What convergence/divergence test can I use on this series? - Calculus II

[opticaltempest]

What test can I use on the following series in order to determine
if it converges or diverges? Looking at it graphically it appears to
diverge but I cannot show it analytically.

$$\sum\limits_{n = 1}^\infty {\frac{{n!}}{{3n! - 1}}}$$

Using the Ratio Test, here is I got thus far

$$\left| {\frac{{a_{n + 1} }}{{a_n }}} \right| = \left| {\frac{{\left( {\frac{{(n + 1)!}}{{3(n + 1)! - 1}}} \right)}}{{\left( {\frac{{n!}}{{3n! - 1}}} \right)}}} \right|$$

$$= \frac{{(n + 1)!}}{{3(n + 1)! - 1}} \cdot \frac{{3n! - 1}}{{n!}}$$

$$= \frac{{(n + 1)n!}}{{3(n + 1)n! - 1}} \cdot \frac{{3n! - 1}}{{n!}}$$

$$= \frac{{(n + 1)(3n! - 1)}}{{3(n + 1)n! - 1}}$$

At this point, I don't see how the Ratio test will work. What test can I use on this series?

Should I start off with long division to reduce the improper rational
expression?

http://img164.imageshack.us/img164/4656/longdiv7nr.jpg

So the equivalent sum would be

$$\sum\limits_{n = 1}^\infty {\left( {\frac{1}{3} + \frac{1}{{3(3n! - 1)}}} \right)}$$

Which must diverge because of the 1/3 always being summed from
1 to infinity.

Is that the correct way to do it? If so, is it the only way? Am I
missing a simple test that could be used?

 

[KoGs]

oops nevermind I looked again and it seems u already did that.

 

[benorin]

Maybe comparison test. But the work have looks good to me.

 

[HallsofIvy]

I would be inclined to use the fact that is a_n does not go to 0 then $\Sum a_n$ does not converge!