Given as true f(x) = (1+1/x)^x is strictly increasing for x>=1

[mebigp]

Homework Statement

Given as true f(x) = (1+1/x)^x is strictly increasing for x>=1 and that f(x) has horizontal asymptote y=e.

Prove that (n/3)^n< n! <(n/2)^n for all integers n>=6 ?

Homework Equations

The Attempt at a Solution

f(x)=(1+1/x)^x is increasing and approach e
prove (n/3)^n< n! <(n/2)^n for all n>=6

So I attempt to use f(x) to replace n but that will not work for the base case 1 because n>6

 

[Mark44]

Homework Statement

Given as true f(x) = (1+1/x)^x is strictly increasing for x>=1 and that f(x) has horizontal asymptote y=e.

Prove that (n/3)^n< n! <(n/2)^n for all integers n>=6 ?

Homework Equations

The Attempt at a Solution

f(x)=(1+1/x)^x is increasing and approach e
prove (n/3)^n< n! <(n/2)^n for all n>=6

So I attempt to use f(x) to replace n but that will not work for the base case 1 because n>6

The base case doesn't have to be n = 1. Use n = 6 for your base case.

 

[mebigp]

Thanks Mark44

So I use k=F(x) from first function

Base k=6;

(k/3)^k< k! <(k/2)^k

64<180<729 true

K+1->;
((k+1)/3)^(k+1)< (k+1)! <((k+1)/2)^(k+1)

But why was I given as true f(x) = (1+1/x)^x all I did was make K=f(x)
Do I have to work out how to get a 6 from that function.( I don't think its possible)