# Thread: Absolute Convergence and the Ratio Test

1. ## Absolute Convergence and the Ratio Test

$\sum(-1)^{1+k}\frac{k^2\cdot4^k}{k!}$

Determine if the series is absolutely convergent, conditionally convergent or divergent. We're using the Ratio and Root Test and absolute convergence in this section.

2. Originally Posted by kl.twilleger
$\sum(-1)^{1+k}\frac{k^2\cdot4^k}{k!}$

Determine if the series is absolutely convergent, conditionally convergent or divergent. We're using the Ratio and Root Test and absolute convergence in this section.
First lets check for absolute convergence. So

$\sum\bigg|\frac{(-1)^{n+1}n^2\cdot{4^n}}{n!}\bigg|$

$=\sum\frac{n^2\cdot{4^n}}{n!}$

So using the root test we see that

$\lim_{n\to\infty}\left(\left|\frac{n^24^n}{n!}\rig ht|\right)^{\frac{1}{n}}$

$=\lim_{n\to\infty}\left(\frac{n^{\frac{2}{n}}4}{(n !)^{\frac{1}{n}}}\right)$

Using the fact that

$n!\sim\sqrt{2\pi{n}}n^ne^{-n}$

We can see that our limit is equal to zero, thus the series converges. And I'm sure you know that if

$\sum|a_n|\text{ converges}\Rightarrow\sum{a_n}\text{ converges}$

$\therefore\sum\frac{(-1)^{n+1}n^2\cdot{4^n}}{n!}$

converges absolutely.