# For every integer n > 1, the inequality (n1)1/n

Navjyot Kalra
10 years ago
Consider n numbers, namely 1, 2, 3, 4, . . . . . . n.
KEY CONCEPT: Now using A. M. > G. M. for distinct numbers, we get
1 + 2 + 3 + 4 . . . . . . . + n/n > (1. 2. 3. 4 . . . . n )1/n
⇒ n(n + 1)/2n > (n!)1/n ⇒ (n!)1/n < n + 1/2 ∴ True
ALTERNATE SOLUTION :
The given inequality can be written as
N! < (n + 1/2)n n > 1.
Let us use mathematical induction to check the validity of given inequality.
For n = 2, we have 2! < (3/2)2 = 9/4 which is true
∴ Inequality is valid for n = 2.
Let it be valid for n = k then k ! < (k + 1/2)k . . . . . . . . . . . . . . (1)
Consider
(k + 1)! = (k + 1) k! < (k + 1) ( k + 1/2)k (Using (1))
Now we will try to check
(k + 1) (k + 1/2)k < (k + 2/2)k + 1 . . . . . . . . . . . . . (2)
Which is equivalent to write 2 < (k + 2/k + 1)k + 1 . . . . . . . . . . . . (3)
Now, (k + 2/k + 1)k + 1 = (1 + 1/k + 1)k + 1
= 1 + (k + 1) 1/k + 1 + (k + 1)k/2! (1/k + 1)2 + . . . . . . . .(Using Binomial expansion)
∴ (k + 2/k + 1)k + 1 > 2 ⇒ (3) holds and hence (2) holds
⇒ (k + 1)! < (k + 2/2)k +1 . Thus statement is true for
n = k + 1
Hence, by Principle of Mathematical Induction given statement is true, ∀ n > 1.