Use ratio test to determine convergence or divergence. Find general term first. I have this series:
$$\frac{2}{3} + \frac{2*5}{3*5} + \frac{2*5*8}{3*5*7} ...$$
I'm confused how to find the general term here to then apply the ratio test.
So it starts with a $$\frac{2}{3}$$ and the next term is $\frac{2*5}{3*5}$
So each time we multiply by what appears to be another term that is $3(n-1)$ bigger than the original 2 in the numerator, and $2(n-1)$ bigger than the original 3 in the denominator. But how do I capture this in a general formula?
But the general term is not:
$$\frac{2 + 3(n-1)}{3 + 2(n-1)}$$
because that would be:
$$\frac{2}{3} + \frac{2*5}{3*5} + \frac{2*8}{3*7}$$
and would leave out all the middle terms. What am I doing wrong?

Find formulae for the product terms in the numerator and the denominator. The questions seems to suggest that they are arithmetic progressions. In particular, the $n$th term (not the $n$th partial product) in the product $2 \cdot 5 \cdot 8 \cdot \ldots$ is supposedly $2 + 3(n - 1)$. Similarly, the $n$th term in the product $3 \cdot 5 \cdot 7 \cdot \ldots$ is supposed to be $3 + 2(n - 1)$. Therefore, to get from the $n$th term to the $(n + 1)$th term, you must multiply by the $(n + 1)$th term of the former product, divided by the $(n + 1)$th term of the latter product. More succinctly, if $a_n$ is the $n$th series term, then
$$a_{n+1} = a_n \cdot \frac{2 + 3(n + 1 - 1)}{3 + 2(n + 1 - 1)} = a_n \cdot \frac{2 + 3n}{3 + 2n}.$$
The ratio you want is therefore
$$\frac{a_{n+1}}{a_n} = \frac{2 + 3n}{3 + 2n}.$$
