1. ## Limit Comparison Test

$\displaystyle \sum^ \infty _{n=1} \frac{2}{3^n-5}$ = $\displaystyle \lim_{n\rightarrow \infty} \frac{6^n}{6^n-25} = 1 > 0$

This is what I think:

The original function is divergent since $\displaystyle \sum^ \infty _{n=1} \frac{2}{3^n}$ is divergent.

Is this right? Could you explain to me how to tell if the function is div. or conv. based on my Comparison Test???

2. Originally Posted by saiyanmx89
$\displaystyle \sum^ \infty _{n=1} \frac{2}{3^n-5}$ = $\displaystyle \lim_{n\rightarrow \infty} \frac{6^n}{6^n-25} = 1 > 0$

This is what I think:

The original function is divergent since $\displaystyle \sum^ \infty _{n=1} \frac{2}{3^n}$ is divergent.

Is this right? Could you explain to me how to tell if the function is div. or conv. based on my Comparison Test???
This series is convergent

$\displaystyle \sum^ \infty _{n=1} \frac{2}{3^n}=2\cdot \frac{\frac{1}{3}}{1-\frac{1}{3}}=1$

It is a geometric series with $\displaystyle r=\frac{1}{3}$

How will this change your conclusion