inequality

[chrislav]

Given :$$A>0$$ find $$k\in N$$ such that:

$$\forall n[n\geq k\implies\frac{1}{(n+1)!}<B]$$

 

[Deleted member 4993]

Given :$$A>0$$ find $$k\in N$$ such that:

$$\forall n[n\geq k\implies\frac{1}{(n+1)!}<B]$$
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[chrislav]

cannot recall any particular $$k$$such that
$$n\geq k\implies\frac{1}{(n+1)!}<B$$

 

[Deleted member 4993]

cannot recall any particular $$k$$such that
$$n\geq k\implies\frac{1}{(n+1)!}<B$$

What is B ? Please post the entire problem

 

[Dr.Peterson]

Given :$A>0$ find $k\in N$ such that:

$\forall n[n\geq k\implies\frac{1}{(n+1)!}<A]$

I will assume that B was meant to be A, as I've changed it above.

You are looking for a value $k$ beyond which $(n+1)!>\frac{1}{A}$, for any given $A>0$. This doesn't have to be a tight bound, so you might just use the fact that $(n+1)!>n+1$. Can you see how to use that?

But also, I wonder if you gave us only part of a bigger problem, and there might be another way to accomplish the ultimate goal.

 

[chrislav]

yes there was a typo B should be A and $$k >\frac{1}{A}\implies\frac{1}{(n+1)!}\leq\frac{1}{n+1}<\frac{1}{n}\leq\frac{1}{k}<A$$

 

[chrislav]

$$\forall n[n\geq k]$$