Determine if the series $\sum_{n=1}^∞ \frac{(8^n)}{n!}$ converges

Question

I know that it converges for n > 16 by the ratio test because I looked at the answer key but am completely lost by the answer keys work. Any help would be much appreciated.

Answer 1

So you have $a_n = 8^n/n!$ and so $$ \frac{a_{n+1}}{a_n} = \frac{8^{n+1}/(n+1)!}{8^n/n!} = \frac{8^{n+1}}{8^n} \times \frac{n!}{(n+1)!} = \frac{8}{1} \times\frac{1}{n+1} = \frac{8}{n+1} \to 0... $$

Answer 2

Since $$e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$$

Your series is $$-1+\sum_{n=0}^{\infty}\frac{8^n}{n!}=e^8-1$$

Answer 3

As gt6989b pointed out: $$\frac{a_{n+1}}{a_n}=\frac{8}{n+1}$$ When this subjected to ratio test: $$\lim_{n\to\infty}\frac{\vert a_{n+1}\vert}{\vert a_n\vert}=\lim_{n\to\infty}\frac{8}{n+1}\to 0$$ This tells us that $\sum_{n=1}^{\infty} \left(\frac{8^n}{n!}\right)$ converges.