find the interval of convergence for the power series $sum_{n = 0}^{infty}(-1)^{n}\frac{n}{4^{n}}x^{2n}$…

Explanation:

Step1: Apply ratio - test

Let $a_n=(-1)^n\frac{n}{4^n}x^{2n}$. Then $a_{n + 1}=(-1)^{n+1}\frac{n + 1}{4^{n+1}}x^{2(n + 1)}$. Calculate the ratio $\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(-1)^{n + 1}\frac{n+1}{4^{n+1}}x^{2(n + 1)}}{(-1)^n\frac{n}{4^n}x^{2n}}\right|$. Simplify it: [ \begin{align*} \left|\frac{a_{n+1}}{a_n}\right|&=\left|\frac{(-1)^{n + 1}}{(-1)^n}\right|\cdot\left|\frac{n + 1}{n}\right|\cdot\left|\frac{4^n}{4^{n+1}}\right|\cdot\left|\frac{x^{2n+2}}{x^{2n}}\right|\ &=1\cdot\left(1+\frac{1}{n}\right)\cdot\frac{1}{4}\cdot|x^2| \end{align*} ]

Step2: Find the limit as $n\to\infty$

$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)\cdot\frac{|x^2|}{4}=\frac{|x^2|}{4}$. For convergence, $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1$. So $\frac{|x^2|}{4}<1$, which implies $|x^2|<4$, or $|x|<2$.

Step3: Check the endpoints

When $x = 2$, the series becomes $\sum_{n = 0}^{\infty}(-1)^n\frac{n}{4^n}\cdot2^{2n}=\sum_{n=0}^{\infty}(-1)^n\frac{n}{4^n}\cdot4^n=\sum_{n = 0}^{\infty}(-1)^n n$. This series diverges by the divergence - test (since $\lim_{n\to\infty}(-1)^n n$ does not exist). When $x=-2$, the series becomes $\sum_{n = 0}^{\infty}(-1)^n\frac{n}{4^n}\cdot(-2)^{2n}=\sum_{n=0}^{\infty}(-1)^n\frac{n}{4^n}\cdot4^n=\sum_{n = 0}^{\infty}(-1)^n n$, which also diverges by the divergence - test.

Answer:

$(-2,2)$