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The problem asks to **determine the radius and interval of convergence** for the following series:

1. \(\sum_{k=1}^{\infty} \frac{(x - 3)^k}{k}\)  
2. \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\)  
3. \(\sum_{n=0}^{\infty} n \cdot x^n\)

Let's solve these step by step.

---

### **1. Series: \(\sum_{k=1}^{\infty} \frac{(x - 3)^k}{k}\)**

This is a power series of the form:  
\[
\sum_{k=1}^{\infty} a_k (x-c)^k
\]
Here, \(a_k = \frac{1}{k}\) and \(c = 3\). To determine the radius of convergence, we use the **Ratio Test**. The Ratio Test states that the series converges when:
\[
\lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| |x - c| < 1
\]

Let us compute:
\[
\left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{\frac{1}{k+1}}{\frac{1}{k}} \right| = \frac{k}{k+1}
\]

So, the ratio becomes:
\[
\lim_{k \to \infty} \frac{k}{k+1} |x - 3| = |x - 3| \cdot \lim_{k \to \infty} \frac{k}{k+1}
\]

Since \(\lim_{k \to \infty} \frac{k}{k+1} = 1\), the radius of convergence is determined by:
\[
|x - 3| < 1
\]

### **Radius of Convergence: \(R = 1\)**  
The interval of convergence is determined by testing the endpoints \(x = 3 \pm 1\), i.e., \(x = 2\) and \(x = 4\).  

- **At \(x = 2\):**  
The series becomes:
\[
\sum_{k=1}^\infty \frac{(2 - 3)^k}{k} = \sum_{k=1}^\infty \frac{(-1)^k}{k}
\]
This is the **alternating harmonic series**, which converges.

- **At \(x = 4\):**  
The series becomes:
\[
\sum_{k=1}^\infty \frac{(4 - 3)^k}{k} = \sum_{k=1}^\infty \frac{1}{k}
\]
This is the **harmonic series**, which diverges.

### **Final Answer:**
- Radius of Convergence: \(R = 1\)  
- Interval of Convergence: \([2, 4)\)

---

### **2. Series: \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\)**

This is a **Taylor series expansion** of \(e^x\), which converges for all real numbers \(x\). Thus:

### **Final Answer:**
- Radius of Convergence: \(R = \infty\)  
- Interval of Convergence: \((-\infty, \infty)\)

---

### **3. Series: \(\sum_{n=0}^{\infty} n \cdot x^n\)**

To determine the radius of convergence, we again use the **Ratio Test**.  

Let \(a_n = n \cdot x^n\). The ratio is:
\[
\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(n+1) \cdot x^{n+1}}{n \cdot x^n} \right| = \left| \frac{n+1}{n} \cdot x \right|
\]

As \(n \to \infty\):
\[
\lim_{n \to \infty} \frac{n+1}{n} = 1
\]

Thus, the Ratio Test gives:
\[
|x| < 1
\]

### **Radius of Convergence: \(R = 1\)**  
To find the interval of convergence, test the endpoints \(x = \pm 1\):  

- **At \(x = 1\):**  
The series becomes:
\[
\sum_{n=0}^\infty n \cdot 1^n = \sum_{n=0}^\infty n
\]
This diverges.

- **At \(x = -1\):**  
The series becomes:
\[
\sum_{n=0}^\infty n \cdot (-1)^n
\]
This diverges because the terms \(n \cdot (-1)^n\) do not tend to 0.

### **Final Answer:**
- Radius of Convergence: \(R = 1\)  
- Interval of Convergence: \((-1, 1)\)

---

### Recap of Answers:
1. \(R = 1\), Interval: \([2, 4)\)  
2. \(R = \infty\), Interval: \((-\infty, \infty)\)  
3. \(R = 1\), Interval: \((-1, 1)\)

---

Would you like further details on any step or topic?

---

### Related Questions:
1. What is the general form of a power series and its convergence criteria?  
2. How does the Ratio Test work for power series?  
3. Why does the series \(\sum \frac{x^n}{n!}\) converge for all \(x\)?  
4. Can the Alternating Series Test be applied to determine endpoint convergence?  
5. What is the difference between radius and interval of convergence?

### **Tip:**  
When solving power series problems, always analyze the **endpoints** of the interval separately after applying the Ratio Test!

Determine the radius and interval of convergence of the following series:
1. \(\sum_{k=1}^{\infty} \frac{(x - 3)^k}{k}\)
2. \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\)
3. \(\sum_{n=0}^{\infty} n \cdot x^n\)

Step-by-step solution for determining the radius and interval of convergence for the given power series. This involves the Ratio Test, Alternating Series Test, and convergence of Taylor series. Suitable for Grades 11-12 or introductory college-level courses.

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