# Maximising $\frac{x^n}{n!}$

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.

• 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 $x$.) – Minus One-Twelfth Mar 16 at 23:19
• Your way is very good for it. Good job – Jakobian Mar 16 at 23:24

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 $$n and decreasing when $$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$$.