# Another Divergence/Convergence Test

• Jul 7th 2008, 11:22 AM
auslmar
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: $\sum_{n=0}^{\infty}\frac{(4^n-2n^6)}{(8(3^n)+5n^4)}$

converges or diverges by finding a p-series,

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

or geometric series,

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

to use in the limit comparison test. Your answer should consist of three items separated by commas.
1. The first item should be an expression for the term $b_n$ of the series used in the limit comparison test.
2. The second item should be a non-zero number L so that the limit of $\frac{a_n}{b_n}$ is $L$.
3. 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 $\frac{4^n}{8(3^n)}$

and I have:

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

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

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

:)
• Jul 7th 2008, 11:53 AM
Isomorphism
Quote:

Originally Posted by auslmar

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

:)

However $\frac{(4^n)}{8(3^n)}$ diverges since $\frac43 > 1$. Thus by limit comparison test the answer is diverges...
• Jul 7th 2008, 11:57 AM
auslmar
Quote:

Originally Posted by Isomorphism
Its wrong because in ratio test, if L = 1 or if the limit does not exist, then the test is inconclusive. So you cant conclude it converges by that sequence...

Oh yes, I forgot about this! Hmm yes, well. Something else is wrong, because the system doesn't allow me to answer "inconclusive" or anything similar. And, it neither accepts "converges" or "diverges" in conjunction with my previous answers.
• Jul 7th 2008, 12:02 PM
Isomorphism
Quote:

Originally Posted by auslmar
Oh yes, I forgot about this! Hmm yes, well. Something else is wrong, because the system doesn't allow me to answer "inconclusive" or anything similar. And, it neither accepts "converges" or "diverges" in conjunction with my previous answers.

Its not wrong actually... but lets fiddle a little...

Try
$\left(\frac{4}{3}\right)^n,\frac18,$diverges
• Jul 7th 2008, 12:04 PM
auslmar
Quote:

Originally Posted by Isomorphism
Its not wrong actually... but lets fiddle a little...Try
$\left(\frac{4}{3}\right)^n,\frac18,$diverges

Well, I'll be. The system took that one! Thanks for the help.