Determine whether the series converges or diverges. Determine whether the series converges or diverges. 
$$
 \sum _{n=1}^{\infty }\:\left(\frac{19}{n!}\right)
$$
I know that this question a lot easier if I use ratio test but I  have not learned ratio test yet. The only option I have is divergence, comparison, limit comparison, and integral test. How can I prove that this series converges by using the limited tests. 
Thanks in advance. 
 A: Then use the fact that $(\forall n\in\mathbb{N}\setminus\{2,3\}):\frac{19}{n!}\leqslant\frac{19}{n^2}$ and apply the integral test in order to prove that $\sum_{n=1}^\infty\frac{19}{n^2}$ converges.
A: $$\sum_{n=1}^{\infty}\frac{19}{n!}=19\left(1+\sum_{n=2}^{\infty}\frac{1}{n!}\right)<19\left(1+\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\right)=$$
$$=19\left(1+\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)\right)=19(1+1)=38.$$
Thus, our series converges.
A: Forget about the constant $19.$ Note that in $n!$ you have $n-1$ factors, each of which is $\ge 2.$ Thus $n! \ge 2^{n-1} \implies 1/n! \le 1/2^{n-1}.$ Use the comparison test.
A: Use inequality
$$ n! > 2^n, \ n\geqslant 4.$$
Then for $n\geqslant 4$
$$\frac{1}{n!} < \frac{1}{2^n},$$ 
so
$$\sum_{n=1}^{\infty }\frac{19}{n!} = 19\sum_{n=1}^{\infty }\frac{1}{n!} < 19\left(1 + \frac{1}{2} + \frac{1}{6} + \sum_{n=4}^{\infty }\frac{1}{2^n}  \right)=19\left(1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{8}\right).$$
A: $$\sum _{n=1}^{\infty }\frac{19}{n!}=19\sum _{n=1}^{\infty }\frac{1}{n!}=19\Bigl(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dotsb\Bigr)\\=
19\Bigr(-\frac{1}{0!}+\underbrace{\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dotsb}_{e}\Bigr)=19(-1+e)$$
