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In his problem you must attempt to use the Ratio Test 0 decide whether the seri

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:ix ed un gs
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E. The Ratio Testis inconclusive, but the series diverges by another test or ests J a
F. The Ratio Test is inconclusive, but the series converges conditionally by another tes or tests
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VIPSolutions · Accepted Answer

Let's solve the problem step by step:

---

### **Step 1: Write the Ratio Test Expression**

Given the series:
$$
\sum_{n=1}^{\infty} \frac{11^n}{(n+1)7^{2n+1}}
$$
Let $ a_n = \frac{11^n}{(n+1)7^{2n+1}} $.

The ratio test requires us to compute:
$$
L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|
$$

Let's compute $ a_{n+1} $:

$$
a_{n+1} = \frac{11^{n+1}}{(n+2)7^{2(n+1)+1}} = \frac{11^{n+1}}{(n+2)7^{2n+3}}
$$

So,
$$
\frac{a_{n+1}}{a_n} = \frac{11^{n+1}}{(n+2)7^{2n+3}} \cdot \frac{(n+1)7^{2n+1}}{11^n}
$$

---

**Explanation:**  
We set up the ratio of consecutive terms and substitute the expressions for $ a_n $ and $ a_{n+1} $.

---

### **Step 2: Simplify the Ratio and Take the Limit**

Let's simplify:
$$
\frac{a_{n+1}}{a_n} = \frac{11^{n+1}}{11^n} \cdot \frac{(n+1)}{(n+2)} \cdot \frac{7^{2n+1}}{7^{2n+3}}
$$

Now, simplify each part:
- $ \frac{11^{n+1}}{11^n} = 11 $
- $ \frac{7^{2n+1}}{7^{2n+3}} = 7^{(2n+1)-(2n+3)} = 7^{-2} = \frac{1}{49} $
- $ \frac{n+1}{n+2} $ approaches 1 as $ n \to \infty $

So,
$$
\left| \frac{a_{n+1}}{a_n} \right| = 11 \cdot \frac{1}{49} \cdot \frac{n+1}{n+2}
$$

Take the limit as $ n \to \infty $:

$$
L = 11 \cdot \frac{1}{49} \cdot 1 = \frac{11}{49}
$$

---

**Explanation:**  
We simplified the ratio and found the limit as $ n \to \infty $. The value is $ \frac{11}{49} $.

---

### **Final Answer**

- The limit $ L = \frac{11}{49} $.
- Because $ L < 1 $, by the Ratio Test, the series **converges absolutely**.

**The answer is:**
- **L = 11/49**
- **A. The Ratio Test says that the series converges absolutely.**

VIPSolutions · Answer

# Solution to the Series Convergence Problem

Let’s go through the problem step by step to determine the convergence of the series using the Ratio Test.

---

## **Step 1: Write the Ratio Test Expression**

Given the series:
$$
\sum_{n=1}^{\infty} \frac{11^n}{(n+1)7^{2n+1}}
$$
Define:
$$
a_n = \frac{11^n}{(n+1)7^{2n+1}}
$$

Using the Ratio Test, we need to compute:
$$
L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|
$$

Now, calculate $ a_{n+1} $:
$$
a_{n+1} = \frac{11^{n+1}}{(n+2)7^{2(n+1)+1}} = \frac{11^{n+1}}{(n+2)7^{2n+3}}
$$

Thus, the ratio becomes:
$$
\frac{a_{n+1}}{a_n} = \frac{11^{n+1}}{(n+2)7^{2n+3}} \cdot \frac{(n+1)7^{2n+1}}{11^n}
$$

---

### **Explanation of Step 1**
We defined the series terms and expressed the ratio needed for the Ratio Test. This setup prepares us to analyze the limit.

---

## **Step 2: Simplify the Ratio and Take the Limit**

Now, we can simplify the ratio:
$$
\frac{a_{n+1}}{a_n} = \frac{11^{n+1}}{11^n} \cdot \frac{(n+1)}{(n+2)} \cdot \frac{7^{2n+1}}{7^{2n+3}} 
$$

Breaking this down:
- $ \frac{11^{n+1}}{11^n} = 11 $
- $ \frac{7^{2n+1}}{7^{2n+3}} = 7^{(2n+1)-(2n+3)} = 7^{-2} = \frac{1}{49} $
- $ \frac{n+1}{n+2} $ approaches 1 as $ n \to \infty $

Combining these results gives:
$$
\left| \frac{a_{n+1}}{a_n} \right| = 11 \cdot \frac{1}{49} \cdot \frac{n+1}{n+2}
$$

Taking the limit:
$$
L = \lim_{n \to \infty} \left( 11 \cdot \frac{1}{49} \cdot \frac{n+1}{n+2} \right) = 11 \cdot \frac{1}{49} \cdot 1 = \frac{11}{49}
$$

---

### **Explanation of Step 2**
We simplified the ratio and calculated the limit to find $ L $. The result indicates the behaviour of the series.

---

## **Final Answer**

- The computed limit is $ L = \frac{11}{49} $.
- Since $ L < 1 $, by the Ratio Test, the series **converges absolutely**.

### **Conclusion**
- The final answer is:
  - **L = 11/49**
  - **A. The Ratio Test says that the series converges absolutely.**

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