# Math Help - Convergance

1. How can i decide if these sequence converge, then go on to state its limit?

$a_n = \frac {9 + 8(0.6)^n}{6(0.3)^n - 6}$ $(n=1,2,3,...)$

Thanks guys

do i start like this:

$a_n = \frac {9 + (4.8)^n}{(1.8)^n - 6}$

2. Originally Posted by ADY
do i start like this:

$a_n = \frac {9 + (4.8)^n}{(1.8)^n - 6}$
You seem to have the misunderstanding that $a (b)^n = (ab)^n$.

Both (0.6)^n and (0.3)^n approach zero as n approaches infinity. What does that tell you?

3. They are divergent?

4. Originally Posted by ADY
They are divergent?
it tells you that $\lim\limits_{n \to \infty} a_n \neq 0$. And that implies......?

5. Originally Posted by Sampras
it tells you that $\lim\limits_{n \to \infty} a_n \neq 0$. And that implies......?
It is convergent?

6. Originally Posted by ADY
It is convergent?
no, divergent.

7. Ok, thank you - i really need to understand this further. Ill try another example

$b_n = \frac{2 - 7n^2}{3n + 4}$

8. so do we say

$b_n = \frac{2 - 7n^2}{3n + 4} \rightarrow b_n = \frac{2 - 7n}{3 + 4/n}$

so 3 + 4/n converges to 3, so it convergent? Whats it limit

9. Originally Posted by ADY
How can i decide if these sequence converge, then go on to state its limit?

$a_n = \frac {9 + 8(0.6)^n}{6(0.3)^n - 6}$ $(n=1,2,3,...)$

Thanks guys

do i start like this:

$a_n = \frac {9 + (4.8)^n}{(1.8)^n - 6}$
$\lim_{n \rightarrow + \infty} a_n = \lim_{n \rightarrow + \infty}\frac {9 + 8(0.6)^n}{6(0.3)^n - 6} = \frac{9 + 0}{0 - 6} = ....$

Therefore the sequence (NB: not series) converges.

Originally Posted by Sampras
it tells you that $\lim\limits_{n \to \infty} a_n \neq 0$. And that implies......?
It is a sequence not a series.