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For $x > 1$, $n > 2$ with $2 \mid x+1$ and $n \mid x+1$, does it then follow that:

$$\log(\lfloor\frac{x}{2}\rfloor!) - \log(\lfloor\frac{x}{n}\rfloor!) \le \log(\lfloor\frac{x+1}{2}\rfloor!) - \log(\lfloor\frac{x+1}{n}\rfloor!)$$

I am especially interested how I would evaluate whether this is always true.

Thanks very much,

-Larry

Larry Freeman
  • 10,189

1 Answers1

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Using $\log(ab)=\log(a)+\log(b)$ you can derive $\log(n!)=\sum_{i=1}^n\log(i)$ and write the left hand side as $$ \sum_{i=1}^{\lfloor \frac{x}{2}\rfloor}\log(i)-\sum_{i=1}^{\lfloor \frac{x}{n}\rfloor}\log(i). $$ The right hand side is the same except that you add one term (for $i=\frac{x+1}{2}$ and $i=\frac{x+1}{n}$) to each of the sums. It remains to show that $$ \log(\frac{x+1}{2})-\log(\frac{x+1}{n})>0. $$ This is true since the logarithm is strictly monotonly increasing and by your assumptions $\frac{x+1}{2}>\frac{x+1}{n}$. So you always have a strict inequality.

MichalisN
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