O limita frumoasa, insa dificila a unui sir (own).

[Virgil Nicula]

Se considera un sir \( (x_n)\ ,\ n\in\mathbb{N}^* \) pentru care \( \lim_{n\to\infty}\ \frac {x_{n+1}-x_n}{n}=l\in (0,\infty )\ . \)

Sa se arate ca \( \lim_{n\to\infty}\ \left(\frac {x_{n+1}}{\sqrt[n+1]{(n+1)!}}-\frac {x_n}{\sqrt[n]{n!}}\right)=\frac {el}{2}\ . \)

Caz particular. Pentru \( x_n=n^2\ ,\ n\in\mathbb{N}^* \) se obtine sirul lui D.M. Batinetu-Giurgiu.

[aleph]

Problema nu este prea dificilă dacă se cunoaşte formula lui Stirling, de unde se obţine
\( \frac{1}{\sqrt[n]{n!}}=\frac{e}{n}-\frac{e\ln n}{2n^{2}}-\frac{\alpha }{n^{2}}+\frac{\varepsilon _{n}}{n^{2}}
\)
\( \frac{1}{\sqrt[n+1]{(n+1)!}}=\frac{e}{n}-\frac{e}{n^{2}}-\frac{e\ln n}{2n^{2}}-\frac{\alpha }{n^{2}}+\frac{\varepsilon _{n}^{\prime }}{n^{2}}
\)

unde \( \alpha =\ln (2\pi )/2 \), \( \varepsilon _{n}\rightarrow 0 \), \( \varepsilon ^{\prime} _{n}\rightarrow 0 \).

[Virgil Nicula]

Problema nu este prea dificilă dacă se cunoaşte formula lui Stirling ...

Evident, asa nimic nu ar fi dificil daca vin cu "bombe atomice", adica "teorie inalta". Limita este dificila in ipoteza pe care am considerat-o implicita ca nu am invatat decat cel mult capitolul "limite de functii/continuitate" (imediat dupa semestrul I, adica etapa locala O.M.). Deci fara derivate, Taylor si ... formula lui Stirling. Poate doar daca elevul demonstreaza elementar formula lui Stirling (vezi cartea lui Iaglom & Iaglom - "Probleme neelementare tratate elementar").

[aleph]

Deci fara derivate, Taylor si ... formula lui Stirling.

OK, se poate folosi şiru lui Lalescu ( \( \sqrt[n+1]{(n+1)!} -\sqrt[n]{n!}\rightarrow \frac{1}{e} \) ), acesta figurând în unele manuale de liceu.
Notând \( a_{n}=\frac{1}{\sqrt[n]{n!}} \) rezultă
\( na_{n}\rightarrow e,\ \ n^{2}(a_{n}-a_{n+1})\rightarrow e
\)
şi deci
\( a_{n+1}x_{n+1}-a_{n}x_{n} =
(n+1)a_{n+1}\cdot \frac{x_{n+1}-x_{n}}{n}\cdot \frac{n}{n+1}
-n^{2}(a_{n}-a_{n+1})\cdot \frac{x_{n}}{n^{2}}\rightarrow e\cdot l\cdot
1-e\cdot \frac{l}{2}=\frac{el}{2} \).

[Virgil Nicula]

Frumoasa solutie, dl. Aleph ! Asta si urmaresc prin asemenea exercitii, nu numai
tehnica, ci si putina inspiratie/imaginatie bizuita pe cunostintele curriculare.
Va multumesc pentru interesul pe care l-ati manifestat acestui exercitiu.

[Virgil Nicula]

Fie sirul \( (x_n)\ ,\ n\in\mathbb{N}^* \) pentru care \( \frac {x_{n+1}-x_n}{n}\ \rightarrow\ l\in \mathbb R^*_+\ . \) Sa se arate ca

\( \lim_{n\to\infty}\ \left[\frac {x_{n+1}}{\sqrt[n+1]{(n+1)!}}-\frac {x_n}{\sqrt[n]{n!}}\right]=\frac {el}{2}\ . \) Pentru \( x_n=n^2\ ,\ n\in\mathbb{N}^* \) se obtine

sirul lui D.M. Batinetu-Giurgiu , adica \( \lim_{n\to\infty}\ \left[\frac {(n+1)^2}{\sqrt[n+1]{(n+1)!}}-\frac {n^2}{\sqrt[n]{n!}}\right]=e\ . \)

Demonstratie.

\( \odot\ \ \frac {x_{n+1}-x_n}{(n+1)^2-n^2}=\frac {x_{n+1}-x_n}{n}\cdot\frac {n}{2n+1}\rightarrow \frac l2\ \stackrel{(\mathrm{ST})}{\ \Longrightarrow\ }\underline{\overline{\left\|\ \lim_{n\to\infty}\ \frac {x_n}{n^2}=\frac l2\ \right\|}} \) . In particular,

\( x_n\right\infty\ \ ,\ \ \frac {x_n}{n}\rightarrow\infty\ \ ,\ \frac {x_{n+1}}{x_n}\rightarrow 1\ \ ,\ \ \frac {x_n}{\sqrt[n]{n!}}\rightarrow\infty \) deoarece \( \frac {n}{\sqrt[n]{n!}}\rightarrow e \) .

Asadar avem cazul de nedeterminare (exceptie) \( \underline{\overline{\left\|\ \infty -\infty\ \right\|}} \) .

\( \odot\ \ \) Se observa ca \( \frac {x_{n+1}}{\sqrt[n+1]{(n+1)!}}-\frac {x_n}{\sqrt[n]{n!}}=\frac {x_n}{n^2}\cdot\frac {n}{\sqrt[n]{n!}}\cdot n\left(a_n-1\right) \) , unde \( a_n=\frac {x_{n+1}\sqrt[n]{n!}}{x_n\sqrt[n+1]{(n+1)!}}\ \rightarrow\ 1 \) .

Deci \( \underline{\overline{\left\|\ \lim_{n\to\infty}\ \left[\frac {x_{n+1}}{\sqrt[n+1]{(n+1)!}}-\frac {x_n}{\sqrt[n]{n!}}\right]=\frac {le}{2}\cdot\lim_{n\to\infty}\ n\left(a_n-1\right)\ \right\|}} \) . Insa

\( \lim_{n\to\infty}\ n\left(a_n-1\right)=\lim_{n\to\infty}\ n\ln a_n=\lim_{n\to\infty}\ n\cdot\left[\ln\frac {x_{n+1}}{x_n}+\frac {\ln n!}{n}-\frac {\ln (n+1)!}{n+1}\right]= \)

\( \lim_{n\to\infty}\ n\cdot\ln\frac {x_{n+1}}{x_n}+\lim_{n\to\infty}\ \frac {(n+1)\ln n!-n\ln (n+1)!}{n+1}= \) \( \lim_{n\to\infty}\ n\cdot\ln\frac {x_{n+1}}{x_n}+\lim_{n\to\infty}\ \frac {\ln n!-n\ln (n+1)}{n+1} \) .

Insa \( \left\|\ \begin{array}{c}
\lim_{n\to\infty}\ n\cdot\ln\frac {x_{n+1}}{x_n}=\lim_{n\to\infty}\ n\cdot\left(\frac {x_{n+1}}{x_n}-1\right)=\lim_{n\to\infty}\ \frac {n^2}{x_n}\cdot\frac {x_{n+1}-x_n}{n}=\frac 2l\cdot l=2\\\\\\\\
\lim_{n\to\infty}\ \frac {\ln n!-n\ln (n+1)}{n+1}\stackrel{(\mathrm{ST})}{\ \ =\ \ }\lim_{n\to\infty}\ (n+1)\ln\frac {n+1}{n+2}=\lim_{n\to\infty}\ \ln\left(1-\frac {1}{n+2}\right)^{n+1}=-1\end{array}\ \right\| \) .

Asadar, \( \underline{\overline{\left\|\ \lim_{n\to\infty}\ n\left(a_n-1\right)=1\ \right\|}}\ \Longrightarrow\ \overline{\underline{\left\|\ \lim_{n\to\infty}\ \left[\frac {x_{n+1}}{\sqrt[n+1]{(n+1)!}}-\frac {x_n}{\sqrt[n]{n!}}\right]=\frac {el}{2}\ \right\|}} \) .