Prove that n ! > ( n / e ) ^ n?

Question

Dr. D,

In your step f (k + 1) - f ( k ) < ln (k + 1 ), you have subtracted two inequalities. This is not correct.
For example, if a < b and c < d, you cannot say that a - c < b - d
Verify with a = 9, b = 10, c = 2, d =5

Answer 1

Short proof: Converting your question we ask whether
e^n > (n^n)/n!, yet this term is in the expansion of e^n:

e^n = 1 + n + n^2/2! + n^3/3! +....+ (n^n)/n! +.......

Converting back we get n! > (n/e)^n as required.

Answer 2

Try induction
Let's prove that (n/e)^n < n!
Take logs
n * (ln(n) - 1) < ln1 + ln2 + ... + ln(n)

We can verify that it holds for n = 1.
Let's let's try to prove it holds for n = k+1 assuming that it holds for n = k.

Let f(k) = k*[ln(k) - 1]
f(k+1) = (k+1) * [ln(k+1) - 1]

Trying to prove that f(k+1) < ln1 + ln2 + ... + lnk + ln(k+1)
is the same as proving that
f(k+1) - f(k) < ln(k+1)
LHS = k * ln[(k+1) / k] - 1 + ln(k+1)
after simplifying.

So all we have to do is prove that
k * ln[(k+1) / k] -1 < 0
or k * ln(1 + 1/k) -1 < 0
Using Maclaurin series
LHS = - 1 + k * [1/k - 1/2 * (1/k)^2 + ...]
= -1/2 (1/k) + 1/3 (1/k)^2 - 1/4 (1/k)^3 + 1/5 (1/k)^4 + ...

This series consists of alternating -ve and +ve terms, so if we can show that each pair of successive terms sums to a -ve value, we are done.

Consider the 2i th and 2i+1 th terms.
-(1/k)^(2i) / (2i+1) + (1/k)^(2i+1) / (2i+2)
= (1/k)^(2i) / [(2i+1)(2i+2)] * [-(2i+2) + (2i+1)(1/k) ]
This is negative if
1/k < (2i+2) / (2i+1) > 1
Since k >= 1, 1/k <= 1 satisfying the above condition.
Thus k * ln[(k+1) / k] -1 < 0 for all k >= 1

Thus n ! > ( n / e ) ^ n for all n
Proved by induction.

*EDIT*
Yes cdmill, it was a typo which I corrected. But I have to agree that Knasha's solution is the best.

Answer 3

Dr D:
In your inductive case, you state as an assumption that k>=1. So essentially, you've proved that (a) it's true for n=0, and (b) if it's true for n>0, then it's true for n+1, but nowhere do you prove that it's true for n>0. It's easily rectified by including n=1 as a base case.

knashha:
Assuming that you're right about the expansion, that's simply elegant.

Answer 4

As you let us know, decrease (a million+a million/n)^n = e, as n -> ? So right here we've a series of approximations of e En = (a million+a million/n)^n converging to e. needless to say partial geometric potential of such series ?m = ?(n=a million to m) (a million+a million/n)^(n/m) = (?(n=a million to m) En) ^(a million/m) additionally converge to e. it particularly is real for incredibly much any series An converging to A, if geometric potential sound no longer convincing adequate, then persist with logarithms, and perform in terms of partial arithmetic potential.

Answer 5

Look the link!

Answer 6

Well, you see. n! is infinity. and n is zero... yup!! so therefore infinity is greater than one.