How to determine if the sequence $a_n= (30+12\arctan(n!))/6^n$ is divergent or convergent $$a_n= \frac{30+12\arctan(n!)}{6^n}$$
Not sure where to start, I know at infinity arctangent tends towards $\frac{\pi}{2}$. I also know I'm supposed to find the limit but not sure how to start, the factorial and the arctangent are throwing me off. limit= 0.
 A: HINT: 
$$
0<a_n<\frac{50}{6^n}.
$$
(Why?)
A: The factorial is not an issue. Since, $n!\to\infty$ as $n\to\infty$, $\arctan n!\to\frac{\pi}{2}$. On the other hand, $6^n\to\infty$, so $\lim_{n\to\infty}\frac{30+12\arctan n!}{6^n}=0$.
A: Convergence of the sequence
$$a_n= \frac{30+12\arctan(n!)}{6^n}\Rightarrow \mbox{converges}$$
Because the limit as $n\to\infty$ exists and is finite as shown below 
$$ \lim\limits_{n\to\infty}\frac{30+12\arctan(n!)}{6^n}= \frac{30+12\left(\frac{\pi}{2}\right)}{\infty} $$
$$ = \frac{30+6\pi}{\infty}=0 $$
Convergence of the series
Assuming that $a_n$ is the $n^{th}$ term of a series, we have
$$\sum\limits_{k=0}^n \frac{30+12\arctan(k!)}{6^k}$$
The divergence test is inconclusive because
 $$\lim_{n\to\infty}\frac{30+12\arctan n!}{6^n}=0$$
So now let's check for absolute convergence. Note that
$$ \left|\arctan\left(n!\right)\right|\lt \frac{\pi}{2} $$
$$ 12\left|\arctan\left(n!\right)\right|\lt 12\left(\frac{\pi}{2}\right) $$
$$ 30+12\left|\arctan\left(n!\right)\right|\lt 30+6\pi $$
$$ \frac{30+12\left|\arctan\left(n!\right)\right|}{|6^n|}\lt \frac{30+6\pi}{|6^n|} $$
$$ \left|\frac{30+12\arctan\left(n!\right)}{6^n}\right|\lt (30+6\pi)\left|\frac{1}{6}\right|^n $$
Since $ \left|\frac{1}{6}\right|\lt 1 $, by the geometric series test we have
$$ (30+6\pi)\sum\limits_{k=0}^n\left(\frac{1}{6}\right)^k\Rightarrow \mbox{absolutely convergent} $$
Therefore by the direct comparison test
$$ \sum\limits_{k=0}^n \frac{30+12\arctan\left(k!\right)}{6^k} \Rightarrow \mbox{absolutely convergent} $$
A: Hint: Since you know $\arctan$ is bounded you're halfway there. Do you see a way to use the squeeze theorem?
A: To help erasing blanks drawn:
$$0\xleftarrow[\infty\leftarrow n]{}\frac{30-12\frac\pi2}{6^n}\le\color{red}{\frac{30+12\arctan n!}{6^n}}\le\frac{30+12\frac\pi2}{6^n}\xrightarrow[n\to\infty]{}0$$
