Ratio Test Determines Divergence: 11.6.1

[karush]

Determine Convergence or divergence and test used
$\displaystyle\sum_{n=1}^{\infty} \dfrac{1+4^n}{1+3^n}$
W|A says diverges using ratio test so
$\therefore L=\lim_{n \to \infty}\left|\dfrac{a_n+1}{a_n}\right|>1$
Steps
$\displaystyle L=\lim_{n \to \infty}\left| \dfrac{1+4^{n+1}}{1+3^{n+1}}\cdot\dfrac{1+3^n}{1+4^n}\right|$ ok just seeing if I have this first step set up ok... before I run it thru the grinder..
I assume ratio test is a limit test...

View attachment 9345

 

[skeeter]

I would show divergence using the nth term test, i.e. ...

$$\text{if } \lim_{n \to \infty} a_n \ne 0 \text{ , then }\sum a_n \text{ diverges}$$

 

[karush]

oh... that would save a ton of calculation!