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?
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: $$ \sum_{n\geq 1}\frac{1}{(3n+8)!} = \color{red}{-\frac{20497}{40320}+\frac{e}{3}-\frac{2}{3\sqrt{e}}\sin\left(\frac{\sqrt{3}}{2}+\frac{\pi}{6}\right)}.\tag{4}$$
$$\lim_{n\to\infty}\frac{1}{(3n+11)(3n+10)(3n+9)} = 0 < 1$$
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 $$\frac{(3n+11)!}{(3n+8)!}=(3n+11)(3n+10)(3n+9)$$
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.
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.
