Problem 8
Question
Find the radius of convergence and interval of conver- gence of the series. $$ \sum_{n=1}^{\infty} \frac{x^{n}}{n 3^{n}} $$
Step-by-Step Solution
Verified Answer
The radius of convergence is 3, and the interval of convergence is [-3, 3).
1Step 1: Identify the Series Form
The given series is \( \sum_{n=1}^{\infty} \frac{x^n}{n \, 3^n} \). This is a power series of the form \( \sum_{n=1}^{\infty} a_n x^n \) where \( a_n = \frac{1}{n \, 3^n} \).
2Step 2: Apply the Ratio Test for Convergence
The Ratio Test is used to determine the convergence of a power series by examining the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). For our series, this is \( L = \lim_{n \to \infty} \left| \frac{x^{n+1}}{(n+1) \, 3^{n+1}} \times \frac{n \, 3^n}{x^n} \right| \). Simplifying gives \( \left| \frac{x}{n+1} \times \frac{n}{3} \right| = \left| \frac{x}{3} \right| \times \frac{n}{n+1} \).
3Step 3: Simplify the Ratio and Calculate the Limit
The limit simplifies as \( \lim_{n \to \infty} \left| \frac{x}{3} \right| \times \frac{n}{n+1} = \left| \frac{x}{3} \right| \) because \( \lim_{n \to \infty} \frac{n}{n+1} = 1 \).
4Step 4: Determine the Radius of Convergence
The series converges if \( L < 1 \). Thus, we have \( \left| \frac{x}{3} \right| < 1 \), leading to \( \left| x \right| < 3 \). Therefore, the radius of convergence \( R \) is 3.
5Step 5: Identify the Interval of Convergence
The interval of convergence before checking endpoints is \( -3 < x < 3 \). We must check the endpoints \( x = -3 \) and \( x = 3 \) separately.
6Step 6: Check Convergence at Endpoints
At \( x = -3 \), the series becomes \( \sum_{n=1}^{\infty} \frac{(-3)^n}{n \, 3^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \), which is the alternating harmonic series and converges.At \( x = 3 \), the series becomes \( \sum_{n=1}^{\infty} \frac{3^n}{n \, 3^n} = \sum_{n=1}^{\infty} \frac{1}{n} \), which diverges.
7Step 7: Write the Final Interval of Convergence
Since the series converges at \( x = -3 \) and diverges at \( x = 3 \), the interval of convergence is \( [-3, 3) \).
Key Concepts
Power SeriesRatio TestInterval of ConvergenceConvergence Tests
Power Series
A power series is a type of infinite series that takes the form \( \sum_{n=0}^{\infty} a_n (x-c)^n \). It’s a method of representing functions in terms of variable powers. In this formula, \( a_n \) are the coefficients, \( c \) is the center of the series, and \( x \) is the variable.
- Power series can represent functions over a certain range of \( x \) values.
- The series converges depending on the values of \( x \) relative to the center \( c \).
- Determining the radius of convergence helps identify the range of \( x \) where the series is valid.
Ratio Test
The Ratio Test is a method used to determine the convergence of a series, particularly useful for power series. It involves evaluating the limit:\[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]
- If \( L < 1 \), the series converges absolutely.
- If \( L > 1 \) or is infinite, the series diverges.
- If \( L = 1 \), the test is inconclusive.
Interval of Convergence
The interval of convergence is the set of \( x \) values for which a power series converges. Once the radius of convergence is determined, we can define the interval:
- For a series centered at \( c \) with radius \( R \), the basic interval is \((c-R, c+R)\).
- Endpoints \( c-R \) and \( c+R \) need to be checked separately to see if the series converges at these points.
Convergence Tests
Convergence tests are essential tools that help determine whether a series converges or diverges. Various tests exist, each applicable under different circumstances:
- Ratio Test: As mentioned, effective for power series.
- Root Test: Considers the nth root of terms, useful for exponential terms.
- Alternating Series Test: Applied to series whose terms alternate in sign, checking for conditional convergence.
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