Question

A power series is given. (a) Find the radius of convergence. (b) Find the interval of convergence. $$\sum_{n=1}^{\infty} \frac{(x+2)^{n}}{n^{3}}$$

Short answer

(a) Radius of convergence: 1. (b) Interval of convergence: \([-3, -1]\).

Step-by-step solution

Identify the General Form

Given the power series \(\sum_{n=1}^{\infty} \frac{(x+2)^{n}}{n^{3}}\), we can see it is in the form \(\sum a_n (x+2)^n\) with \(a_n = \frac{1}{n^3}\).

Apply the Ratio Test

The ratio test involves computing the limit \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\). For our series, this becomes \[ L = \lim_{n \to \infty} \left| \frac{(x+2)^{n+1}}{(n+1)^3} \cdot \frac{n^3}{(x+2)^n} \right| = \lim_{n \to \infty} \left| \frac{(x+2) n^3}{(n+1)^3} \right|. \]

Simplify the Ratio Limit

Notice that \(\frac{(x+2) n^3}{(n+1)^3}\) can be approximated as \(\frac{(x+2) n^3}{n^3}\) for large \(n\). This simplifies to \(x+2\). Thus, \[ L = |x+2| \cdot \lim_{n \to \infty} \frac{n^3}{(n+1)^3} = |x+2|. \]

Establish the Radius of Convergence

The ratio test states the series converges when \(L < 1\). Therefore, \(|x+2| < 1\). Solving this gives \(-3 < x < -1\). Here, the radius of convergence is \(|x+2| < 1\), so the radius is 1.

Determine the Interval of Convergence

The interval suggested by the ratio test is \(-3 < x < -1\). We need to check convergence at the endpoints: \(x = -3\) and \(x = -1\).

Check Endpoint \(x = -3\)

At \(x = -3\), the series becomes \(\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}\), which is an alternating series. This converges, as \(\frac{1}{n^3}\) is a decreasing sequence and its limit approaches 0.

Check Endpoint \(x = -1\)

At \(x = -1\), the series becomes \(\sum_{n=1}^{\infty} \frac{1^n}{n^3} = \sum_{n=1}^{\infty} \frac{1}{n^3}\). This is a p-series with \(p = 3 > 1\) and thus converges.

Conclusion for Interval of Convergence

Since the series converges at both endpoints, \(x = -3\) and \(x = -1\), the interval of convergence is \([-3, -1]\).

Key concepts

Radius of Convergence

The radius of convergence actually refers to how far you can move away from a given point, typically called the center of the series, and still maintain convergence of the power series. For the series in our problem, \[ \sum_{n=1}^{\infty} \frac{(x+2)^{n}}{n^{3}}, \]the center is at \( x = -2 \), since we're dealing with \( (x+2) \) raised to powers. In order to find this radius, the Ratio Test is used.
The calculation of the radius involves determining when the limit, or value \( L \), based on the terms of the series becomes less than 1. So, the basic idea of the radius of convergence is to figure out how far (in terms of distance) the series can stay valid from its center.
For this particular series, this distance turned out to be 1, giving us a radius of convergence of 1. It's like drawing a circle around the center \( -2 \) with 1 unit of length in any direction where the series remains convergent.
Knowing the radius is crucial because it tells you the span in which you can plug in values for \( x \) and expect the series to converge.

Interval of Convergence

After determining the radius of convergence, you begin to explore within it to find out where exactly the series converges. Think of this as akin to identifying the breadth of values on a real number line for which your power series will be well-behaved.
In the solution, this power series gives us the inequality: \(|x+2| < 1\). If we unravel this, the inequality simplifies to \(-3 < x < -1\).
This is called the interval of convergence. More specifically, it suggests that without any further checks, the series appears to converge between these two points.
To ensure this interval is accurate, we perform additional checks at the endpoints, \(x = -3\) and \(x = -1\).

• At \(x = -3\), the series transforms into an alternating series, which converges through the alternating series test.
• At \(x = -1\), the series converts into a p-series which is known to converge when \(p > 1\).

Thus, despite being initially given as \(-3 < x < -1\), with the checks at the endpoints, the interval becomes inclusive from \([-3, -1]\). This ensures that both endpoints are part of the interval where the power series will indeed converge.

Ratio Test

The Ratio Test is a method used to determine whether a series converges or diverges. This test is quite handy, especially with series where terms involve factorials or powers.
For the given series, the Ratio Test requires calculating the limit of the absolute ratio of consecutive terms. This is expressed mathematically as \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \]In our example, this equates to simplifying \( \left| \frac{(x+2)^{n+1}}{(n+1)^3} \cdot \frac{n^3}{(x+2)^n} \right| \).
Through algebraic manipulation and simplification, it narrows down to \(|x+2| \cdot \lim_{n \to \infty} \frac{n^3}{(n+1)^3}\), which further simplifies to \(|x+2|\).
This reveals that for convergence, \(L < 1\), hence \(|x+2| < 1\).

• If \(L < 1\), the series converges.
• If \(L > 1\), the series diverges.
• If \(L = 1\), the test is inconclusive, and more steps would be needed.

Thus, the Ratio Test not only helped to determine the radius but also provided insight into which values of \(x\) the series holds true within our defined interval.