
Does the sum converge?
There are two different questions but I am hoping that if I get help with the first one, I will be able to do the second one on my own.
Find the sum of the series if it converges or show that it diverges
∑ n=0 to infinity of [2^(3n)1] / [ 3^(2n)]
Do I do the nth term test and take the limit of it? I am not good at limits either.
Thanks!

Is this the series? $\displaystyle \sum_{n=0}^{\infty} \frac{2^{3n}  1}{3^{2n}}$
In any case, note that this is equivalent to:
$\displaystyle = \sum_{n=0}^{\infty} \left[ \frac{\left(2^3\right)^n}{\left(3^2\right)^{n}}  \frac{1}{\left(3^2\right)^n}\right] \ = \ \sum_{n=0}^{\infty} \left[ \left( \frac{8}{9}\right)^n  \left(\frac{1}{9}\right)^n\right]$
by simple exponent rules. We can see this is a sum of two geometric series ($\displaystyle r < 1$) which of course converges.

thanks o_O. But wow, I definitely need to review my notes because that still doesn't make that much sense! I will work through what you gave me and with my notes to see if I understand it
Thanks again!