Problem 9 Find the range of values of \(x\... [FREE SOLUTION]

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Chapter 24: Problem 9

Find the range of values of $x$ for which the series $$ \frac{x}{27}+\frac{x^{2}}{125}+\ldots+\frac{x^{n}}{(2 n+1)^{3}}+\ldots $$ is absolutely convergent.

Short Answer

Expert verified

The given series is absolutely convergent for values of $x$ in the range $-1 < x < 1$.

Step by step solution

Define the General Term of the Series

First, we define the general term of the series as $a_n$: $$ a_n = \frac{x^{n}}{(2n + 1)^3} $$

Apply the Ratio Test

The Ratio Test requires finding the limit of the ratio of consecutive terms as follows: $$ L = \lim_{n\to\infty} \frac{a_{n+1}}{a_n} $$ So, let's compute $\frac{a_{n+1}}{a_n}$: $$ \frac{a_{n+1}}{a_n} = \frac{\frac{x^{n+1}}{(2(n+1) + 1)^3}}{\frac{x^{n}}{(2n + 1)^3}} $$

Simplify the Ratio

Next, let's simplify the fraction: $$ \frac{a_{n+1}}{a_n} = \frac{x^{n+1}(2n + 1)^3}{x^{n}(2(n+1) + 1)^3} $$ Factor out $x^n$: $$ \frac{a_{n+1}}{a_n} = x \cdot \frac{(2n + 1)^3}{(2(n+1) + 1)^3} $$

Find the Limit of the Ratio

Now, we will find the limit as $n \to \infty$: $$ L = \lim_{n\to\infty} x \cdot \frac{(2n + 1)^3}{(2(n+1) + 1)^3} $$ Let's find the limit of the fraction itself first: $$ \lim_{n\to\infty} \frac{(2n + 1)^3}{(2(n+1) + 1)^3} = \lim_{n\to\infty} \frac{(2n + 1)^3}{(2n + 3)^3} $$ Now, factor $8n^3$ from both numerator and denominator: $$ \lim_{n\to\infty} \frac{(2n + 1)^3}{(2n + 3)^3} = \lim_{n\to\infty} \frac{1 + \frac{1}{2n}}{1 + \frac{3}{2n}} = 1 $$ Thus, we have: $$ L = |x| \cdot 1 = |x| $$

Determine the Convergence of the Series

According to the Ratio Test, the series converges if $L < 1$. Therefore, we have the inequality: $$ |x| < 1 $$ This inequality indicates that the series is absolutely convergent for $x$ in the range: $$ -1 < x < 1 $$ Thus, the given series is absolutely convergent for values of $x$ in the range $(-1, 1)$.

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Key Concepts

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

Ratio Test

The Ratio Test is a widely used method for determining the convergence of series. It's an easy tool to use when you're faced with a series where each term can be expressed as a ratio of consecutive terms. Essentially, this test involves looking at the limit of the absolute value of the terms.Here's how it works:

First, find the ratio of the consecutive terms in your series. This will often involve dividing the $n+1$th term by the $n$th term.
Simplify as much as possible to make the calculation easy.
Then calculate the limit of this ratio as $n$ approaches infinity.

If this limit $L$ satisfies $L < 1$, the series converges absolutely. If $L > 1$, or if $L$ is infinite, the series series diverges. If $L = 1$, the test is inconclusive, and other methods must be used.

Absolute Convergence

Absolute convergence is an important concept in the study of series. If a series is absolutely convergent, it is guaranteed to converge.Here's the process for defining absolute convergence:

Take your series and form a new series from the absolute value of its terms, that is, replace each term $a_n$ with $|a_n|$.
Now, would this new series, made up of positive terms only, converge?

If the answer is yes, then your original series is absolutely convergent.Absolute convergence provides a stronger kind of convergence, ensuring that not only does the series converge, but it will do so regardless of the arrangement of terms, which can be useful in understanding the behavior of series particularly when dealing with alternating signs or complex numbers.

Limit Calculation

Limit calculation is a fundamental aspect of testing convergence with the Ratio Test. Calculating limits involves determining what value, if any, a function or sequence approaches as the input or index goes to infinity.In the context of the Ratio Test for the given series:

We started by simplifying the ratio $\frac{a_{n+1}}{a_n}$ as \[ x \cdot \frac{(2n + 1)^3}{(2n + 3)^3} \]
Next, we focus on the expression inside the limit, \[ \lim_{n\to\infty} \frac{(2n + 1)^3}{(2n + 3)^3} \]
Expanding the cubes and factoring out the highest power of $n$, both in the numerator and the denominator, led us to conclude that this fraction simplifies to 1 as $n$ approaches infinity.

Thus, we find that the Ratio Test ultimately revolves around calculating these kinds of limits, and that $L = |x|$ provides the boundary condition for absolute convergence.

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91影视

Find the range of values of \(x\) for which the series $$ \frac{x}{27}+\frac{x^{2}}{125}+\ldots+\frac{x^{n}}{(2 n+1)^{3}}+\ldots $$ is absolutely convergent.

Short Answer

Step by step solution

Define the General Term of the Series

Apply the Ratio Test

Simplify the Ratio

Find the Limit of the Ratio

Determine the Convergence of the Series

Key Concepts

Ratio Test

Absolute Convergence

Limit Calculation

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