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

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 $$[-\frac{1}{2},\frac{1}{2}]$$

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