_n \to 0 \). **Conclusion:** **The series converges by the Alternating Series Test.** --- ### **(4) \(\sum_{n=1}^{+\infty} \frac{(-10)^n}{n^{10}}\)** Let's use the **root test**: \[ \lim_{n \to \infty} \sqrt[n]{\left| \frac{(-10)^n}{n^{10}} \right|} = \lim_{n \to \infty} \frac{10}{n^{10/n}} \] But \( n^{10/n} \to 1 \) as \( n \to \infty \), so limit is 10. - Limit \( > 1 \) ⇒ **series diverges by the root test**. --- ### **(5) \(\sum_{n=1}^{+\infty} \frac{(-10)^n}{n!}\)** Again, **root test**: \[ \lim_{n \to \infty} \sqrt[n]{\left| \frac{(-10)^n}{n!} \right|} = \lim_{n \to \infty} \frac{10}{(n!)^{1/n}} \] But \( n! \) grows much faster than any exponential, so this limit is 0. - Limit \( < 1 \) ⇒ **series converges absolutely**. --- ### **(6) \(\sum_{n=1}^{+\infty} \frac{(-10)^n}{n^n}\)** Again, **root test**: \[ \lim_{n \to \infty} \sqrt[n]{\left| \frac{(-10)^n}{n^n} \right|} = \lim_{n \to \infty} \frac{10}{n} = 0 \] - Limit \( < 1 \) ⇒ **series converges absolutely**. --- ## **Summary Table** | Series | Converges or Diverges | Reason | |--------|----------------------|--------| | (3) \( \sum_{n=0}^{+\infty} \frac{(-1)^n}{3n+7} \) | **Converges** | Alternating Series Test | | (4) \( \sum_{n=1}^{+\infty} \frac{(-10)^n}{n^{10}} \) | **Diverges** | Root Test (\(L=10>1\)) | | (5) \( \sum_{n=1}^{+\infty} \frac{(-10)^n}{n!} \) | **Converges** | Root Test (\(L=0<1\)) | | (6) \( \sum_{n=1}^{+\infty} \frac{(-10)^n}{n^n} \) | **Converges** | Root Test (\(L=0<1\)) | If you need a more detailed explanation for any particular series, let me know!

Let's analyze each series in detail, including the calculations and reasoning. --- ### **Series (3): \(\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{3n+7}\)** **Type:** Alternating series with \(a_n = \frac{1}{3n+7}\). **Step 1: Check \(a_n \ge 0\):** Yes, since numerator is positive and denominator is positive for all \(n\). **Step 2: Monotonic decreasing \(a_n\):** \[ a_{n+1} = \frac{1}{3(n+1) + 7} = \frac{1}{3n + 10} \quad \text{vs} \quad a_n = \frac{1}{3n+7} \] Since \(3n+10 > 3n+7\), it follows that: \[ a_{n+1} < a_n \] Thus, \(\{a_n\}\) is decreasing. **Step 3: Limit of \(a_n\):** \[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{3n+7} = 0 \] **Conclusion:** By the **Alternating Series Test** (Leibniz criterion): \[ \boxed{ \text{Series (3) converges} } \] --- ### **Series (4): \(\displaystyle \sum_{n=1}^\infty \frac{(-10)^n}{n^{10}}\)** **Step 1: Absolute value of general term:** \[ \left| \frac{(-10)^n}{n^{10}} \right| = \frac{10^n}{n^{10}} \] **Step 2: Use the root test:** \[ L = \lim_{n \to \infty} \sqrt[n]{\left| a_n \right|} = \lim_{n \to \infty} \sqrt[n]{\frac{10^n}{n^{10}}} = \lim_{n \to \infty} \frac{10}{n^{10/n}} \] Because: \[ n^{10/n} = e^{(10/n) \ln n} \] and as \(n \to \infty\), \[ (10/n) \ln n \to 0 \] so \[ n^{10/n} \to e^0 = 1 \] Therefore: \[ L = \frac{10}{1} = 10 \] **Step 3: Conclusion:** Since \(L = 10 > 1\), the series **diverges** by the root test. --- ### **Series (5): \(\displaystyle \sum_{n=1}^\infty \frac{(-10)^n}{n!}\)** **Step 1: Analyze the absolute value:** \[ |a_n| = \frac{10^n}{n!} \] **Step 2: Root test:** \[ L = \lim_{n \to \infty} \sqrt[n]{\frac{10^n}{n!}} = \lim_{n \to \infty} \frac{10}{(n!)^{1/n}} \] **Step 3: Behavior of \((n!)^{1/n}\):** By Stirling's approximation: \[ n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \] then: \[ (n!)^{1/n} \sim \left(\sqrt{2\pi n}\right)^{1/n} \left(\frac{n}{e}\right) \] As \(n \to \infty\), \[ \left(\sqrt{2\pi n}\right)^{1/n} \to 1 \] and so: \[ (n!)^{1/n} \sim \frac{n}{e} \] Thus: \[ L = \lim_{n \to \infty} \frac{10}{n/e} = \lim_{n \to \infty} \frac{10 e}{n} = 0 \] Since \(L=0 < 1\), by the root test, the **series converges absolutely**. **Conclusion:** \[ \boxed{ \text{Series (5) converges absolutely} } \] --- ### **Series (6): \(\displaystyle \sum_{n=1}^\infty \frac{(-10)^n}{n^n}\)** **Step 1: Absolute value:** \[ |a_n| = \frac{10^n}{n^n} \] **Step 2: Use root test:** \[ L = \lim_{n \to \infty} \sqrt[n]{\frac{10^n}{n^n}} = \lim_{n \to \infty} \frac{10}{n} \] **Step 3: Limit:** \[ \lim_{n \to \infty} \frac{10}{n} = 0 \] Since \(L=0<1\), the series **converges absolutely**. --- ### **Final Summary:** | Series | Final Result | Justification | |---------|----------------|----------------| | (3) \(\sum_{n=0}^\infty \frac{(-1)^n}{3n+7}\) | **Converges** | Alternating Series Test (terms decrease to 0) | | (4) \(\sum_{n=1}^\infty \frac{(-10)^n}{n^{10}}\) | **Diverges** | Root test limit \(L=10>1\) | | (5) \(\sum_{n=1}^\infty \frac{(-10)^n}{n!}\) | **Converges absolutely** | Root test limit \(L=0<1\) | | (6) \(\sum_{n=1}^\infty \frac{(-10)^n}{n^n}\) | **Converges absolutely** | Root test limit \(L=0<1\) | Let me know if you'd like further clarification!