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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)

$${Prove}/{disprove}/{prove}\:{for}\:{an} \\ $$ $${interval}\:{as}\:{the}\:{case}\:{may}\:{be}: \\ $$ $$\left({x}!\right)^{\frac{\mathrm{1}}{{x}}} \:\overset{?} {<}\:\left\{\left({x}+\mathrm{1}\right)!\right\}^{\frac{\mathrm{1}}{{x}+\mathrm{1}}} \:\:,\:{x}\in\mathbb{N}\:\left[{x}\neq\mathrm{0}\right] \\ $$ $$\left({Generalization}\:{of}\:{Q}\:\mathrm{1700}\right) \\ $$

Commented by123456 last updated on 04/Sep/15

x∈N^∗ (x≠0)

$${x}\in\mathbb{N}^{\ast} \left({x}\neq\mathrm{0}\right) \\ $$

Commented by123456 last updated on 04/Sep/15

f(x)=(x!)^(1/x)   lim_(x→0^+ ) f(x)=?

$${f}\left({x}\right)=\left({x}!\right)^{\frac{\mathrm{1}}{{x}}} \\ $$ $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{f}\left({x}\right)=? \\ $$

Commented byRasheed Ahmad last updated on 04/Sep/15

Thanks! Question is corrected.

$${Thanks}!\:{Question}\:{is}\:{corrected}. \\ $$

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