Question

A power series is given. (a) Find the radius of convergence. (b) Find the interval of convergence. $$\sum_{n=1}^{\infty} \frac{(-1)^{n}(x-3)^{n}}{n}$$

Short answer

(a) Radius of convergence is 1; (b) Interval of convergence is \((2, 4]\).

Step-by-step solution

Identify the Power Series

The given power series is \( \sum_{n=1}^{\infty} \frac{(-1)^{n}(x-3)^{n}}{n} \). It is centered at \(x = 3\).

Apply the Ratio Test

For power series, the ratio test is commonly used to determine the radius of convergence. Calculate:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(-1)^{n+1}(x-3)^{n+1}/(n+1)}{(-1)^n (x-3)^n/n} \right|\] Simplify:\[= \lim_{n \to \infty} \left| \frac{(-1)(x-3)}{n+1} \right| \cdot \frac{n}{1} = |x-3| \cdot \lim_{n \to \infty} \frac{n}{n+1}\] \[= |x-3| \cdot 1 = |x-3|\]

Determine the Radius of Convergence

The ratio test gives convergence when \(|x-3| < 1\). Thus, the radius of convergence \(R\) is 1.

Find the Interval of Convergence Endpoint Test

The interval formed from \(|x-3| < 1\) is \(2 < x < 4\). Now, test the endpoints \(x = 2\) and \(x = 4\) by substituting into the series.

Test x = 2

Substitute \(x = 2\):\[\sum_{n=1}^{\infty} \frac{(-1)^{n}(2-3)^{n}}{n} = \sum_{n=1}^{\infty} \frac{(-1)^{n}(-1)^{n}}{n} = \sum_{n=1}^{\infty} \frac{1}{n}\]This is the harmonic series, which diverges.

Test x = 4

Substitute \(x = 4\):\[\sum_{n=1}^{\infty} \frac{(-1)^{n}(4-3)^{n}}{n} = \sum_{n=1}^{\infty} \frac{(-1)^{n}(1)^{n}}{n} = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\]This is the alternating harmonic series, which converges.

Conclude the Interval of Convergence

Since the series diverges at \(x = 2\) and converges at \(x = 4\), the interval of convergence is \((2, 4]\).

Key concepts

Radius of Convergence

The radius of convergence of a power series helps to understand how far from its center the series converges. For any power series \( \sum_{n=0}^{\infty} a_n (x - c)^n \), the radius of convergence \( R \) determines the interval around \( c \) where the series is guaranteed to converge. The radius tells you the width of the interval centered at \( c \) over which the series remains convergent.
To find the radius of convergence, we often use the *ratio test*. By applying the ratio test to \( a_n = \frac{(-1)^{n}(x-3)^{n}}{n} \), we calculate:

• \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \)
• Solving this gives \( \lim_{n \to \infty} |x-3| \cdot \frac{n}{n+1} \)
• This simplifies to \( |x-3| \)

Since the series converges when \( |x-3| < 1 \), the radius of convergence \( R \) is 1.

Interval of Convergence

Knowing the radius of convergence helps us find where the series converges, but it's essential to test the endpoints to know the complete interval. Given the radius \( R = 1 \), the inequality \( |x-3| < 1 \) provides the interval \( 2 < x < 4 \). This interval is centered at \( x = 3 \) and extends one unit in both directions. But wait! We must also check the endpoints \( x = 2 \) and \( x = 4 \) for convergence.

• At \( x = 2 \), substituting into the series yields the harmonic series, which diverges.
• At \( x = 4 \), the substitution results in the alternating harmonic series, which converges.

Therefore, the interval of convergence is \( (2, 4] \), because the series converges at \( x = 4 \) but not at \( x = 2 \).

Ratio Test

The ratio test is a powerful tool for determining the convergence of series, particularly power series. It provides a simple method to identify the radius of convergence for series expressions. When encountering a series \( \sum a_n \), the ratio test examines the limit of the absolute value of the ratio of consecutive terms:
\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]Here's how we used it for our power series:

• The expression simplifies to \( |x-3| \).
• The ratio test tells us that the series converges if this value is less than 1, providing the criterion \( |x-3| < 1 \).
• This criterion directly gives us the radius of convergence, \( R = 1 \).

This simple approach is why the ratio test is favored for series involving variable terms.

Alternating Harmonic Series

The alternating harmonic series is a fascinating concept in series and convergence. It is defined as \( \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \).
The series is called *alternating* because its terms alternate in sign. This change in sign makes a significant difference in convergence properties.While the regular harmonic series \( \sum \frac{1}{n} \) is divergent, the alternating version converges. This disparity showcases the importance of alternating signs.
There is a special test for series that alternate, known as the *Alternating Series Test*. This test states that a series \( \sum (-1)^{n} b_n \) converges if:

• The absolute values \( b_n \) are decreasing.
• The limit \( b_n \to 0 \) as \( n \to \infty \).

In our example, the alternating harmonic series at \( x = 4 \) meets these conditions, hence it converges. This different behavior due to alternating signs is a vital concept in series evaluation.