Problem 7 Find the interval of convergence... [FREE SOLUTION]

Chapter 1: Problem 7

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

Short Answer

Expert verified

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

Step by step solution

- Identify the General Form and Apply Root Test

Given the power series $sum_{n=1}^{\infty} \frac{x^{3 n}}{n}$, we identify the general term as $a_n = \frac{x^{3n}}{n}$. To find the radius of convergence, apply the Root Test: \[\lim_{n \to \infty} \sqrt[n]{|a_n|}.\]

- Calculate the Root

The Root Test requires the calculation \[\lim_{n \to \infty} \sqrt[n]{\left| \frac{x^{3n}}{n} \right|} = \lim_{n \to \infty} \frac{|x^{3}|}{n^{1/n}}.\] Since $\lim_{n \to \infty} n^{1/n} = 1,$ this simplifies to \[\lim_{n \to \infty} |x^3| \cdot 1 = |x^3|.\]

- Determine the Radius of Convergence

By the Root Test, the series converges when $|x^3| < 1,$ which simplifies to $|x| < 1.$ Therefore, the radius of convergence is 1, and the interval of convergence is initially $-1 < x < 1.$

- Check Convergence at Endpoints

To investigate the convergence at the endpoints $x = -1$ and $x = 1,$ substitute these values into the original series. \[\text{At } x = 1: \sum_{n=1}^{\infty} \frac{(1)^{3n}}{n} = \sum_{n=1}^{\infty} \frac{1}{n}, \text{ which diverges (harmonic series).}\] \[\text{At } x = -1: \sum_{n=1}^{\infty} \frac{(-1)^{3n}}{n} = \sum_{n=1}^{\infty} \frac{-1}{n}, \text{ which also diverges (harmonic series with alternating signs simplifies to harmonic series).}\]

- Conclusion

Since the power series does not converge at the endpoints, the interval of convergence is $(-1, 1).$

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Key Concepts

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

interval of convergence

Power series have an interval of convergence where they sum to a finite value. For the series $\sum_{n=1}^{\infty} \frac{x^{3 n}}{n}$, this interval helps us understand the values of $\text{x}$ for which the series converges.
The steps to find the interval involve calculating the radius of convergence, then checking endpoints. We start with determining boundaries where the series remains meaningful.
We achieve this by using tests such as the Root Test or Ratio Test to ensure accurate results.

root test

The Root Test is a handy tool to determine the convergence of a series. It involves taking the $\text{n}$-th root of the terms in the series and analyzing the limit behavior.
\[\lim_{n \to \infty} \sqrt[n]{|a_n|}\]
For our series, $\text{a}_n = \frac{x^{3n}}{n}$. Applying Root Test gives us:
\[\lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} \frac{|x^{3}|}{n^{1/n}} = |x^3|.\]
The key concept here is recognizing how $\text{|x|}$ affects the series' behavior, making the Root Test an essential convergence tool.

radius of convergence

This radius is a critical value defining the 'circle' within which a power series converges. By applying the Root Test, we find it.
For our series $\frac{x^{3n}}{n}$, the limit simplifies as:
$\text{R} = \lim_{n \to \infty} |x^3| < 1 \rightarrow |x| < 1 \text { there is Radius of Convergence is 1 } $.
This means the series converges when $\text{|x|} = 1$ and gives us an interval $[-1, 1]$. The radius dictates how far we can stretch the series before it diverges.

endpoint convergence

To precisely define the interval of convergence, we assess series behavior at endpoints. For $\text{x = -1}$ or $\text{x = 1}$, we substitute these in the original series.
- At $\text{x = 1}$: Series $\textsum_{n=1}^{\infty} \frac{(1)^{3n}}{n}$ sums as the harmonic series, which diverges.
- At $\text{x = -1}$: Series $\textsum_{n=1}^{\infty} \frac{(-1)^{3n}}{n}$ also diverges as it simplifies to a harmonic series.
By investigating endpoints, we conclude that the series doesn't converge at these bounds, clarifying that interval of convergence is $\text{(-1, 1)}$.

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