Problem 6 - Essential Calculus Early Transcendentals

Question

Find the radius of convergence and interval of convergence of the series. $$\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{n}}{n^{2}}$$

Step-by-Step Solution

Answer

The radius of convergence is 1, and the interval of convergence is [-1, 1].

Step 1: Choose the Convergence Test

To find the radius and interval of convergence, we'll use the Ratio Test for the series \(\sum_{n=1}^{\infty} a_n \), where \( a_n = \frac{(-1)^n x^n}{n^2} \). The Ratio Test involves finding \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

Step 2: Set Up the Ratio of Consecutive Terms

Calculate \( a_{n+1} \) for the given series: \( a_{n+1} = \frac{(-1)^{n+1} x^{n+1}}{(n+1)^2} \). We then find the ratio:\[\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(-1)^{n+1} x^{n+1}}{(n+1)^2} \times \frac{n^2}{(-1)^n x^n} \right| = \left| \frac{x \cdot n^2}{(n+1)^2} \right|.\]

Step 3: Compute the Limit of the Ratio

Now, take the limit as \( n \to \infty \) of the expression found in the previous step:\[\lim_{n \to \infty} \left| \frac{x \cdot n^2}{(n+1)^2} \right| = \lim_{n \to \infty} \left| x \cdot \frac{n^2}{n^2 + 2n + 1} \right| = \lim_{n \to \infty} \left| x \cdot \frac{1}{1 + \frac{2}{n} + \frac{1}{n^2}} \right| = |x|.\]

Step 4: Apply the Ratio Test Result

According to the Ratio Test, the series converges if \( |x| < 1 \). So the radius of convergence \( R \) is 1. Thus, \(|x| = 1\) represents the boundaries for checking convergence (i.e., \(-1 \leq x \leq 1\)).

Step 5: Check Convergence at Endpoints

For \( x = 1 \), the series becomes \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \), which converges by the Alternating Series Test. For \( x = -1 \), the series becomes \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), which converges by the p-series test with \( p = 2 > 1 \).

Step 6: Determine the Interval of Convergence

Since the series converges for both endpoints, the interval of convergence is \(-1 \leq x \leq 1\).

Key Concepts

Ratio Test

The Ratio Test is a powerful tool to find out if a series converges or diverges. It's particularly useful when dealing with series involving terms that have powers or factorials. The test examines the limit of the absolute value of the ratio of consecutive terms in the series. For a series \(\sum a_n\), you check the following limit:

• \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\)

If this limit is less than 1, then the series converges absolutely. If the limit is greater than 1, or infinity, the series diverges. If the limit equals 1, the test is inconclusive, and we might need other tests to determine convergence.
In the original problem, applying the Ratio Test helped find the radius of convergence: if \(|x| < 1\), the series converges.

Interval of Convergence

Once the radius of convergence is determined, the interval of convergence examines where exactly the series converges on the real number line. Generally, you find the interval by solving \(|x| < R\), where \(R\) is the radius of convergence. The endpoints \(-R\) and \(+R\) of this interval must be checked separately; a series may converge or diverge at these points.
In the given exercise, after applying the Ratio Test, the radius of convergence was found to be 1. Thus, the series converges within \(-1 < x < 1\). Further tests on the endpoints show whether the interval extends to include the endpoints (\[-1, 1\]).

Alternating Series Test

The Alternating Series Test is used to determine the convergence of series whose terms alternate in sign. For a series \(\sum (-1)^n b_n\), it can converge if:

• The absolute value of the terms \(b_n\) decreases steadily, \(b_{n+1} \leq b_n\).
• The limit of the term as \(n\) approaches infinity is zero, \(\lim_{n\to\infty} b_n = 0\).

In our problem, for \(x = 1\), the test showed convergence by alternating signs to \(\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}\), confirming the convergence for this endpoint.

P-Series Test

The p-series test is specific to series that look like \(\sum \frac{1}{n^p}\). Whether the series converges depends on the value of \(p\):

• If \(p > 1\), the series converges.
• If \(p \leq 1\), the series diverges.

For \(x = -1\), our series turned into \(\sum \frac{1}{n^2}\), which is a p-series with \(p = 2\). Here, \(p > 1\), so the series converges. This check for \(x = -1\) is crucial, confirming that the series also converges at this endpoint, making the interval of convergence \([-1, 1]\).