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$\forall n\ge 2, n \in \mathbb N$ ; define $f$ as

$$f(n) = \binom{n}{2} - (n!)^{\frac{1}{n}} + n.$$

I have easily proved that $f(n) > 0$ in the given domain using A.M.-G.M. inequality.

However, I ran a program for curiosity and observed that $f$ is increasing for first $1000$ natural numbers (obviously except for $n = 1)$.

I want to determine if this is true or not for all $n \in \mathbb N$ that $f$ is an increasing function. Can you give me a hint?

Thanks.

Bill Dubuque
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An_Elephant
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  • Welcome to MSE. Here's a guide for MathJax. Please, use it to improve your your knowledge so you can write proper MathJax. For example, use \forall to get $\forall$ instead the character ∀, or use \binom{x}{y} to get $\binom{x}{y}$. – jjagmath Apr 06 '23 at 12:14
  • Maybe you need to apply induction or assume to values $n_1$, $n_2$ where the function value is the same. However, all these are very hard. – NoChance Apr 06 '23 at 13:01
  • @jjagmath Okay thanks. I did searched for mathJax equivalent on google but couldn't find it directly. Thanks for link. – An_Elephant Apr 06 '23 at 19:00
  • @NoChance I have only always used induction for proving simple equalities , let far go inequalities and if they are hard. Anyway, thanks for another method. I hope I'll learn it in future. – An_Elephant Apr 06 '23 at 19:02

1 Answers1

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We have $n! = 1\cdot 2\cdot 3\cdots n < n\cdot n\cdots n = n^n$ or $(n!)^{\frac{1}{n}} < n$ or $n - (n!)^{\frac{1}{n}} > 0$ hence,

$$ f(n+1) = {n+1\choose 2} - (n+1)!^{\frac{1}{n+1}} + (n+1) > {n+1\choose 2} = \frac{n(n+1)}{2} \tag 1 $$

Also

$$ f(n) < {n\choose 2} + n = \frac{n(n-1)}{2} + n \tag 2 $$

Now, $ f(n+1) - f(n) > \textit{something less than $f(n+1)$} - \textit{something more than $f(n)$} $

Hence $$ f(n+1) - f(n) > \frac{n(n+1)}{2} - \frac{n(n-1)}{2} - n = 0 $$

or $f(n+1) > f(n)$.

jjagmath
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Nilotpal Sinha
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