Question

Find the interval of convergence, including end-point tests: \(\sum_{n=1}^{\infty} \frac{(x+2)^{n}}{(-3)^{n} \sqrt{n}}\)

Short answer

The interval of convergence is \([ -5, 1 ] \).

Step-by-step solution

Identify the series form

Recognize that the given series is \(\frac{(x+2)^{n}}{(-3)^{n} \sqrt{n}}\). Note that it is a power series and can be written as \(\frac{(x+2)^{n}}{(-3)^{n} \sqrt{n}}\).

Apply the Ratio Test

To find the radius and interval of convergence, use the Ratio Test: \(\frac{a_{n+1}}{a_{n}}\). Compute \(\frac{a_{n+1}}{a_{n}} = \frac{\frac{(x+2)^{n+1}}{(-3)^{n+1} \sqrt{n+1}}}{\frac{(x+2)^{n}}{(-3)^{n} \sqrt{n}}} = \frac{(x+2) \sqrt{n}}{-3} \frac{1}{\sqrt{n+1}} = \frac{(x+2)}{-3} \frac{\sqrt{n}}{\sqrt{n+1}}\).

Simplify the Ratio

Simplify the expression for the ratio: \[ \left| \frac{(x+2)}{-3} \frac{\sqrt{n}}{\sqrt{n+1}} \right| = \frac{|x+2|}{3} \frac{\sqrt{n}}{\sqrt{n+1}} \approx \frac{|x+2|}{3} as n \rightarrow \infty \].

Set Up the Inequality

For convergence, we require \(\frac{|x+2|}{3} < 1 \Rightarrow |x+2| < 3 \).

Solve the Inequality

Solve \[ |x+2| < 3 \Rightarrow -3 < x+2 < 3 \Rightarrow -5 < x < 1 \]. This gives the interval of convergence before testing the endpoints.

Test the Endpoint x = -5

Substitute \( x = -5 \) into the original series: \(\frac{(-5+2)^{n}}{(-3)^{n} \sqrt{n}} = \frac{(-3)^{n}}{(-3)^{n} \sqrt{n}} = \frac{1}{\sqrt{n}}\), which is a convergent series by the p-series test (p = 1/2 < 1).

Test the Endpoint x = 1

Substitute \( x = 1 \) into the original series: \(\frac{(1+2)^{n}}{(-3)^{n} \sqrt{n}} = \frac{3^{n}}{(-3)^{n} \sqrt{n}} = \frac{(-1)^{n}}{\sqrt{n}}\), which is conditionally convergent by the Alternating Series Test.

Combine Results

From both endpoint tests, \(x = -5\) and \(x = 1\), we find that the series converges at both endpoints, so the interval of convergence is \([ -5, 1 ] \).

Key concepts

Power Series

A power series is an infinite series of the form \(\sum_{n=0}^{\infty} c_n (x - a)^n\), where \(c_n\) are coefficients and \(a\) is the center of the series. The series converges for values of \(x\) within a certain radius around \(a\), known as the radius of convergence. This creates an interval centered at \(a\) within which the series converges. Power series are important due to their ability to represent functions as infinite polynomials, making them handy for solving differential equations and analyzing functions.

Ratio Test

The Ratio Test helps determine the convergence of a series. For a series \(\sum_{n=1}^{\infty} a_n\), we analyze the limit \(\lim_{{n \to \infty}} \frac{|a_{n+1}|}{|a_n|}\). If this limit is less than 1, the series converges absolutely. If greater than 1, the series diverges. If equal to 1, the test is inconclusive. For the given problem, \(\frac{(x+2)^{n+1}}{(-3)^{n+1} \sqrt{n+1}} \frac{(-3)^n \sqrt{n}}{(x+2)^n}\) simplifies to \(\frac{|x+2|}{3}\frac{\sqrt{n}}{\sqrt{n+1}}\). As \(n\) approaches infinity, \(\frac{|x+2|}{3}\) determines convergence.

Endpoint Tests

After finding the interval excluding end points \([-5, 1]\), it's essential to check if the series converges at the endpoints. This is done using specific tests:

• For \(x = -5\): the series becomes \(\frac{1}{\sqrt{n}}\), which converges by the p-series test since \(p = 1/2 < 1\)
• For \(x = 1\): the series becomes \(\frac{(-1)^n}{\}\), which converges by the Alternating Series Test.

Thus, both endpoints \([-5, 1]\) are included in the interval of convergence.

Convergence Criteria

The convergence criteria help determine where a power series will converge. Use the Ratio Test to find the radius of convergence and then test the endpoints:
- If \(\frac{|x+2|}{3} < 1\), it simplifies to \(-5 < x < 1\), indicating potential convergence between \(-5\) and \(1\).
- At \(x = -5\) and \(x = 1\) test using p-series and Alternating Series Tests.
This will clarify if the series converges at these points. Thus the interval of convergence is [ -5, 1 ].
Understanding these tests and criteria ensures you accurately find intervals where series like \(\frac{(x+2)^n}{(-3)^n \sqrt{n}}\) converge.