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!