11.8.4 Find the radius of convergence and interval of convergence

In summary, the radius of convergence for the given series is 1 and the interval of convergence is -1 < x < 1. These values were found using the ratio test, where the absolute value of x was used to determine the limit as n approaches infinity. The endpoints were then checked to determine the endpoint values of the interval of convergence. The example provided by W|A was also followed to obtain these results.
  • #1
karush
Gold Member
MHB
3,269
5
Find the radius of convergence and interval of convergence
of the series.
$$\sum_{n=1}^{\infty}\dfrac{(-1)^n x^n}{\sqrt[3]{n}}$$
(1)
$$a_n=\dfrac{(-1)^n x^n}{\sqrt[3]{n}}$$
(2)
$$\left|\dfrac{a_{a+1}}{a_n}\right|
=\left|\dfrac{(-1)^{n+1} x^{n+1}}{\sqrt[3]{n+1}}
\cdot\dfrac{\sqrt[3]{n}}{(-1)^n x^n}\right|
=-\frac{\sqrt[3]{n}x\left(n+1\right)^{\frac{2}{3}}}{n+1}$$
(3) W|A Convergence Interval is
$$-1\le \:x\le \:1$$

ok on (2) I was expecting a different result to take $\infty$ to
on (3) the example I was trying to follow was ? on how W|A got this interval
 
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  • #2
$\displaystyle \lim_{n \to \infty} \bigg| \dfrac{(-1)^{n+1}x^{n+1}}{\sqrt[3]{n+1}} \cdot \dfrac{\sqrt[3]{n}}{(-1)^n x^n} \bigg| < 1$

$\displaystyle \lim_{n \to \infty} \sqrt[3]{\dfrac{n}{n+1}} \cdot |x| < 1$

$\displaystyle |x| \cdot \lim_{n \to \infty} \sqrt[3]{\dfrac{n}{n+1}} < 1$

$|x| \cdot 1 < 1 \implies -1 < x < 1$

checking the endpoints in the original power series ...

$x = 1$ ...

$\displaystyle \sum \dfrac{(-1)^n \cdot 1^n}{\sqrt[3]{n}} = \sum \dfrac{(-1)^n}{\sqrt[3]{n}}$, an alternating series whose nth term approaches zero $\implies$ convergence.

$x=-1$ ...

$\displaystyle \sum \dfrac{(-1)^n \cdot (-1)^n}{\sqrt[3]{n}} = \sum \dfrac{(-1)^{2n}}{\sqrt[3]{n}} = \sum \dfrac{1}{\sqrt[3]{n}} $, a series whose nth term approaches zero, but does not satisfy the p-test for convergence, therefore $\implies$ divergence.

interval of convergence is $-1 < x \le 1$
 
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  • #3
skeeter said:
$\displaystyle \lim_{n \to \infty} \sqrt[3]{\dfrac{n}{n+1}} \cdot |x| < 1$

where does $|x|$ come from?
 
  • #4
karush said:
where does $|x|$ come from?

from the ratio test ...

\(\displaystyle \lim_{n \to \infty} \bigg|\dfrac{a_{n+1}}{a_n} \bigg| = \lim_{n \to \infty} \bigg|\sqrt[3]{\dfrac{n}{n+1}} \cdot x \bigg| = \lim_{n \to \infty} \sqrt[3]{\dfrac{n}{n+1}} \cdot |x| = |x| \cdot \lim_{n \to \infty} \sqrt[3]{\dfrac{n}{n+1}}
= |x| \cdot 1 = |x| < 1\)
 

Related to 11.8.4 Find the radius of convergence and interval of convergence

1. What is the purpose of finding the radius of convergence and interval of convergence?

The radius of convergence and interval of convergence are used to determine the values of x for which a given power series will converge. This information is important in understanding the behavior and convergence of a series, and can be useful in solving mathematical problems.

2. How do you find the radius of convergence and interval of convergence?

The radius of convergence can be found using the ratio test, which involves taking the limit of the ratio of consecutive terms in the series. The interval of convergence can be determined by testing the endpoints of the interval and checking for convergence or divergence.

3. What is the significance of the radius of convergence?

The radius of convergence is the distance from the center of the series where the series will converge. It indicates the range of values for x where the series will converge, and any values of x outside of this range will result in divergence.

4. Can the radius of convergence be negative?

No, the radius of convergence cannot be negative. It is always a positive value, representing the distance from the center of the series where the series will converge.

5. How does the radius of convergence affect the convergence of a series?

The radius of convergence determines the range of values for x where the series will converge. If the value of x is within this range, the series will converge. If the value of x is outside of this range, the series will diverge. Therefore, the radius of convergence directly affects the convergence of a series.

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