#### K1ngbadlex

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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...

- Thread starter K1ngbadlex
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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...

- Joined
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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.

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Thanks for your reply

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.

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

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

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.

Thank you

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

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.

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Thank you very much, I have prove it thanks to your help. Have a nice day | ||||||