/sci/ - Science & Math

[Anonymous No.5245437]

Hey /sci/, I need to prove by induction and with following help:
e = [an, bn]
an = (1+1/n)^n
bn = (1+1/n)^n+1

e(n/e)^n <= n! <= n*e(n/e)^n

- for n=1
e*1/e = 1 <= 1 <= 1e*1/e

So:
e((n+1)/e)^n+1 <= (n+1)! <= (n+1)*e((n+1)/e)^n+1

What now ?
I could
e * (n+1)/e * ((n+1)/e)^n <= (n+1)n! <= (n+1)e * ((n+1)/e) * ((n+1)/e)^n

n+1 ((n+1)/e)^n <= (n+1)n! <= (n+1)² * ((n+1)/e)^n

[Anonymous No.5245457]

bump

[Anonymous No.5245467]

Write it in a readable form if you want help.

[Anonymous No.5245505]

>>5245437
Jesus Christ I can't even read that crap.

Rewrite that shit and I'll try to help you out.l

[Anonymous No.5245512]

Simply: Show by induction

e* (n/e)^n <= n! <= n*e* (n/e)^n

[Anonymous No.5245528]

>>5245512
For which inequality do you need help?
e* (n/e)^n <= n!
or
n! <= n*e* (n/e)^n

[Anonymous No.5245555]

both ?

[Anonymous No.5245601]

>>5245555
Ok, here is the first assuming you have already checked the trivial base case.

<div class="math">(n+1)! = (n+1) \, n! </div><div class="math"> \geq (n+1) e \left( \frac{n}{e}\right)^n </div><div class="math"> = e \left( \frac{n+1}{e}\right)^{n+1} e \left( \frac{n}{n+1}\right)^n </div><div class="math"> = e \left( \frac{n+1}{e}\right)^{n+1} \left(e^{1/n} \right)^n \left( \frac{n}{n+1}\right)^n </div><div class="math"> \geq e \left( \frac{n+1}{e}\right)^{n+1} \left( 1 + {1 \over n} \right)^n \left( \frac{n}{n+1} \right)^n </div><div class="math"> = e \left( \frac{n+1}{e}\right)^{n+1}</div>

[Anonymous No.5245647]

Thanks alot, but I don't get the third step