Consider the power series:
∞X
n=1
(2x)n
n2
1. Find the interval of convergence for the given series.
2. Determine the function represented by the series within the interval of convergence.
3. Use the function representation to evaluate the series at x = 1.

We are given the power series:

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

Let's go through the steps to solve each part of the problem.

### 1. Interval of Convergence
To find the interval of convergence, we use the **ratio test**. Consider the general term of the series:

\[
a_n = \frac{(2x)^n}{n^2}
\]

Now apply the ratio test:

\[
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(2x)^{n+1}}{(n+1)^2} \cdot \frac{n^2}{(2x)^n} \right| = \lim_{n \to \infty} \left| 2x \cdot \frac{n^2}{(n+1)^2} \right|
\]

As \(n \to \infty\), \(\frac{n^2}{(n+1)^2} \to 1\), so we get:

\[
\lim_{n \to \infty} \left| 2x \right| = |2x|
\]

For the ratio test to imply convergence, we require:

\[
|2x| < 1 \quad \Rightarrow \quad |x| < \frac{1}{2}
\]

Thus, the radius of convergence is \( R = \frac{1}{2} \).

To determine the interval of convergence, we must check the endpoints \(x = \pm \frac{1}{2}\).

- **At \( x = \frac{1}{2} \):**
  The series becomes:

  \[
  \sum_{n=1}^{\infty} \frac{(2 \cdot \frac{1}{2})^n}{n^2} = \sum_{n=1}^{\infty} \frac{1^n}{n^2} = \sum_{n=1}^{\infty} \frac{1}{n^2}
  \]

  This is a **convergent p-series** with \(p = 2\).

- **At \( x = -\frac{1}{2} \):**
  The series becomes:

  \[
  \sum_{n=1}^{\infty} \frac{(2 \cdot -\frac{1}{2})^n}{n^2} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}
  \]

  This is the **alternating series**:

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

  The alternating series converges by the **alternating series test**.

Therefore, the series converges for \(x \in \left[-\frac{1}{2}, \frac{1}{2}\right]\).

### 2. Function Represented by the Series
The given series resembles the power series for the polylogarithm function. Specifically, we recognize that:

\[
\text{Li}_2(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^2}
\]

Thus, our series can be written as:

\[
\sum_{n=1}^{\infty} \frac{(2x)^n}{n^2} = \text{Li}_2(2x)
\]

Within the interval of convergence \(x \in \left[-\frac{1}{2}, \frac{1}{2}\right]\), the function represented by the series is the **dilogarithm function** evaluated at \(2x\), i.e., \(\text{Li}_2(2x)\).

### 3. Evaluate the Series at \(x = 1\)
To evaluate the series at \(x = 1\), we first check whether \(x = 1\) is within the interval of convergence. Since \(x = 1\) is outside the interval \( \left[-\frac{1}{2}, \frac{1}{2}\right] \), the series does **not converge** at \(x = 1\).

Therefore, the series **diverges** at \(x = 1\).

---

### Summary:
1. The interval of convergence is \(\left[-\frac{1}{2}, \frac{1}{2}\right]\).
2. The function represented by the series is \(\text{Li}_2(2x)\).
3. The series diverges at \(x = 1\).

Do you want more details or have any questions? Here are 5 related questions:

1. What is the polylogarithm function and how is it used in series representations?
2. Can you explain the alternating series test in detail?
3. How does the ratio test determine the radius of convergence?
4. What is a p-series and how do you identify its convergence?
5. How would you evaluate \(\text{Li}_2(z)\) for other values of \(z\)?

**Tip:** When testing endpoints for convergence, remember to use different convergence tests like the alternating series test or p-series comparison!

Consider the power series: ∞Σ n=1 (2x)^n / n^2. 1. Find the interval of convergence for the given series. 2. Determine the function represented by the series within the interval of convergence. 3. Use the function representation to evaluate the series at x = 1.

Explore the solution to a power series problem, including finding the interval of convergence, determining the dilogarithm function represented by the series, and evaluating convergence at x = 1. Detailed explanation of the ratio test, alternating series test, and p-series. Suitable for university-level calculus students.

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