Determine the radius of convergence and the interval of convergence for each power series. sum_{n=1}^inftyfrac{(-2)^nx^n}{sqrt[4]{n}}

Determine the radius of convergence and the interval of convergence for each power series. sum_{n=1}^inftyfrac{(-2)^nx^n}{sqrt[4]{n}}

Question
Series
asked 2021-02-10
Determine the radius of convergence and the interval of convergence for each power series.
\(\sum_{n=1}^\infty\frac{(-2)^nx^n}{\sqrt[4]{n}}\)

Answers (1)

2021-02-11
Given series is: \(\sum_{n=1}^\infty a_nx^n=\sum_{n=1}^\infty\frac{(-2)^nx^n}{\sqrt[4]{n}}\)
By Applying ratio test , we have
\(\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|=\lim_{n\to\infty}|\frac{(-2)^{n+1}x^{n+1}\sqrt[4]{n}}{\sqrt[4]{n+1}(-2)^nx^n}|\)
\(=\lim_{n\to\infty}|\frac{(-2)x\sqrt[4]{n}}{\sqrt[4]{n}(1+\frac{1}{n})^\frac{1}{4}}|\)
\(=\lim_{n\to\infty}|\frac{(-2)x}{(1+\frac{1}{n})^\frac{1}{4}}|\)
\(=|-2x|\)
\(=2|x|\) for radius of convergence \(2|x|=1\)
\(\Rightarrow|x|=\frac{1}{2}\)
so \(r=\frac{1}{2}\) is the radius of convergence.
Now, when \(x=\frac{1}{2}\), given series becomes \(\sum_{n=1}^\infty\frac{(-2)^n(\frac{1}{2})^n}{\sqrt[4]{n}}\)
\(\sum_{n=1}^\infty(-1)^na_n=\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt[4]{n}}\)
as \(\lim_{n\to\infty}a_n=0\) and seq. \(a_n\) is decreasing, so by Leibnitz alternating test, this series is convergent at \(x=\frac{1}{2}\)
Now, when \(x=-\frac{1}{2}\), given series becomes \(\sum_{n=1}^\infty\frac{(-2)^n(\frac{1}{2})^n}{\sqrt[4]{n}}\)
\(=\sum_{n=1}^\infty\frac{1}{\sqrt[4]{n}}\)
By, p-test this series is convergent.
so, interval of convergence is \([-\frac{1}{2},\frac{1}{2}]\)
0

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