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


Note that$$\frac{a_{n+1}}{a_n}=\frac2{n+2}<1$$and that therefore $a_{n+1}<a_n$.


$$a_{n+1} = \frac{2^{n+1}}{(n+2)!} = \underbrace{\frac{2}{n+2}}_{< 1}\cdot\underbrace{\frac{2^n}{(n+1)!}}_{=a_n} < a_n$$


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

$$a_n-a_{n+1}=\frac{2^n}{(n+1)!}-\frac{2^{n+1}}{(n+2)!}=\frac{2^n(n+2)-2^{n+1}}{(n+2)!}=2^n\frac{n}{(n+2)!}> 0$$

  • $\begingroup$ Be careful: $(n + 2)! = (n + 2)(n + 1)!$. $\endgroup$ – N. F. Taussig May 13 '18 at 21:54
  • $\begingroup$ @N.F.Taussig Opsss, thanks I fix. $\endgroup$ – gimusi May 13 '18 at 21:55
  • $\begingroup$ In the first calculation, I get $\frac{2}{n + 2}$. In the second, I get $2^n \cdot \frac{n}{(n + 2)!}$. $\endgroup$ – N. F. Taussig May 13 '18 at 21:57
  • $\begingroup$ @N.F.Taussig Oh God! I revise all! Thanks again! $\endgroup$ – gimusi May 13 '18 at 21:58

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