ISI2017-MMA-7

Question

Let $n \geq 3$  be an integer.Then the statement  $$(n!)^{1/n} \leq \frac{n+1}{2}$$ is

• A. true for every $n \geq 3$
• B. true if and only if $n \geq 5$
• C. not true for $n \geq 10$
• D. true for even integers $n \geq 6$, not true for odd $n \geq 5$

Comments

just go with some values w.r.t the options

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simply use Use the AM-GM inequality on the numbers 1,......,n. It would me much easier!

Answer 1

Assume (n!)^1/n <= (n+1)/2 for all n>=1 .................(1)

<=> n! <= ( (n+1)/2 )ⁿ for all n>=1

<=> 2ⁿ * n! <= (n+1)ⁿ for all n>=1

So, if we can prove 2ⁿ * n! <= (n+1)ⁿ is true for all integer n >=1 then the inequality 1 will also be proved.

Statement P(n) : 2ⁿ * n! <= (n+1)ⁿ 

We have to prove P(n) is true for all integer n>=1

Base case : P(1) : 2<=2

So, P(1) is true

Inductive Hypothesis : We assume P(k) is true i.e 2^k*k! <= (k+1)^k for some integer k>=1 ................(2)

Inductive Step : Assuming the statement (2) we have to prove P(k+1) is true.

P(k+1) : 2^k+1 * (k+1)! <= (k+2)^(k+1)

From statement (2) we get 

2^k *k! <= (k+1)^k

Multiplying both sides of the inequality with 2(k+1) we get

2^(k+1) * (k+1)! <= 2(k+1)^k+1 <= (k+2)^k+1 ( Since 2nⁿ <= (n+1)ⁿ (THINK) )

Hence 2^(k+1) * (k+1)! <= (k+2)^k+1 ( Proved )

That means whenever P(k) is true P(k+1) is also true.

So, P(n) is true for all integer n>=1

Hence the inequality (n!)^1/n <= (n+1)/2 is true for all integer n>=1

So, Option A is the correct answer.

Answer 2

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

(A) is correct