Factorial inequation

[K1ngbadlex]

Hello,i hope everybody on this forum is fine, I'm actually in grade 12 and I need
Help for a inequation with factorial :
(K^(n+1))/K(n!) <= (K^K) / K!
Thank you
Ps sorry for my English it is not my mother langage...

 

[Dr.Peterson]

What is the goal? Are you trying to solve for K in terms of n, or for n in terms of K, or to prove it is always true, or what? I assume n and K are both natural numbers (positive integers).

Also, please tell us your context (what subject you are studying if this is for a class, and what you have learned that might be useful), and show us your thinking, so we can tell what help you need.

 

[K1ngbadlex]

What is the goal? Are you trying to solve for K in terms of n, or for n in terms of K, or to prove it is always true, or what? I assume n and K are both natural numbers (positive integers).

Also, please tell us your context (what subject you are studying if this is for a class, and what you have learned that might be useful), and show us your thinking, so we can tell what help you need.

Thanks for your reply
You are right I'm sorry, I want to prove that this is always true, I forgot to add, like you say, that N is a natural number and K also and K > n
I'm a student in Japan and I am doing Exercice in recurrence thinking (I don't know if my translation is true) but it is basically when you suppose a statement for a n number and you prove it with n+1 usually used in sequence...
Thank you

 

[K1ngbadlex]

What is the goal? Are you trying to solve for K in terms of n, or for n in terms of K, or to prove it is always true, or what? I assume n and K are both natural numbers (positive integers).

Also, please tell us your context (what subject you are studying if this is for a class, and what you have learned that might be useful), and show us your thinking, so we can tell what help you need.

Edit : I changed my previous reply because I did a mistake it is k>n not k>0!
Thank you

 

[ksdhart2]

To me this almost feels like a bit of a "trick" question, in that it seems really complex but it's actually not. You don't even need to do a proof by induction (the concept you were describing in post #3), as you can directly demonstrate it's true for any pair \( (K, n) \) that satisfies \(K \gt n\). Work on the left-hand side and divide top and bottom by \(K\):

\(\displaystyle \frac{K^{n+1}}{K \times n!} = \frac{\left( \frac{K^{n+1}}{K} \right)}{\left( \frac{K \times n!}{K} \right)}\)

For this step to be valid, you do have to assume that \( K \neq 0 \), but that's already known because \(K\) and \(n\) are natural numbers and \(K \gt n\) ensures \(K \geq 1\) anyway. Once you've done this step, the rest of the proof should be clear.

 

[K1ngbadlex]

To me this almost feels like a bit of a "trick" question, in that it seems really complex but it's actually not. You don't even need to do a proof by induction (the concept you were describing in post #3), as you can directly demonstrate it's true for any pair \( (K, n) \) that satisfies \(K \gt n\). Work on the left-hand side and divide top and bottom by \(K\): \(\displaystyle \frac{K^{n+1}}{K \times n!} = \frac{\left( \frac{K^{n+1}}{K} \right)}{\left( \frac{K \times n!}{K} \right)}\) For this step to be valid, you do have to assume that \( K \neq 0 \), but that's already known because \(K\) and \(n\) are natural numbers and \(K \gt n\) ensures \(K \geq 1\) anyway. Once you've done this step, the rest of the proof should be clear. Thank you very much, I have prove it thanks to your help. Have a nice day