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Let $x$ be a number such that $x\gt 0$ and $x\in\mathbb{R}$. Is it true that the maximum value of the expression $$\frac{x^n}{n!}$$ occurs for $n\in\mathbb{N}$ where $n=\lceil x \rceil - 1$? If true, how does one prove this?

My attempt was to find conditions such that $$\frac{x^{n+1}}{(n+1)!} \gt \frac{x^n}{n!}$$ $$\frac{x}{n+1} \gt 1$$ $$x\gt n+1$$ So the consecutive values of $\frac{x^n}{n!}$ are only increasing if $x\gt n+1$ which will be true for all $n+1\lt x$ - the maximum of which is $n+1=\lfloor x \rfloor$. This leads to the value given above.

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    $\begingroup$ What you want is related to this: math.stackexchange.com/questions/246496/…. (There, they are maximising $\frac{\lambda^k}{k!}$ as a function of $k\in\mathbb{N}$. Their $\lambda$ is your $x$.) $\endgroup$ – Minus One-Twelfth Mar 16 at 23:19
  • $\begingroup$ Your way is very good for it. Good job $\endgroup$ – Jakobian Mar 16 at 23:24
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To find the maximum value in the sequence $\left(\tfrac{x^n}{n!}\right)_{n\in\Bbb{N}}$, note that for any $n\in\Bbb{N}$ the ratio of two consecutive terms is $$\frac{\tfrac{x^n}{n!}}{\tfrac{x^{n+1}}{(n+1)!}}=\frac{n+1}{x},$$ so the sequence is increasing as long as $n<x-1$ and decreasing when $n>x-1$. This means the sequence is maximal at $n=\lceil x-1\rceil=\lceil x\rceil-1$. If $x$ is an integer then it is also maximal at $n=\lceil x\rceil$.

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  • $\begingroup$ Inequalities are wrong here, it should be like in the question $\endgroup$ – Jakobian Mar 16 at 23:23
  • $\begingroup$ @Jakobian You are right, I have corrected them. $\endgroup$ – Inactive - avoiding CoC Mar 16 at 23:25

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