To solve for the radius and interval of convergence for the two given power series, we apply the Ratio Test to determine when the series converges. Let us address each part.

Part (a)

The given series is: $\sum_{n=1}^\infty \frac{(2x)^n}{n}.$

1. Applying the Ratio Test:

The general term is $a_n = \frac{(2x)^n}{n}$. Using the Ratio Test: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1.$

Substituting $a_n$: $\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{(2x)^{n+1}}{n+1}}{\frac{(2x)^n}{n}} \right| = \left| \frac{(2x)^{n+1} \cdot n}{(n+1) \cdot (2x)^n} \right| = \left| \frac{2x \cdot n}{n+1} \right|.$

Simplifying: $\lim_{n \to \infty} \left| \frac{2x \cdot n}{n+1} \right| = \lim_{n \to \infty} \left| 2x \cdot \frac{n}{n+1} \right| = \left| 2x \right|.$

For convergence: $\left| 2x \right| < 1 \implies |x| < \frac{1}{2}.$

2. Radius and Interval of Convergence:

• Radius of convergence: $R = \frac{1}{2}$.
• Interval of convergence: To test endpoints:
  • At $x = \frac{1}{2}$, the series becomes $\sum_{n=1}^\infty \frac{1}{n}$, which diverges (harmonic series).
  • At $x = -\frac{1}{2}$, the series becomes $\sum_{n=1}^\infty \frac{(-1)^n}{n}$, which converges (alternating harmonic series).

Thus, the interval of convergence is: $\boxed{[-\frac{1}{2}, \frac{1}{2})}.$

Part (b)

The given series is: $\sum_{n=1}^\infty \frac{10^n (x-2)^n}{n!}.$

1. Applying the Ratio Test:

The general term is $a_n = \frac{10^n (x-2)^n}{n!}$. Using the Ratio Test: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1.$

Substituting $a_n$: $\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{10^{n+1} (x-2)^{n+1}}{(n+1)!}}{\frac{10^n (x-2)^n}{n!}} \right| = \left| \frac{10^{n+1} (x-2)^{n+1} n!}{10^n (x-2)^n (n+1)!} \right|.$

Simplify: $\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{10 \cdot (x-2)}{n+1} \right|.$

Taking the limit as $n \to \infty$: $\lim_{n \to \infty} \left| \frac{10 \cdot (x-2)}{n+1} \right| = 0.$

Since the limit is always 0, the series converges for all $x$.

2. Radius and Interval of Convergence:

• Radius of convergence: $R = \infty$ (converges for all $x$).
• Interval of convergence: $(-\infty, \infty)$.

Final Answers:

1. Part (a):

   • Radius: $\frac{1}{2}$
   • Interval: $\boxed{[-\frac{1}{2}, \frac{1}{2})}$

2. Part (b):

   • Radius: $\infty$
   • Interval: $\boxed{(-\infty, \infty)}$

Would you like further explanation or visualizations?
Here are some related questions:

1. What does the Ratio Test measure in power series?
2. Can alternating series always converge at endpoints?
3. How does factorial growth affect convergence in part (b)?
4. How would results change if part (b) had $n^2$ instead of $n!$?
5. What is the difference between conditional and absolute convergence?

Tip: Factorial terms $n!$ in denominators often lead to convergence for all $x$, because factorials grow very fast compared to exponential terms.