### 2 Answers

Relevance

- Barry GLv 72 years agoBest Answer
Using the AM-GM inequality :

(1+2+3+...+n)/n > [1*2*3*...n]^(1/n)

Equality does not hold because 1 ≠ 2 ≠ 3 ≠ etc. for n >= 2.

LHS = [n(n+1)/2][1/n] = (n+1)/2

RHS = [n!]^(1/n)

Therefore

[(n+1)/2]^n > n!

- Anonymous2 years ago
(1 + 2 + 3 + ... + n)/n > ⁿ√(n!)

[(n(n + 1))/2]/n > √(n!)

(n + 1)/2 > ⁿ√(n!)

((n + 1)/2)ⁿ > n! edit.

Still have questions? Get your answers by asking now.