Question Number 1729 by Rasheed Ahmad last updated on 04/Sep/15
![Prove/disprove/prove for an interval as the case may be: (x!)^(1/x) <^(?) {(x+1)!}^(1/(x+1)) , x∈N [x≠0] (Generalization of Q 1700)](https://www.tinkutara.com/question/Q1729.png)
$${Prove}/{disprove}/{prove}\:{for}\:{an} \\ $$$${interval}\:{as}\:{the}\:{case}\:{may}\:{be}: \\ $$$$\left({x}!\right)^{\frac{\mathrm{1}}{{x}}} \:\overset{?} {<}\:\left\{\left({x}+\mathrm{1}\right)!\right\}^{\frac{\mathrm{1}}{{x}+\mathrm{1}}} \:\:,\:{x}\in\mathbb{N}\:\left[{x}\neq\mathrm{0}\right] \\ $$$$\left({Generalization}\:{of}\:{Q}\:\mathrm{1700}\right) \\ $$
Commented by 123456 last updated on 04/Sep/15
$${x}\in\mathbb{N}^{\ast} \left({x}\neq\mathrm{0}\right) \\ $$
Commented by 123456 last updated on 04/Sep/15
$${f}\left({x}\right)=\left({x}!\right)^{\frac{\mathrm{1}}{{x}}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{f}\left({x}\right)=? \\ $$
Commented by Rasheed Ahmad last updated on 04/Sep/15
$${Thanks}!\:{Question}\:{is}\:{corrected}. \\ $$