# Math Help - determine whether the following series converge or diverge

1. ## determine whether the following series converge or diverge

Determine whether the following series converge or diverge:

(3n^4+6n^2-17)/(2n^6-9n^4+n+1). I compared it with (5/6)^n, which converges

5^n/n!. Converges by ratio test

1/l(og(n))^n converges by ratio test

(-1)^n/n^(1/3) converges by alternating series test since 1/n^(1/3) is a decreasing sequence which tends to 0

I'm a bit suspicious of my answers since I got that they all converge

2. Originally Posted by poirot
I'm a bit suspicious of my answers since I got that they all converge

(3n^4+6n^2-17)/(2n^6-9n^4+n+1). I compared it with (5/6)^n, which converges

5^n/n!. Converges by ratio test

1/l(og(n))^n converges by ratio test

(-1)^n/n^(1/3) converges by alternating series test since 1/n^(1/3) is a decreasing sequence which tends to 0

if $a_n \rightarrow a$ and $b_n \rightarrow b$ with $b_n,b \neq 0$ for all $n$ then $\frac{a_n}{b_n} \rightarrow \frac{a}{b}$.
define $a_n=\frac{3}{n^2}+\frac{6}{n^4}-\frac{17}{n^6}, \, b_n=2-\frac{9}{n^2} + \frac{1}{n^5} +\frac{1}{n^6}$. then $a_n \rightarrow 0, \, b_n \rightarrow 2$. also $b_n \neq 0$. so $\frac{a_n}{b_n} \rightarrow 0$

now $\frac{a_n}{b_n}= \frac{3n^4+6n^2-17}{2n^6-9n^4+n+1}$

3. They are series not sequences and I just need to tell whether they converge or diverge. Thanks anyway.

4. Originally Posted by poirot
Determine whether the following series converge or diverge:

(3n^4+6n^2-17)/(2n^6-9n^4+n+1). I compared it with (5/6)^n, which converges

use the limit comparison test with the known convergent series 1/n^2

5^n/n!. Converges by ratio test

correct

1/l(og(n))^n converges by ratio test

correct, but try root test

(-1)^n/n^(1/3) converges by alternating series test since 1/n^(1/3) is a decreasing sequence which tends to 0

correct
...

5. Thanks for suggesting the root test. It reminds me of the ratio test in that if c<1 it converges. Anyway so I was correct they are converge?