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I would like to find $\min_{n\in\mathbb{N}} (n!\cdot x^n)$ for $0<x\ll1$, $x$ fixed. My initial thought was to write $n!x^n=\Gamma(n+1)x^n$, and then set the derivative to zero, giving:

$$ 0=x^{n-1}\Gamma(n+1)\left(n+x\left[\sum_{k=1}^n\frac{1}{k}-\gamma\right] \right) $$

however this doesn't look very good.

Does anybody know a better way of tackling this problem?

ocstat
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1 Answers1

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Hint:

Set $u_n=n! x^n$. The sequence $(u_n)$ is non-increasing as long as $\;\dfrac{u_n}{u_{n-1}}=nx\le1$, i.e. $\; n\le\biggl\lfloor\dfrac1x\biggr\rfloor$. Thus the minimum is attained at $N=\biggl\lfloor\dfrac1x\biggr\rfloor$.

Bernard
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