Let's solve the radius and interval of convergence for each of the series in parts (a) and (g) using the Ratio Test, which is commonly used to find the radius of convergence for power series.

Part (a)

Given: $\sum_{k=1}^{\infty} \left( \frac{x - 1}{2} \right)^k$

This series is of the form: $\sum_{k=1}^{\infty} a_k = \sum_{k=1}^{\infty} \left( \frac{x - 1}{2} \right)^k$

To find the radius of convergence, $R$, we can use the Ratio Test, which states that for a power series $\sum a_k$, the radius of convergence $R$ is given by: $\lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| < 1$

In this case:

1. We can write $a_k = \left( \frac{x - 1}{2} \right)^k$.
2. Then, $a_{k+1} = \left( \frac{x - 1}{2} \right)^{k+1}$.

Applying the Ratio Test: $\left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{\left( \frac{x - 1}{2} \right)^{k+1}}{\left( \frac{x - 1}{2} \right)^k} \right| = \left| \frac{x - 1}{2} \right|$

For convergence, we need: $\left| \frac{x - 1}{2} \right| < 1$

This inequality gives us: $-1 < \frac{x - 1}{2} < 1$

Multiplying through by 2: $-2 < x - 1 < 2$

Adding 1 to all parts: $-1 < x < 3$

Thus, the interval of convergence is $(-1, 3)$. To determine whether the endpoints $x = -1$ and $x = 3$ are included, we would substitute each endpoint back into the series and test for convergence.

The radius of convergence $R$ is the distance from the center point $x = 1$ to either endpoint, which is $R = 2$.

Part (g)

Given: $\sum_{k=0}^{\infty} \left( \frac{3}{4} \right)^k (x + 5)^k$

This can be rewritten as: $\sum_{k=0}^{\infty} \left( \frac{3(x + 5)}{4} \right)^k$

This is a geometric series with ratio $r = \frac{3(x + 5)}{4}$.

For a geometric series $\sum r^k$, it converges when $|r| < 1$.

Therefore, we have: $\left| \frac{3(x + 5)}{4} \right| < 1$

Solving this inequality:

1. Multiply both sides by 4: $|3(x + 5)| < 4$
2. Divide by 3: $|x + 5| < \frac{4}{3}$

This gives: $-\frac{4}{3} < x + 5 < \frac{4}{3}$

Subtracting 5 from all parts: $-\frac{4}{3} - 5 < x < \frac{4}{3} - 5$

Simplifying further: $x \in \left(-\frac{19}{3}, -\frac{11}{3}\right)$

So the interval of convergence is $\left(-\frac{19}{3}, -\frac{11}{3}\right)$.

The radius of convergence $R$ is the distance from the center point $x = -5$ to either endpoint, which is $R = \frac{4}{3}$.

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Would you like further details on these calculations or have any questions?

Here are five related questions that expand on the information:

1. How do we test the convergence at the endpoints for power series?
2. What are the conditions for using the Ratio Test in power series?
3. Can other tests be used to find the radius of convergence?
4. How do you handle series with factorial terms in the ratio test?
5. How does changing the center of a power series affect the interval of convergence?

Tip: When finding the radius of convergence, remember that the Ratio Test is particularly useful for series with powers or factorials in the terms.