I was given a practice question in my math lecture today, it was to prove that the sequence below was decreasing:
$a_n = \frac{2^n}{(n+1)!}$
using the condition that a sequence is decreasing if $a_{n+1} \leq a_n$
So far I have found that
$a_{n+1} = \frac{2^{n+1}}{(n+2)!}$
but I dont know how to prove this other than plugging in numbers and just showing that $a_{n+1}$ is less than $a_n$. Any help would be much appreciated
$\endgroup$3 Answers
$\begingroup$Note that$$\frac{a_{n+1}}{a_n}=\frac2{n+2}<1$$and that therefore $a_{n+1}<a_n$.
$\endgroup$ $\begingroup$$$a_{n+1} = \frac{2^{n+1}}{(n+2)!} = \underbrace{\frac{2}{n+2}}_{< 1}\cdot\underbrace{\frac{2^n}{(n+1)!}}_{=a_n} < a_n$$
$\endgroup$ $\begingroup$Note that
$$\frac{a_{n+1}}{a_n}=\frac{2^{n+1}}{(n+2)!}\frac{(n+1)!}{2^n}=\frac{2}{n+2}< 1$$
or as an alternative
$$a_n-a_{n+1}=\frac{2^n}{(n+1)!}-\frac{2^{n+1}}{(n+2)!}=\frac{2^n(n+2)-2^{n+1}}{(n+2)!}=2^n\frac{n}{(n+2)!}> 0$$
$\endgroup$ 4
