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The radius of Convergence for the function [tex]\frac{\[(-1)^{n} \; x^{n} }{\sqrt[3]{n} }[/tex] is 1.
What is radius of convergence?
The radius of the largest disk at the center of the series in which the series converges.
Perform the ratio test for absolute convergence, which says
if [tex]\lim_{n \to \infty} \frac{a_{n}}{a_{n+1}}[/tex] =L
Then,
1) if L<1, the series is absolutely convergent
2) If L>1, the series is divergent
3) If L= 1, apply a different test.
Now, Applying test
[tex]a_{n+1}[/tex] =[tex]\frac{(-1)^{{n+1} } \; x^{n+1}}{\sqrt[3]{n+1} }[/tex]
So,
[tex]\lim_{n \to \infty} \frac{a_{n+1}}{a_{n}}[/tex] = [tex]\lim_{n \to \infty}|\frac{ (-1)^{{n+1} } \; x^{n+1}/\sqrt[3]{n+1} }{ (-1)^{n} \; x^{n} /\sqrt[3]{n} } |[/tex]
= [tex]\lim_{n \to \infty}|\frac{ (-1)^{{n+1} } \; x^{n+1}\sqrt[3]{n}}{ (-1)^{n} \; x^{n} \sqrt[3]{n+1} } |[/tex]
= [tex]\lim_{n \to \infty}|\frac{ (-1) \; x^{n}\sqrt[3]{n}}{ \sqrt[3]{n+1} } |[/tex]
= [tex]\lim_{n \to \infty}\sqrt[3]{\frac{1}{1+\frac{1}{n}} } } |x|[/tex]
Applying the limit n→∞,
= 1 x |x|
= |x|
By ratio test, the given series will be convergent if |x|< 1.
So, the radius of Convergence = 1
Learn more about radius of Convergence here:
https://brainly.com/question/18763238
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