Question

Determine the radius of convergence of the given power series. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n} n^{2}(x+2)^{n}}{3^{n}} $$

Short answer

Answer: The radius of convergence is 3.

Step-by-step solution

Write down the general term of the series and identify the variable and constant terms

The general term of the series can be written as $$ a_n = \frac{(-1)^{n} n^{2}(x+2)^{n}}{3^{n}} $$ Here, the variable term is \((x+2)^{n}\) and the other terms are constant with respect to \(x\).

Apply the Ratio Test

The Ratio Test states that if the limit as \(n \to \infty\) of the absolute value of the ratio of the \((n+1)\)th term to the \(n\)th term is less than 1, then the series converges. So let us compute this limit $$ \lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{(-1)^{n+1}(n+1)^{2}(x+2)^{n+1}\cdot 3^{n}}{(-1)^{n} n^{2}(x+2)^{n}\cdot 3^{n+1}} $$

Solve for x

The limit can be simplified as $$ \lim_{n\to\infty} \frac{(n+1)^{2}(x+2)}{3n^{2}} $$ For convergence, this limit must be less than 1. $$ \left|\frac{(n+1)^{2}(x+2)}{3n^{2}}\right| < 1 $$ When we take the limit as \(n \to \infty\), the left-hand side will become $$ \left|\frac{(x+2)}{3}\right| $$ Thus, for convergence, $$ \left|\frac{(x+2)}{3}\right| < 1 $$

Determine the radius of convergence

To find the radius of convergence, isolate the absolute value expression and solve for x. The inequality becomes $$ |x+2| < 3 $$ This inequality represents an interval for the values of \(x\) that make the series converge. The radius of convergence is half the length of this interval; in this case, the radius is 3. So the radius of convergence of the given power series is 3.

Key concepts

Power Series

A power series is like a polynomial but with infinitely many terms. It can be written in the form:

• \(\sum_{n=0}^{\infty} a_n (x-a)^{n}\)

This expression means we're adding up squared, cubed, and higher powers of a variable, each multiplied by a coefficient.
Power series often appear in mathematics as a way to represent functions that might be complicated to handle otherwise.
The center of the series is given by the value \(a\), and the variable in the series \(x\) can change to give different sums.
In the exercise, our power series is:

• \(\sum_{n=1}^{\infty} \frac{(-1)^{n} n^{2}(x+2)^{n}}{3^{n}}\)

The expression involves powers of \((x+2)\), starting from 1.

Ratio Test

The Ratio Test is a way to determine whether a power series converges or diverges.
It simplifies checking convergence when dealing with complicated series.
The rule is to take the ratio of the \((n+1)\)th term to the \(n\)th term and then to compute the limit of this ratio as \(n\) goes to infinity:

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

If this limit is less than 1, the series converges.
If it is greater than 1, the series diverges, and if it equals 1, this test fails.
In the step-by-step solution provided, the Ratio Test was used to observe the behavior of:

• \(\frac{a_{n+1}}{a_{n}} = \frac{(-1)^{n+1}(n+1)^{2}(x+2)^{n+1}\cdot 3^{n}}{(-1)^{n} n^{2}(x+2)^{n}\cdot 3^{n+1}}\)

By simplifying and taking the limit, convergence can be established when this value is less than 1.

General Term

The general term of a series is crucial as it helps determine the series' behavior.
For the power series we are considering, the general term is:

• \(a_n = \frac{(-1)^{n} n^{2}(x+2)^{n}}{3^{n}}\)

Breaking this down:

• The \((-1)^n\) creates alternating positive and negative terms.
• The \(n^2\) factors affect the growth rate of the coefficients.
• \((x+2)^n\) incorporates the variable part, affecting when the series converges or diverges.
• Finally, the denominator \(3^n\) tends to reduce the term’s magnitude as \(n\) increases.

Understanding this format helps apply tests like the Ratio Test effectively and predict convergence behavior.

Interval of Convergence

The interval of convergence tells us the range of \(x\)-values for which the power series converges.
For a series centered at \((x+2)\), it converges within the bounds determined by applying the Ratio Test and finding a key inequality:

• \(|x+2| < 3|\)

To understand the limits of convergence, solve this inequality to find the values for \(x\):

• \(-5 < x < 1\)

This means that for any \(x\) within this range, the series will add up to a finite value.
The length of this interval, 6 (from -5 to 1), divided by 2, gives the radius of convergence, which is 3.
The interval test is vital because it not only tells when the power series converges but also helps us understand the functional behavior in the region around the center value.