11.6.8 determine convergent or divergence by Ratio Test

[karush]

Use the Ratio Test to determine whether the series is convergent or divergent
$$\sum_{n=1}^{\infty}\dfrac{(-2)^n}{n^2}$$
If $\displaystyle\lim_{n \to \infty}
\left|\dfrac{a_{n+1}}{a_n}\right|=L>1
\textit{ or }
\left|\dfrac{a_{n+1}}{a_n}\right|=\infty
\textit{ then the series } \sum_{n=1}^{\infty}a_n \textit{ is divergent}$

ok I spent about half hour trying to do this but not sure what a is

also this one if can...

$$\sum_{n=1}^{\infty} \frac{(2n)!}{9^n n!}$$

 

[skeeter]

Use the Ratio Test to determine whether the series is convergent or divergent
$$\sum_{n=1}^{\infty}\dfrac{(-2)^n}{n^2}$$
If $\displaystyle\lim_{n \to \infty}
\left|\dfrac{a_{n+1}}{a_n}\right|=L>1
\textit{ or }
\left|\dfrac{a_{n+1}}{a_n}\right|=\infty
\textit{ then the series } \sum_{n=1}^{\infty}a_n \textit{ is divergent}$

ok I spent about half hour trying to do this but not sure what a is

also this one if can...

$$\sum_{n=1}^{\infty} \frac{(2n)!}{9^n n!}$$

$a_{n+1} = \dfrac{(-2)^{n+1}}{(n+1)^2}$
$a_n = \dfrac{(-2)^n}{n^2}$

$\dfrac{a_{n+1}}{a_n} = \dfrac{(-2)^{n+1}}{(n+1)^2} \cdot \dfrac{n^2}{(-2)^n} = \dfrac{-2n^2}{(n+1)^2}$

$$\lim_{n \to \infty} \bigg|\dfrac{-2n^2}{(n+1)^2} \bigg| = \lim_{n \to \infty} \dfrac{2}{1 + \frac{2}{n} + \frac{1}{n^2}}$$

so ... what next?

-----------------------------------------------------------------------------

$a_{n+1} = \dfrac{[2(n+1)]!}{9^{n+1} \cdot (n+1)!}$

$a_n = \dfrac{(2n)!}{9^n \cdot n!}$$$\lim_{n \to \infty} \bigg|\dfrac{[2(n+1)]!}{9^{n+1} \cdot (n+1)!} \cdot \dfrac{9^n \cdot n!}{(2n)!} \bigg| = $$

$$\lim_{n \to \infty} \bigg|\dfrac{(2n+2)!}{9^{n+1} \cdot (n+1)!} \cdot \dfrac{9^n \cdot n!}{(2n)!} \bigg|$$

$$\lim_{n \to \infty} \bigg|\dfrac{(2n+2)(2n+1)}{9 \cdot (n+1)} \bigg|$$

now what ... ?

 

[karush]

$\displaystyle\lim_{n \to \infty} \dfrac{2}{1 + \frac{2}{n} + \frac{1}{n^2}}=\dfrac{2}{1+0+0}=2$
so $L>1$ divergent.and$\displaystyle\lim_{n \to \infty} \bigg|\dfrac{(2n+2)(2n+1)}{9 \cdot (n+1)} \bigg|
=\bigg|\dfrac{2(n+1)(2n+1)}{9(n+1)}\bigg|=\dfrac{4n+2}{9}=\infty$
so $L=\infty$ divergent.

 

[karush]

Mahalo much...