Problem 15 Find the interval of convergence... [FREE SOLUTION]

Chapter 1: Problem 15

Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case. $$\sum_{n=1}^{\infty} \frac{(x-2)^{n}}{3^{n}}$$

Short Answer

Expert verified

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

Step by step solution

Title - Identify the power series and its form

Given power series: $$ \frac {(x-2)^{n}}{3^{n}} $$. It can be written in the general form as: $$a_{n}(x)^{n}$$ where $$a_{n}=\frac{1}{3^{n}}$$ . To find the interval of convergence, the ratio test will be used.

Title - Apply the Ratio Test

The Ratio Test is given by: \frac{a_{n+1}}{a_{n}} < 1 The terms of our series give: $$a_{n}=\frac{(x-2)^{n}}{3^{n}}$$ $$a_{n+1}=\frac{(x-2)^{n+1}}{3^{n+1}}$$. Now, compute the limit of $$\frac{a_{n+1}}{a_{n}}$$: $$a_{n}$$= $$\frac{(x-2)^{n}}{3^{n}}$$, $$a_{n+1}= \frac{(x-2)^{n+1}}{3^{n+1}}$$ Apply the ratio test: \frac{a_(n+1)}{a_(n)} =\frac{(x-2)^{n+1}}{3^{n+1}} \frac{3^{n}}{(x-2)^{n}} =\frac{(x-2)}{3} $$=|\frac {x-2}{3}|.$$ to be less than 1.

Title - Solve the Inequality

Now solve the inequality obtained in Step 2: $$|\frac{x-2}{3}| < 1$$. This simplifies to: $$| x-2| < 3$$. which gives the inequality $$-3 < x-2 < 3.$$ Adding 2 to all parts: i.e., $$ -3 + 2 < x -2 + 2 < 3 + 2$$ $$ -1 < x < 5$$ So, the interval of convergence is $$(-1, 5).$$

Title - Check the Endpoints

Now check the endpoints, $$x= -1 $$ and $$x=5$$ separately. Set x=-1 and compute the series: $$(\frac{-3}{3})^{n} = (\frac{(-1)^n)}.$$ This is clearly divergent. For x=5, the series is $${(\frac{3}{3})^{n}}.$$ This is also divergent. Thus the interval of convergence is $$ ( -1, 5 )$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Series

A power series is an infinite series of the form: $$\text{S}(x) = \frac{a_{n}(x-c)^{n}}{b_{n}}$$ where:

$x$ is the variable.
$a_n$ is the coefficient of the power $n$.
$c$ is the center of the series.
$b_n$ is some term in the denominator.

Power series aren't relevant everywhere. They have a radius of convergence that shows where they work. The interval where the series converges is called the interval of convergence.

Ratio Test

The ratio test helps us find the radius and interval of convergence. For any series $\text{S}(x) = \frac{a_{n}(x-c)^{n}}{b_{n}}$, it's done using the following limit: $$L = \frac{a_{n+1}(x-c)^{n+1}}{a_{n}(x-c)^{n}}$$ If

$L < 1$, the series converges.
$L > 1$, the series diverges.
$L = 1$, more tests are needed.

Let's apply the ratio test to our problem: Given series: $$\frac{(x-2)^{n}}{3^{n}}$$ We get $a_{n} = \frac{(x-2)^{n}}{3^{n}}$, $a_{n+1} = \frac{(x-2)^{n+1}}{3^{n+1}}$. Now find the limit: $L = |\frac{(x-2)3^{n}}{3^{n+1}}| = |\frac{x-2}{3}| < 1$.

Inequality Solving

To find where the series converges to, we solve this inequality: Simplify the modulus and write:

$$|\frac{x-2}{3}| < 1$$ Rewriting it becomes:

$$ |x-2| < 3 $$
which means: $$ -3 < x-2 < 3 $$

Add $2$ to each part:

$$ -3 + 2 < x-2 + 2 < 3 + 2$$ This simplifies to:

$$-1 < x < 5$$ So, the series converges when $x$ is in the interval $(-1,5)$.

Convergence of Series

To confirm the interval $(-1, 5)$ as the convergence range, substitute the endpoints ($x = -1$ and $x = 5$).

For $x = -1$: Since $(x-2) = -3$,

$$\frac{(-3)^{n}}{3^{n}} = (-1)^{n}$$ This oscillates and does not settle at any value 鈥� hence, it diverges.
For $x = 5$: Since $(x-2) = 3$,

$$\frac{(3)^{n}}{3^{n}} = 1$$ This sum grows indefinitely, and hence it diverges too.

So, after checking the endpoints, we confirm the interval of convergence is: $$ (-1, 5) $$

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