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.
Best Answer
$$\sum _{n=1}^{\infty }\frac{19}{n!}=19\sum _{n=1}^{\infty }\frac{1}{n!}=19\Bigl(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dotsb\Bigr)\\= 19\Bigr(-\frac{1}{0!}+\underbrace{\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dotsb}_{e}\Bigr)=19(-1+e)$$