MHB 11.8.4 Find the radius of convergence and interval of convergence

[karush]

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

 

[skeeter]

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

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

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

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

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 $-1 < x \le 1$

 

[karush]

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

where does $|x|$ come from?

 

[skeeter]

where does $|x|$ come from?

from the ratio test ...

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