Math Problem Statement

Find the radius of convergence and the interval of convergence of the power series Σ (-1)^n (x - 3)^n / √n.

Solution

To determine the radius of convergence RR and the interval of convergence II for the given power series:

∑n=1∞(−1)n(x−3)nn\sum_{n=1}^{\infty} \frac{(-1)^n (x-3)^n}{\sqrt{n}}

we use the Ratio Test or Root Test to analyze convergence. Here's the step-by-step solution:

Step 1: Apply the Ratio Test

Let an=(−1)n(x−3)nna_n = \frac{(-1)^n (x-3)^n}{\sqrt{n}}.

The Ratio Test examines the limit:

lim⁡n→∞∣an+1an∣\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|

Compute the ratio:

= \left| \frac{(x-3)^{n+1}}{(x-3)^n} \cdot \frac{\sqrt{n}}{\sqrt{n+1}} \right| = \left| (x-3) \right| \cdot \frac{\sqrt{n}}{\sqrt{n+1}}$$ Simplify as $$n \to \infty$$: $$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \left| x-3 \right| \cdot \lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n+1}}$$ Since $$\frac{\sqrt{n}}{\sqrt{n+1}} \to 1$$ as $$n \to \infty$$, we get: $$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \left| x-3 \right|$$ For convergence, this must be less than 1: $$\left| x-3 \right| < 1$$ --- ### Step 2: Determine the Radius of Convergence From $$\left| x-3 \right| < 1$$, we find: $$R = 1$$ --- ### Step 3: Find the Interval of Convergence The inequality $$\left| x-3 \right| < 1$$ implies: $$2 < x < 4$$ We now check the endpoints $$x=2$$ and $$x=4$$: 1. **At $$x=2$$:** Substituting $$x=2$$ into the series, it becomes: $$\sum_{n=1}^\infty \frac{(-1)^n (-1)^n}{\sqrt{n}} = \sum_{n=1}^\infty \frac{1}{\sqrt{n}}$$ This is the harmonic series with terms $$\frac{1}{\sqrt{n}}$$, which diverges. 2. **At $$x=4$$:** Substituting $$x=4$$ into the series, it becomes: $$\sum_{n=1}^\infty \frac{(-1)^n (1)^n}{\sqrt{n}} = \sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$$ This is the alternating harmonic series with terms $$\frac{1}{\sqrt{n}}$$, which converges by the Alternating Series Test. --- ### Final Answer: - Radius of Convergence: $$R = 1$$ - Interval of Convergence: $$I = [2, 4)$$ Correct option: **$$R = 1, I = [2, 4)$$** --- Would you like further explanation of the steps or the tests used? ### Related Questions: 1. What is the Ratio Test, and how is it used to find convergence? 2. How does the Alternating Series Test determine convergence? 3. Why does the series diverge at $$x=2$$? 4. How does the radius of convergence differ from the interval of convergence? 5. What other methods can be used to find the radius of convergence? **Tip:** Always verify endpoint behavior separately when determining the interval of convergence.

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Math Problem Analysis

Mathematical Concepts

Power Series
Radius of Convergence
Interval of Convergence
Ratio Test
Alternating Series

Formulas

lim (n→∞) |a_(n+1) / a_n|
|x - c| < R

Theorems

Ratio Test
Alternating Series Test

Suitable Grade Level

Undergraduate (Calculus II or III)

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