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This is actually one part of a larger problem I am solving. I have so far used Calculus to show that the sequence converges and I have made a conjecture that the limit of the sequence is 0. Now I need to show that this sequence is monotone decreasing before proving that the sequence does indeed converge to my conjectured limit of 0.

For the base case, I have that $x_1=2$ and $x_2=2$ so this follows the definition of monotone decreasing, that $x_n \geq x_{n+1}$. Now I am trying to use induction and I am a bit unsure how to begin this process. I tried setting $2^n/n! \geq 2^{n+1}/(n+1)!$ and then breaking down $2^{n+1}/(n+1)!$ into $2^n/n!*2/(n+1)$ and then multiplying both sides by $(n+1)/2$. I guess my goal was to show that the LHS would still be greater than the RHS after doing this.

At this point I am kind of stuck and I have no idea if this is what I should even be doing here. Any help would be greatly appreciated.

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you have done it! more confidence needed?

clearly for $n \gt 1$ $$ \frac2{n+1} \lt 1 \tag{1} $$ multiplying both sides by $\frac{2^n}{n!}$ gives: $$ \frac{2^n}{n!}\frac2{n+1} \lt \frac{2^n}{n!} 1 $$ i.e. $$ \frac{2^{n+1}}{(n+1)!} \lt \frac{2^n}{n!} $$

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  • $\begingroup$ I see now! Thank you. Does it matter that I worked backwards though? For some reason, I felt it was wrong to start out showing what I was trying to prove, even though I didn't know what else to do. I was just trying to get some ideas. $\endgroup$ – user554608 Oct 17 '18 at 5:25
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    $\begingroup$ To the proposer: It is often useful to look at the implications of what you want to prove in order to get some ideas. $\endgroup$ – DanielWainfleet Oct 17 '18 at 6:36
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Denote $a_n=\frac {2^n}{n!}$, we have $$\frac{a_{n+1}}{a_n}=\frac{2^{n}n!2/(n+1)}{2^nn!}=\frac2{n+1}<1$$ when $n>2$.
Since $a_n>0$ for all $n\in\mathbb{N}$, $a_{n+1}<a_n$.

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