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To find the radius of convergence \( R \) of the power series 

\[
\sum_{n=0}^{\infty} (-1)^n \frac{(x - 1)^{2n + 1}}{2n + 1},
\]

we can use the **Ratio Test**. The Ratio Test states that for a series \(\sum a_n\), the series converges if

\[
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1.
\]

Here, our terms are given by

\[
a_n = (-1)^n \frac{(x - 1)^{2n + 1}}{2n + 1}.
\]

### Step 1: Apply the Ratio Test
We calculate the limit of \(\left| \frac{a_{n+1}}{a_n} \right|\) as \(n \to \infty\).

1. Compute \( a_{n+1} \):
   \[
   a_{n+1} = (-1)^{n+1} \frac{(x - 1)^{2(n+1) + 1}}{2(n+1) + 1} = (-1)^{n+1} \frac{(x - 1)^{2n + 3}}{2n + 3}.
   \]

2. Find \(\left| \frac{a_{n+1}}{a_n} \right|\):
   \[
   \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(-1)^{n+1} \frac{(x - 1)^{2n + 3}}{2n + 3}}{(-1)^n \frac{(x - 1)^{2n + 1}}{2n + 1}} \right|.
   \]

3. Simplify this expression:
   \[
   = \left| \frac{(x - 1)^2}{\frac{2n + 3}{2n + 1}} \right| = |x - 1|^2 \cdot \frac{2n + 1}{2n + 3}.
   \]

4. Take the limit as \(n \to \infty\):
   \[
   \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = |x - 1|^2.
   \]

### Step 2: Set up the Ratio Test inequality for convergence
For convergence, we need

\[
|x - 1|^2 < 1.
\]

Taking the square root of both sides, we find

\[
|x - 1| < 1.
\]

### Conclusion
The radius of convergence \( R \) is therefore \( R = 1 \).

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Would you like more details on this solution or have further questions?

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**Related Questions:**

1. How would you find the interval of convergence for this series?
2. What would happen if the exponent on \((x - 1)\) was different, say \(2n\) instead of \(2n + 1\)?
3. Can the root test also be used here? Why or why not?
4. What is the effect of the \((-1)^n\) term on convergence?
5. How does the Ratio Test differ from the Root Test for determining convergence?

**Tip:** When determining the radius of convergence for power series, the Ratio Test is often more straightforward than the Root Test due to easier simplification of terms.

Determine the radius of convergence of the power series 
\(
\sum_{n=0}^{\infty} (-1)^n \frac{(x - 1)^{2n + 1}}{2n + 1}
\). You do not have to find the interval of convergence.

Learn how to determine the radius of convergence for the power series \(\sum_{n=0}^{\infty} (-1)^n \frac{(x - 1)^{2n + 1}}{2n + 1}\) using the Ratio Test. This detailed step-by-step solution is suitable for college-level students.

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