Nepero numero 5

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Sia

    \[b_n:\mathbb{N}\setminus \{0\}\rightarrow \mathbb{R}\]

la successione definita da

    \[b_n=\left(1+\dfrac{1}{n} \right)^n.\]

Si definisce numero di Nepero

    \begin{equation*} e=\lim_{n \rightarrow +\infty}b_n \end{equation*}

con 2<e<3.
In generale vale quanto segue: data una successione {c_n} infinita per n \rightarrow +\infty, vale che

(1)   \begin{equation*} e=\lim_{n \rightarrow +\infty }\left(1+\dfrac{1}{c_n} \right)^{c_n}. \end{equation*}

    \[\ast \quad \ast \quad \ast\]

Esercizio 1.  (\bigstar\bigstar\largewhitestar\largewhitestar\largewhitestar) Calcolare il seguente limite, applicando (1):

(2)   \begin{equation*} \lim_{n \to +\infty} \left(1+\dfrac{1}{n!}\right) ^{n^n}. \end{equation*}

 

Svolgimento esercizio 1.  Risolviamo (2) applicando (1):

    \[\begin{aligned} \lim_{n \to +\infty} \left(1+\dfrac{1}{n!}\right) ^{n^n} & = \lim_{n \to +\infty} \left( \left( 1+\dfrac{1}{n!}\right) ^{n!}\right) ^{\dfrac{n^n}{n!}}=e^{+\infty}=+\infty. \end{aligned}\]

Si conclude che:

    \[\boxcolorato{analisi}{\lim_{n \to +\infty} \left(1+\dfrac{1}{n!}\right) ^{n^n}=+\infty. }\]

 


Esercizio 2. (\bigstar\bigstar\bigstar\largewhitestar\largewhitestar) Calcolare il seguente limite, applicando (1):

(3)   \begin{equation*} \lim_{n \to +\infty} \left( 1+\dfrac{1}{n^n}\right) ^{(n+1)^n}. \end{equation*}

 

Svolgimento esercizio 2. Risolviamo (3) applicando (1):

    \[\begin{aligned} \lim_{n \to +\infty} \left( 1+\dfrac{1}{n^n}\right) ^{(n+1)^n} & = \lim_{n \to +\infty}\left( \left( 1+\dfrac{1}{n^n}\right) ^{n^n}\right) ^{\dfrac{(n+1)^n}{n^n}}=\\\\ &=\lim_{n \to +\infty}\left(\left( 1+\dfrac{1}{n^n}\right) ^{n^n}\right) ^{\dfrac{n^n\cdot\left( 1+\frac{1}{n}\right) ^n}{n^n}}=e^e. \end{aligned}\]

Si conclude che:

    \[\boxcolorato{analisi}{ \lim_{n \to +\infty} \left( 1+\dfrac{1}{n^n}\right) ^{(n+1)^n}=e^e.}\]

 

Esercizio 3. (\bigstar\bigstar\bigstar\bigstar\largewhitestar) Calcolare il seguente limite, applicando (1):

(4)   \begin{equation*} \lim_{n \to +\infty} \left( 1+\dfrac{1}{\ln(n^n)}\right) ^{\ln(n!)}. \end{equation*}

 

 

Svolgimento esercizio 3. Riscriviamo (4) come segue :

    \[\lim_{n \to +\infty} \left( 1+\dfrac{1}{\ln(n^n)}\right) ^{\ln(n!)}= \lim_{n \to +\infty} \left(\left( 1+\dfrac{1}{\ln(n^n)}\right) ^{\ln(n^n)}\right) ^{\dfrac{\ln(n!)}{\ln(n^n)}}\]

e notiamo che per n \rightarrow +\infty si ha [1]

    \[\begin{aligned} \dfrac{\ln(n!)}{\ln(n^n)}=&\dfrac{\ln\left(\sqrt{2\pi n}\cdot \left( \dfrac{n}{e}\right) ^n\left(1+o\left(1\right) \right) \right) }{\ln(n^n)}=\dfrac{\ln\sqrt{2\pi n}+n\ln n-n+\ln\left(1+o\left(1\right)\right)}{n\ln n}=\\\\ &=\dfrac{\ln\sqrt{2\pi n}}{n\ln n}+\dfrac{n\ln n}{n\ln n}-\dfrac{n}{n\ln n}+o\left(1\right) =1+o\left(1 \right) . \end{aligned}\]

Quindi, tornando ad (4) e applicando (1), otteniamo

    \[\lim_{n \to +\infty} \left( 1+\dfrac{1}{\ln(n^n)}\right) ^{\ln(n!)}= \lim_{n \to +\infty} \left(\left( 1+\dfrac{1}{\ln(n^n)}\right) ^{\ln(n^n)}\right) ^{\dfrac{\ln(n!)}{\ln(n^n)}}=e.\]

Si conclude che:

    \[\boxcolorato{analisi}{ \lim_{n \to +\infty} \left( 1+\dfrac{1}{\ln(n^n)}\right) ^{\ln(n!)}=e.}\]

 

 

1.  Si ricordi l’approsimazione di Stirling.}n!=\sqrt{2\pi n}\cdot\dfrac{n^n}{e^n}\left(1+o(1) \right) per n\to+\infty. ↩

 

 

 

Esercizio 4. (\bigstar\bigstar\bigstar\bigstar\largewhitestar) Calcolare il seguente limite, applicando(1):

(5)   \begin{equation*} \lim_{n \to +\infty} \left( \dfrac{n^{n+1}+n}{n^{n+1}+\sqrt{n}}\right) ^{(n+1)^n}. \end{equation*}

 

Svolgimento esercizio 4. Risolviamo (5) applicando (1):

    \[\begin{aligned} & \lim_{n \to +\infty} \left( \dfrac{n^{n+1}+n}{n^{n+1}+\sqrt{n}}\right) ^{(n+1)^n}= \lim_{n \to +\infty} \left( \dfrac{n^{n+1}+n-\sqrt{n}+\sqrt{n}}{n^{n+1}+\sqrt{n}}\right)^{(n+1)^n}=\\\\ &= \lim_{n \to +\infty}\left[ \left( 1+\dfrac{1}{\dfrac{n^{n+1}+\sqrt{n}}{n-\sqrt{n}}}\right) ^{\dfrac{n^{n+1}+\sqrt{n}}{n-\sqrt{n}}}\right] ^{\dfrac{\left( n-\sqrt{n}\right) \cdot(n+1)^n}{n^{n+1}+\sqrt{n}}}=\\\\ &= \lim_{n \to +\infty}\left[ \left( 1+\dfrac{1}{\dfrac{n^{n+1}+\sqrt{n}}{n-\sqrt{n}}}\right) ^{\dfrac{n^{n+1}+\sqrt{n}}{n-\sqrt{n}}}\right] ^{\dfrac{n\cdot n^n\cdot\left( 1-\dfrac{\sqrt{n}}{n}\right) \cdot \left( 1+\dfrac{1}{n}\right) ^n}{n^{n}\cdot n\left( 1+\dfrac{\sqrt{n}}{n^{n+1}}\right) }}=e^e. \end{aligned}\]

Si conclude che:

    \[\boxcolorato{analisi}{\lim_{n \to +\infty} \left( \dfrac{n^{n+1}+n}{n^{n+1}+\sqrt{n}}\right) ^{(n+1)^n}=e^e.}\]

 

 

Esercizio 5.  (\bigstar\bigstar\bigstar\bigstar\largewhitestar) Calcolare il seguente limite, applicando(1):

(6)   \begin{equation*} \lim_{n \to +\infty} \dfrac{\left( 2^n\cdot n!+n^n\right) ^2}{(2n)!+(n+1)^{2n}}. \end{equation*}

 

Svolgimento esercizio 5.  Risolviamo (6) applicando (1) e notando che
[2]

    \[\lim_{n \to +\infty} \dfrac{\left( 2^n\cdot n!+n^n\right) ^2}{(2n)!+(n+1)^{2n}}= \lim_{n \to +\infty}\dfrac{n^{2n}\left( 1+\dfrac{2^n\cdot n!}{n^n}\right) ^2}{n^{2n}\left( \left( \left( 1+\dfrac{1}{n}\right) ^n\right) ^2+\dfrac{(2n)!}{n^{2n}}\right) }=\dfrac{1}{e^2}.\]

Si conclude che:

    \[\boxcolorato{analisi}{ \lim_{n \to +\infty} \dfrac{\left( 2^n\cdot n!+n^n\right) ^2}{(2n)!+(n+1)^{2n}}=\dfrac{1}{e^2}.}\]

 

 

2. Sia

    \[\lim_{n \to +\infty}\dfrac{b^n\cdot n!}{n^n} \quad\text{con}\quad b>0\;\wedge b\neq 1\]

e ricordando l’approsimazione di Stirling

    \[n!=\sqrt{2\pi n}\cdot\dfrac{n^n}{e^n}\left(1+o(1) \right)\]

otteniamo:

    \[\begin{aligned} \lim_{n \to +\infty} \dfrac{b^n\cdot n!}{n^n}=\lim_{n \to +\infty}\dfrac{b^n\cdot n^n\sqrt{2\pi n}}{e^n\cdot n^n}\left(1+o(1) \right)=\lim_{n \to +\infty}\dfrac{b^n\sqrt{2\pi n}}{e^n}\left(1+o(1) \right)=\begin{cases} +\infty\quad \text{se}\quad b\geq e\\ 0\quad \text{se}\quad 0<b<e\;\wedge b\neq 1, \end{cases} \end{aligned}\]

posto b=2 si ha \lim\limits_{n \to +\infty}\dfrac{2^n\cdot n!}{n^n}=0.

Notiamo che

    \[\begin{aligned} \lim_{n \to +\infty} \dfrac{(2n)!}{n^{2n}}& =\lim_{n \to +\infty}\dfrac{\sqrt{4\pi n}\left( \dfrac{2n}{e}\right) ^{2n}}{n^{2n}}\left(1+o(1) \right)=\lim_{n \to +\infty}\dfrac{\sqrt{4\pi n}\cdot 2^{2n}\cdot n^{2n}}{n^{2n}\cdot e^{2n}}\left(1+o(1) \right)=\\ & = \lim_{n \to +\infty}\sqrt{4\pi n}\left( \dfrac{2}{e}\right) ^{2n}\left(1+o(1) \right)=0.\end{aligned}.\]

↩

 

 

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