Question Number 227964 by mr W last updated on 05/Mar/26
Answered by Lara2440 last updated on 07/Mar/26
![Let′s define Sequence A_n as (((n!)^(1/n) )/n) , n=1,2,3..... for all n>1 and for all x>0 1+nx<(x+1)^n ∙∙∙∙∙(1) (Bernoulli Inequality) for all n<N and N>1 n<n+1<2^n ⇒ N!<(N+1)!<2^((N(N+1))/2) Product a_1 a_2 ...a_(N−1) a_N (N!)^(1/N) <((N+1)!)^(1/N) <2^((N+1)/2) (N!)^(1/(N+1)) <((N+1)!)^(1/(N+1)) <2^(N/2) (N!)^(1/(N+1)) <(N!)^(1/N) <((N+1)!)^(1/(N+1)) <((N+1)!)^(1/N) <2^((1/2)N) <2^((1/2)(N+1)) (when N>5) A_n <A_(n+1) ∴ ∀n>1 , A_n is monotonic increase. Q_n ={1,2,3.....,n} , n∈N (1/n)Σ_(k=1) ^n k (AM)>(n!)^(1/n) (GM) ⇒ ((n+1)/2)>(n!)^(1/n) lim_(n→∞) (1/n)(n!)^(1/n) <lim_(n→∞) ((n+1)/(2n))=(1/2). so, ∀n>1 ,A_n is upper bounded The completeness axiom which states that every nonempty set bounded above has a supremum is equivalent to the monotone convergence theorem and the fact that every cauchy sequence converges to a unique limit..... thus the given sequence is convergent. Let′s assume that lim_(n→∞) (((n!)^(1/n) )/n) convergence to L for all M∈N ,exist for any sufficiently Large M s.t. M>n (((M!)^(1/M) )/M)=L ⇒ ln((1/M)(M!)^(1/M) )=(1/M)ln(M!)−ln(M) lim_(M→∞) ((1/M)Σ_(k=1) ^M ln(k)−ln(M))=lim_(M→∞) ((1/n)Σ_(k=1) ^M [ln(k)−ln(M)]) lim_(M→∞) (1/M)Σ_(k=1) ^M ln((k/M))=∫_0 ^( 1) ln(r)dr=[r∙ln(r)−r+C]_(r=0) ^(r=1) =−1 ∴ln(L)=−1 ⇒ L=(1/e) ■](https://www.tinkutara.com/question/Q227967.png)
$$\mathrm{Let}'\mathrm{s}\:\mathrm{define}\:\mathrm{Sequence}\:{A}_{{n}} \:\mathrm{as}\:\frac{\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} }{{n}}\:,\:{n}=\mathrm{1},\mathrm{2},\mathrm{3}….. \\ $$$$\mathrm{for}\:\mathrm{all}\:{n}>\mathrm{1}\:\mathrm{and}\:\mathrm{for}\:\mathrm{all}\:{x}>\mathrm{0} \\ $$$$\mathrm{1}+{nx}<\left({x}+\mathrm{1}\right)^{{n}} \:\centerdot\centerdot\centerdot\centerdot\centerdot\left(\mathrm{1}\right)\:\:\left(\mathrm{Bernoulli}\:\mathrm{Inequality}\right) \\ $$$$\: \\ $$$$\mathrm{for}\:\mathrm{all}\:{n}<{N}\:\mathrm{and}\:\:{N}>\mathrm{1} \\ $$$${n}<{n}+\mathrm{1}<\mathrm{2}^{{n}} \:\Rightarrow\:{N}!<\left({N}+\mathrm{1}\right)!<\mathrm{2}^{\frac{{N}\left({N}+\mathrm{1}\right)}{\mathrm{2}}} \:\:\mathrm{Product}\:{a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{N}−\mathrm{1}} {a}_{{N}} \\ $$$$\left({N}!\right)^{\frac{\mathrm{1}}{{N}}} <\left(\left({N}+\mathrm{1}\right)!\right)^{\frac{\mathrm{1}}{{N}}} <\mathrm{2}^{\frac{{N}+\mathrm{1}}{\mathrm{2}}} \:\: \\ $$$$\left({N}!\right)^{\frac{\mathrm{1}}{{N}+\mathrm{1}}} <\left(\left({N}+\mathrm{1}\right)!\right)^{\frac{\mathrm{1}}{{N}+\mathrm{1}}} <\mathrm{2}^{\frac{{N}}{\mathrm{2}}} \\ $$$$\left({N}!\right)^{\frac{\mathrm{1}}{{N}+\mathrm{1}}} <\left({N}!\right)^{\frac{\mathrm{1}}{{N}}} <\left(\left({N}+\mathrm{1}\right)!\right)^{\frac{\mathrm{1}}{{N}+\mathrm{1}}} <\left(\left({N}+\mathrm{1}\right)!\right)^{\frac{\mathrm{1}}{{N}}} <\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}{N}} <\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}\left({N}+\mathrm{1}\right)} \\ $$$$\left(\mathrm{when}\:{N}>\mathrm{5}\right)\: \\ $$$${A}_{{n}} <{A}_{{n}+\mathrm{1}} \: \\ $$$$\therefore\:\forall{n}>\mathrm{1}\:,\:{A}_{{n}} \:\mathrm{is}\:\mathrm{monotonic}\:\mathrm{increase}.\: \\ $$$${Q}_{{n}} =\left\{\mathrm{1},\mathrm{2},\mathrm{3}…..,{n}\right\}\:,\:{n}\in\mathbb{N} \\ $$$$\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:{k}\:\left(\mathrm{AM}\right)>\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} \left(\mathrm{GM}\right)\:\Rightarrow\:\frac{{n}+\mathrm{1}}{\mathrm{2}}>\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\left({n}!\right)^{\mathrm{1}/{n}} <\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}=\frac{\mathrm{1}}{\mathrm{2}}.\: \\ $$$$\mathrm{so},\:\forall{n}>\mathrm{1}\:,{A}_{{n}} \:\mathrm{is}\:\mathrm{upper}\:\mathrm{bounded}\: \\ $$$$\: \\ $$$$\mathrm{The}\:\mathrm{completeness}\:\mathrm{axiom}\:\mathrm{which}\:\mathrm{states}\:\mathrm{that}\:\mathrm{every}\:\mathrm{nonempty}\:\mathrm{set} \\ $$$$\mathrm{bounded}\:\mathrm{above}\:\mathrm{has}\:\mathrm{a}\:\mathrm{supremum}\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{monotone}\:\mathrm{convergence}\:\mathrm{theorem}\:\mathrm{and}\:\mathrm{the}\:\mathrm{fact}\:\mathrm{that} \\ $$$$\mathrm{every}\:\mathrm{cauchy}\:\mathrm{sequence}\:\mathrm{converges}\:\mathrm{to}\:\mathrm{a}\:\mathrm{unique}\:\mathrm{limit}….. \\ $$$$\mathrm{thus}\:\mathrm{the}\:\mathrm{given}\:\mathrm{sequence}\:\mathrm{is}\:\mathrm{convergent}.\: \\ $$$$\: \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{assume}\:\mathrm{that}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} }{{n}}\:\mathrm{convergence}\:\mathrm{to}\:{L} \\ $$$$\mathrm{for}\:\mathrm{all}\:{M}\in\mathbb{N}\:,\mathrm{exist}\:\mathrm{for}\:\mathrm{any}\:\mathrm{sufficiently}\:\mathrm{Large}\:{M}\:\mathrm{s}.\mathrm{t}.\:{M}>{n} \\ $$$$\frac{\left({M}!\right)^{\frac{\mathrm{1}}{{M}}} }{{M}}={L}\:\:\Rightarrow\:\:\mathrm{ln}\left(\frac{\mathrm{1}}{{M}}\left({M}!\right)^{\frac{\mathrm{1}}{{M}}} \right)=\frac{\mathrm{1}}{{M}}\mathrm{ln}\left({M}!\right)−\mathrm{ln}\left({M}\right) \\ $$$$\underset{{M}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{M}}\underset{{k}=\mathrm{1}} {\overset{{M}} {\sum}}\:\mathrm{ln}\left({k}\right)−\mathrm{ln}\left({M}\right)\right)=\underset{{M}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{M}} {\sum}}\left[\mathrm{ln}\left({k}\right)−\mathrm{ln}\left({M}\right)\right]\right) \\ $$$$\underset{{M}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{M}}\underset{{k}=\mathrm{1}} {\overset{{M}} {\sum}}\:\mathrm{ln}\left(\frac{{k}}{{M}}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\mathrm{ln}\left({r}\right)\mathrm{d}{r}=\left[{r}\centerdot\mathrm{ln}\left({r}\right)−{r}+{C}\right]_{{r}=\mathrm{0}} ^{{r}=\mathrm{1}} =−\mathrm{1} \\ $$$$\therefore\mathrm{ln}\left({L}\right)=−\mathrm{1}\:\Rightarrow\:{L}=\frac{\mathrm{1}}{{e}}\:\blacksquare \\ $$
Commented by mr W last updated on 06/Mar/26
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Answered by mr W last updated on 06/Mar/26
![n!∼(√(2πn)) ((n/e))^n L=lim_(n→∞) (([(√(2πn)) ((n/e))^n ]^(1/n) )/n) =lim_(n→∞) (((2πn)^(1/(2n)) )/e) =lim_(n→∞) (1/(e[((1/(2πn)))^(1/(2πn)) ]^π )) =lim_(x→0) (1/(e(x^x )^π )) =(1/(e×1^π )) =(1/e)](https://www.tinkutara.com/question/Q227970.png)
$${n}!\sim\sqrt{\mathrm{2}\pi{n}}\:\left(\frac{{n}}{{e}}\right)^{{n}} \\ $$$${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left[\sqrt{\mathrm{2}\pi{n}}\:\left(\frac{{n}}{{e}}\right)^{{n}} \right]^{\frac{\mathrm{1}}{{n}}} }{{n}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{2}\pi{n}\right)^{\frac{\mathrm{1}}{\mathrm{2}{n}}} }{{e}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{{e}\left[\left(\frac{\mathrm{1}}{\mathrm{2}\pi{n}}\right)^{\frac{\mathrm{1}}{\mathrm{2}\pi{n}}} \right]^{\pi} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{e}\left({x}^{{x}} \right)^{\pi} } \\ $$$$=\frac{\mathrm{1}}{{e}×\mathrm{1}^{\pi} } \\ $$$$=\frac{\mathrm{1}}{{e}} \\ $$
Answered by MrAjder last updated on 06/Mar/26
Commented by MrAjder last updated on 06/Mar/26
$$\mathrm{Among}\:\mathrm{them},\psi\left({z}\right)=\frac{\Gamma'\left({z}\right)}{\Gamma\left({z}\right)}\sim\mathrm{ln}\:{z}−\frac{\mathrm{1}}{\mathrm{2}{z}}+{O}\left({z}^{−\mathrm{2}} \right)\wedge{z}\rightarrow\infty \\ $$
Answered by profcedricjunior last updated on 07/Mar/26
$$\Rightarrow\ell=\underset{\boldsymbol{{n}}\rightarrow+\infty} {\mathrm{lim}}\frac{\sqrt[{\boldsymbol{{n}}}]{\boldsymbol{{n}}!}}{\boldsymbol{{n}}}=\underset{\boldsymbol{{n}}\rightarrow+\infty} {\mathrm{lim}}\frac{\left(\sqrt{\mathrm{2}\boldsymbol{\pi{n}}}\right)^{\frac{\mathrm{1}}{\boldsymbol{{n}}}} \left(\frac{\boldsymbol{{n}}}{\boldsymbol{{e}}}\right)}{\boldsymbol{{n}}} \\ $$$$\Rightarrow\ell=\frac{\mathrm{1}}{\boldsymbol{{e}}}\underset{\boldsymbol{{n}}\rightarrow+\infty} {\mathrm{lim}}\boldsymbol{{e}}^{\frac{\boldsymbol{{ln}}\left(\boldsymbol{{n}}\right)}{\boldsymbol{{n}}}+\frac{\boldsymbol{{ln}}\left(\sqrt{\mathrm{2}\boldsymbol{\pi}}\right)}{\boldsymbol{{n}}}} =\frac{\mathrm{1}}{\boldsymbol{{e}}} \\ $$$$\frac{\boldsymbol{{d}}}{\boldsymbol{{dz}}}\left(\boldsymbol{{e}}^{\left(\boldsymbol{{puissantD}}{r}\right)\boldsymbol{{z}}} \right)\mid_{\boldsymbol{{z}}=\mathrm{0}} ={puissantDr} \\ $$