Question Number 219090 by depressiveshrek last updated on 19/Apr/25
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sequence}\:{a}_{{n}} =\frac{\mathrm{1}}{\:\sqrt[{{n}}]{{n}!}}\:\mathrm{is}\:\mathrm{decreasing}. \\ $$
Answered by Frix last updated on 19/Apr/25
![n→∞ ⇒ n!≈((n/e))^n (√(2πn)) ⇒ (1/( ((n!))^(1/n) ))≈(e/((2πx)^(1/(2x)) x)) We have to show that (2πx)^(1/(2x)) x is increasing ((d[(2πx)^(1/(2x)) x])/dx)=(1/2)(2πx)^(1/(2x)) x(2x−ln x −ln 2π +1) Obviously (1/2)(2πx)^(1/(2x)) x>0∀x>0 y=2x−ln x −ln 2π +1 has a minimum at x=(1/2), y=2−ln π >0 ⇒ (2πx)^(1/(2x)) x is increasing ⇒ (1/( ((n!))^(1/n) )) is decreasing](https://www.tinkutara.com/question/Q219094.png)
$${n}\rightarrow\infty\:\Rightarrow\:{n}!\approx\left(\frac{{n}}{\mathrm{e}}\right)^{{n}} \sqrt{\mathrm{2}\pi{n}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\:\sqrt[{{n}}]{{n}!}}\approx\frac{\mathrm{e}}{\left(\mathrm{2}\pi{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} {x}} \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{to}\:\mathrm{show}\:\mathrm{that}\:\left(\mathrm{2}\pi{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} {x}\:\mathrm{is}\:\mathrm{increasing} \\ $$$$\frac{{d}\left[\left(\mathrm{2}\pi{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} {x}\right]}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\pi{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} {x}\left(\mathrm{2}{x}−\mathrm{ln}\:{x}\:−\mathrm{ln}\:\mathrm{2}\pi\:+\mathrm{1}\right) \\ $$$$\mathrm{Obviously}\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\pi{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} {x}>\mathrm{0}\forall{x}>\mathrm{0} \\ $$$${y}=\mathrm{2}{x}−\mathrm{ln}\:{x}\:−\mathrm{ln}\:\mathrm{2}\pi\:+\mathrm{1}\:\mathrm{has}\:\mathrm{a}\:\mathrm{minimum}\:\mathrm{at} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}},\:{y}=\mathrm{2}−\mathrm{ln}\:\pi\:>\mathrm{0} \\ $$$$\Rightarrow\:\left(\mathrm{2}\pi{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}{x}}} {x}\:\mathrm{is}\:\mathrm{increasing} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\:\sqrt[{{n}}]{{n}!}}\:\mathrm{is}\:\mathrm{decreasing} \\ $$
Answered by Ghisom last updated on 19/Apr/25
$${a}_{{n}} \:\mathrm{is}\:\mathrm{decreasing}\:\Rightarrow\:\frac{\mathrm{1}}{{a}_{{n}} }\:\mathrm{is}\:\mathrm{increasing} \\ $$$$\Rightarrow \\ $$$${n}!^{\mathrm{1}/{n}} <\left({n}+\mathrm{1}\right)!^{\mathrm{1}/\left({n}+\mathrm{1}\right)} \\ $$$${n}!^{\left({n}+\mathrm{1}\right)/{n}} <\left({n}+\mathrm{1}\right)! \\ $$$${n}!^{\frac{\mathrm{1}}{{n}}+\mathrm{1}} <{n}!\left({n}+\mathrm{1}\right) \\ $$$${n}!^{\mathrm{1}/{n}} <{n}+\mathrm{1} \\ $$$${n}!<\left({n}+\mathrm{1}\right)^{{n}} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{obvious} \\ $$