-/1.81 points details my notes scalcet9 11.6.004. use the ratio test to determine whether the series is…

Explanation:

Step1: Identify (a_n)

Given the series (\sum_{n = 1}^{\infty}\frac{(-2)^n}{n^2}), we have (a_n=\frac{(-2)^n}{n^2}).

Step2: Find (a_{n + 1})

Replace (n) with (n+1) in (a_n), so (a_{n+1}=\frac{(-2)^{n + 1}}{(n + 1)^2}).

Step3: Calculate (\left|\frac{a_{n+1}}{a_n}\right|)

[ \begin{align*} \left|\frac{a_{n + 1}}{a_n}\right|&=\left|\frac{\frac{(-2)^{n+1}}{(n + 1)^2}}{\frac{(-2)^n}{n^2}}\right|\ &=\left|\frac{(-2)^{n+1}}{(n + 1)^2}\cdot\frac{n^2}{(-2)^n}\right|\ &=\left|-2\cdot\frac{n^2}{(n + 1)^2}\right|\ &=2\cdot\frac{n^2}{(n + 1)^2} \end{align*} ]

Step4: Evaluate the limit

[ \begin{align*} \lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\rightarrow\infty}2\cdot\frac{n^2}{(n + 1)^2}\ &=2\lim_{n\rightarrow\infty}\frac{n^2}{n^2+2n + 1}\ &=2\lim_{n\rightarrow\infty}\frac{1}{1+\frac{2}{n}+\frac{1}{n^2}}\ &=2\cdot1\ &= 2 \end{align*} ]

Answer:

(a_n=\frac{(-2)^n}{n^2}), (\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right| = 2), Since (\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|>1), the series is divergent.