Another Divergence/Convergence Test

Hello,

So, I've got yet again another problem where I'm not sure if I made an error or the system is fickle. The problem is as follows:

"Show that the series: $\displaystyle \sum_{n=0}^{\infty}\frac{(4^n-2n^6)}{(8(3^n)+5n^4)}$

converges or diverges by finding a p-series,

$\displaystyle \sum_{n=1}^{\infty}\frac{c}{n^p}$

or geometric series,

$\displaystyle \sum_{n=1}^{\infty}cr^n$

to use in the limit comparison test. Your answer should consist of three items separated by commas.- The first item should be an expression for the term $\displaystyle b_n$ of the series used in the limit comparison test.
- The second item should be a non-zero number L so that the limit of $\displaystyle \frac{a_n}{b_n}$ is $\displaystyle L$.
- The third item should be the word
**converges** or the word ** diverges** to indicate whether you are saying that the given series converges or diverges."

So, I used $\displaystyle \frac{4^n}{8(3^n)}$

and I have:

$\displaystyle L = $ $\displaystyle \lim_{n \to \infty}\frac{\frac{4^n-2n^6}{8(3^n)+5n^4}}{\frac{4^n}{8(3^n)}}$ = $\displaystyle \lim_{n\to\infty}\frac{(4^n-2n^6)(8(3^n))}{(8(3^n)+5n^4)(4^n)}$ = $\displaystyle \lim_{n\to\infty}\frac{(3^n)(4^n)(8-16n^6(1/4^n))}{(3^n)(4^n)(8+5n^4(1/3^n))}$ = $\displaystyle \lim_{n\to\infty}\frac{8-16n^6(1/4^n))}{8+5n^4(1/(3^n))}$ = $\displaystyle \frac{8}{8} = 1 \ne 0$

therefore, $\displaystyle \sum_{n=0}^{\infty}\frac{(4^n-2n^6)}{(8(3^n)+5n^4)}$ converges.

and my answer looks like:

$\displaystyle \frac{(4^n)}{8(3^n)}$,1,converges

which is not recognized as correct by the system. So, I'm wondering if I've faulted somewhere along the lines or if the system is just being picky. If you can offer any hints or ideas, I would be grateful.

Thanks

:)