Question

Here's a good one.
Prove by induction that (n!)^1/n < (n+1)/2

Answer 1

Let's see if anyone can solve this gem!

Answer 2

\[\sqrt[n]{n!} \le \frac{ n+1 }{ 2 }\]

Answer 3

first step write for n=1 so than will get

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

1! = 1

1^1 =1

1 <= 2/2

1 <= 1 so this is true

now write for n=k and accept it true and on the next step write for k=k+1 and prove it true

for n=k you get

(k!)^(1/k) <= (k+1)/2 so this accept it true

for k=k+1 will get

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

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

so now you need prove this that is true

for k=1 what will get

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

(2!)^(1/2) <= 3/2

2! = 1*2 = 2

(2)^(1/2) = sqrt2 = 1,41

(1+2)/2 = 3/2 = 1,5

1,41 < 1,5 so what is true

q.e.d.

hope you understand it