Question

Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case. $$\sum_{n=1}^{\infty}(-1)^{n} n^{3} x^{n}$$

Short answer

\[ (-1, 1) \text{ (open interval)} \]

Step-by-step solution

- Identify the general term

The general term of the series is \[ a_n = (-1)^{n} n^{3} x^{n} \]

- Apply the Ratio Test

To find the interval of convergence, apply the Ratio Test: \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]

- Compute the ratio

Substitute the general term into the ratio test formula: \[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(-1)^{n+1} (n+1)^{3} x^{n+1}}{(-1)^{n} n^{3} x^{n}} \right| = \left| x \frac{(n+1)^{3}}{n^{3}} \right| \]

- Simplify the ratio

Simplify the expression inside the limit: \[ \left| x \frac{(n+1)^{3}}{n^{3}} \right| = \left| x \frac{1 + 3/n + 3/n^2 + 1/n^3}{1} \right| \]

- Calculate the limit

Evaluate the limit as n approaches infinity: \[ L = \lim_{n \to \infty} \left| x \left(1 + \frac{3}{n} + \frac{3}{n^2} + \frac{1}{n^3} \right) \right| = \left| x \right| \]

- Determine the interval of convergence

For the series to converge, \ L < 1 \: \[ \left| x \right| < 1 \]

- Test the endpoints

Test the endpoints \ x = -1 \ and \ x = 1 \ separately: \[ \text{For } x = 1: \sum_{n=1}^{\infty} (-1)^{n} n^{3} (1)^{n} = \sum_{n=1}^{\infty} (-1)^{n} n^{3} \text{ (diverges)} \] \[ \text{For } x = -1: \sum_{n=1}^{\infty} (-1)^{n} n^{3} (-1)^{n} = \sum_{n=1}^{\infty} n^{3} \text{ (diverges)} \]

- Compile the results

Combine the information from the tests: \[ \left| x \right| < 1 \text{ (internal)} \text{with both endpoints divergent} \]

Key concepts

Power Series

A power series is a type of infinite series expressed in the form: oindent \ \[ \sum_{n=0}^{\infty} a_n x^n \]oindent \ here, \( a_n \) represents the coefficients, and \( x \) is a variable. Power series are important because they can represent many functions, such as polynomials and exponentials, over certain intervals.
The interval where the series converges (i.e., sums up to a finite value) or diverges (i.e., sums up to infinity or does not have a sum) is known as the interval of convergence. To determine this interval, techniques like the Ratio Test can be used.
In a power series, when \( x \) is within this interval, the series converges; outside the interval, it diverges.

Ratio Test

The Ratio Test is a method used to determine the convergence or divergence of an infinite series.
To apply the Ratio Test to a series \( \sum_{n=0}^{\infty} a_n \), compute the limit: \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]oindent \The results of this limit can determine the behavior of the series:

• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \), the series diverges.
• If \( L = 1 \), the test is inconclusive.

For power series, the ratio test can also help in determining the radius of convergence, \( R \), and then the interval of convergence can be expressed as \( -R < x < R \). In the provided exercise, the ratio test revealed \( |x| < 1 \) as our interval of convergence.

Convergence

Convergence in the context of series refers to whether the sum of the series approaches a finite value as the number of terms increases. When a series converges, the partial sums of the series get closer and closer to a specific value rather than oscillating or becoming infinitely large.
There are several important types of convergence:

• Absolute Convergence: If \( \sum_{n=0}^{\infty} |a_n| \) converges, then \( \sum_{n=0}^{\infty} a_n \) also converges absolutely.
• Conditional Convergence: If \( \sum_{n=0}^{\infty} a_n \) converges but \( \sum_{n=0}^{\infty} |a_n| \) diverges, the series convergence is conditional.

In the exercise, we check the endpoints \( x = 1 \) and \( x = -1 \) by substituting them into the series and analyzing whether the series converges for these specific values. Both instances showed divergence, indicating these points are not within the interval.