Convergence of $\sum_{n=1}^{\infty}\frac{1}{(3n+8)!}$ 
Determine whether or not this series converges or diverges $$\sum_{n=1}^{\infty}\dfrac{1}{(3n+8)!}$$

My attempt: I used the ratio test and ended up having 
$$\lim_{n\to\infty}\frac{1}{(3n+11)(3n+10)(3n+9)(3n+8)}$$
which I then plugged in $\infty$ for $n$ and got $1/\infty$ which makes the limit $1$. Since the limit is less than $1$ the series converges according to the ratio test. Is this correct? 
 A: We can do more than prove convergence, we can compute it.
Let $\omega=\exp\left(\frac{2\pi i}{3}\right)$. Then:
$$ \sum_{n\geq 0}\frac{x^{3n}}{(3n)!} = \frac{1}{3}\left(e^{x}+e^{\omega x}+e^{\omega^2 x}\right)\tag{1}$$
hence, by differentiating both sides:
$$ \sum_{n\geq 0}\frac{x^{3n+2}}{(3n+2)!}= \frac{1}{3}\left(e^{x}+\omega\, e^{\omega x}+\omega^2\,e^{\omega^2 x}\right)\tag{2}$$
so, by evaluating at $x=1$:
$$ \sum_{n\geq 0}\frac{1}{(3n+2)!} = \frac{e}{3}-\frac{2}{3\sqrt{e}}\sin\left(\frac{\sqrt{3}}{2}+\frac{\pi}{6}\right)\tag{3}$$
then subtracting the first terms of the LHS:
$$ \sum_{n\geq 1}\frac{1}{(3n+8)!} = \color{red}{-\frac{20497}{40320}+\frac{e}{3}-\frac{2}{3\sqrt{e}}\sin\left(\frac{\sqrt{3}}{2}+\frac{\pi}{6}\right)}.\tag{4}$$
A: $$\lim_{n\to\infty}\frac{1}{(3n+11)(3n+10)(3n+9)} = 0 < 1$$
The limit of the above is $0$ since informally, as you've said $1/\infty = 0$. Take any small number and divide it by increasingly huger numbers on your calculator, you'll see the result approaching $0$. Since obviously $0 < 1$, the series converges by the ratio test. 
Also, as Jack pointed out below, the $3n+8$ should not appear in the denominator, since $$\frac{(3n+11)!}{(3n+8)!}=(3n+11)(3n+10)(3n+9)$$ 
A: You are right.
You may also notice that
$$
(3n+8)!>n^2,\qquad n\geq1,
$$ giving
$$
0<\sum_1^{\infty}\frac1{(3n+8)!}<\sum_1^{\infty}\frac1{n^2}<+\infty
$$ and thus the convergence of your initial series by the comparison test.
A: Yet another way to do this one. See that 
$$
\sum \frac{1}{n!}
$$
is convergent isn't hard. Then note that
$$
\frac{1}{(3n+2)!} < \frac{1}{n!}
$$
and use the Comparison Test.
