Exponential Inequality Proof: n! >or= (n^n)e^(1-n) for Positive Integers

[moo5003]

Homework Statement

Prove: n! >or= (n^n)e^(1-n)

Edit: For some positive integers n. (I'm very sure its all)

The Attempt at a Solution

Proof by Induction:

Base: n=1
1 >or= (1^1)e^0 = 1

Induction Step (n->n+1)

(n+1)! = (n+1)n! >= (n+1)(n^n)e^(1-n) = (n+1)(n^n)ee^(-n)

Now I want to show (n+1)(n^n)e is greater or equal to (n+1)^(n+1)

Taking natural log of both sides we get:

ln(n+1) + nln(n) + 1 >=? (n+1)ln(n+1)

nln(n) + 1 >=? nln(n+1)

ne^(1/n) >=? n+1

e >=? (1+1/n)^n

Here at the very end of my proof is where I get stuck. I know that e can be defined as the limit as n approaches infinity of (1+1/n)^n but I'm unsure how to show that its greater then the sequence at all specific values of n. Any help would be appreciated.

 

[Dick]

Ok, so you want to show (1+1/n)^n not only approaches e, but that it is increasing. You could do this by showing (1+1/x)^x is an increasing function for x>=1. I.e. show it's derivative is positive. The only way I've figured out so far is looking at power series expansions of log(1+x). It's not very elegant. Maybe you can think of something better.

 

[moo5003]

Alright I think I solved the last part.

e >=? (1+1/n)^n

then

[ne^(1/n) - 1]/n >=? 0

which occurs iff

ne^(1/n)-1 >=? 0

e^(1/n) >=? 1/n

1/n >=? ln(1/n)

Since 1>= ln(1) and ln(1/n) becomes negative for higher n it holds that 1/n>=ln(1/n) and therefore e >= (1+1/n)^n. Thus our induction is finished.

Edit: I'll rewrite the proof for class to be more presentable, but I was wondering if this is totally rigorous?

 

[Gib Z]

I have a nice solution, but it uses a more powerful result than your question :(

 

[Dick]

Alright I think I solved the last part.

e >=? (1+1/n)^n

then

[ne^(1/n) - 1]/n >=? 0

which occurs iff

ne^(1/n)-1 >=? 0

e^(1/n) >=? 1/n

1/n >=? ln(1/n)

Since 1>= ln(1) and ln(1/n) becomes negative for higher n it holds that 1/n>=ln(1/n) and therefore e >= (1+1/n)^n. Thus our induction is finished.

Edit: I'll rewrite the proof for class to be more presentable, but I was wondering if this is totally rigorous?

It's worse than not totally rigorous. It's wrong. The second line "[ne^(1/n) - 1]/n >=? 0" should be >=1.

 

[Dick]

Can you use the power series expansion of e^x at x=0? That's a nice way to show e^(1/n)>(1+1/n).

 

[moo5003]

Alright, I fixed my mistake:

e >=? (1+1/n)^n

e^(1/n) >=? 1 + 1/n

n(e^(1/n)-1) >=? 1

But e^x >=? 1+x since if you take the derivative of either side:

e^x >= 1 Thus e^x has the the same tangent line at x=0 and a strictly greater slope for x > 0 and since e^x = x+1 at x=0 we know that e^x >= x+1 for x>0 (True for x<0 but not needed here).

Thus n(e^(1/n)-1) >= n(1/n + 1 - 1) = 1 >= 1. Thus the inequality holds.

BTW: Thanks for catching my mistake, I wouldn't have caught it in all likelihood.

 

[Dick]

I think that's it.

 

[Gib Z]

If you don't mind using the strong result, the inequality is trivial with Stirling's series.