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 $R = 1$.

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

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.