inequality

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chrislav

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Given :[math] A>0[/math] find [math]k\in N[/math] such that:

[math]\forall n[n\geq k\implies\frac{1}{(n+1)!}<B][/math]
 
Given :[math] A>0[/math] find [math]k\in N[/math] such that:

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

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cannot recall any particular [math] k[/math]such that
[math]n\geq k\implies\frac{1}{(n+1)!}<B[/math]
 
chrislav said:
Given :[imath] A>0[/imath] find [imath]k\in N[/imath] such that:

[imath]\forall n[n\geq k\implies\frac{1}{(n+1)!}<A][/imath]
Click to expand...
I will assume that B was meant to be A, as I've changed it above.

You are looking for a value [imath]k[/imath] beyond which [imath](n+1)!>\frac{1}{A}[/imath], for any given [imath]A>0[/imath]. This doesn't have to be a tight bound, so you might just use the fact that [imath](n+1)!>n+1[/imath]. 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.
 
yes there was a typo B should be A and [math]k >\frac{1}{A}\implies\frac{1}{(n+1)!}\leq\frac{1}{n+1}<\frac{1}{n}\leq\frac{1}{k}<A[/math]
 
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