Nepero numero 6

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

(2)   \begin{equation*} \lim_{n \to +\infty} \dfrac{2^{n+\sin n}+3^{n-1}}{100^{\sqrt{n}}+\left( 3+\dfrac{1}{n}\right) ^n}. \end{equation*}

 

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

    \[\begin{aligned} \lim_{n \to +\infty} \dfrac{2^{n+\sin n}+3^{n-1}}{100^{\sqrt{n}}+\left( 3+\dfrac{1}{n}\right) ^n} & = \lim_{n \to +\infty}\dfrac{2^{n+\sin n}+\dfrac{1}{3}\cdot 3^n}{100^{\sqrt{n}}+3^n\left(1+\dfrac{1}{3n} \right) ^n} = \lim_{n \to +\infty}\dfrac{\dfrac{1}{3}+\dfrac{2^{n+\sin n}}{3^n}}{\left(1+\dfrac{1}{3n} \right)^n+\dfrac{100^{\sqrt{n}}}{3^n}} . \end{aligned}\]

Osserviamo che:

    \[\dfrac{2^{n+\sin n}}{3^n}\leq\dfrac{2^{n+1}}{3^n}=2\left(\dfrac{2}{3} \right) ^n \quad \text{per}\,\, n \geq 1\]

e siccome

    \[\lim_{n \rightarrow +\infty}\left(\dfrac{2}{3} \right) ^n=0\]

per il teorema del confronto [1] si ha

    \[\lim_{n \rightarrow +\infty}\dfrac{2^{n+\sin n}}{3^n}=0.\]

Inoltre osserviamo che

    \[\dfrac{100^{\sqrt{n}}}{3^n}=\dfrac{e^{\sqrt{n}\ln100 }}{3^{n \ln3 }}\]

dove

    \[\sqrt{n}\ln100\leq n \ln 3 \quad \text{per}\,\, n \rightarrow +\infty\]

quindi

    \[\lim_{n \rightarrow + \infty}\dfrac{100^{\sqrt{n}}}{3^n}=0.\]

Torniamo ora al limite e tenendo conto dei risultati ottenuti e applicando (1) si ottiene

    \[\begin{aligned} \lim_{n \to +\infty}\dfrac{\dfrac{1}{3}+\dfrac{2^{n+\sin n}}{3^n}}{\left(1+\dfrac{1}{3n} \right)^n+\dfrac{100^{\sqrt{n}}}{3^n}} =\dfrac{\frac{1}{3}}{e^{\frac{1}{3}}}=\dfrac{1}{3\sqrt[3]{e}}. \end{aligned}\]

Si conclude che:

    \[\boxcolorato{analisi}{\lim_{n \to +\infty} \dfrac{2^{n+\sin n}+3^{n-1}}{100^{\sqrt{n}}+\left( 3+\dfrac{1}{n}\right) ^n}=\dfrac{1}{3\sqrt[3]{e}}. }\]

 

 

1 Teorema del confronto. Siano \{b_n\}_ {n\in \mathbb{N}}, \{c_n\}_ {n\in \mathbb{N}} e \{d_n\}_ {n\in \mathbb{N}} successioni in \mathbb{R}. Si assuma che

    \[bn\leq c_n \leq d_n \quad \text{per}\,\, n \geq n_0\]

e che

    \[\lim_{n \rightarrow +\infty}b_n=\lim_{n \rightarrow +\infty}d_n=l \in \mathbb{R}\]

allora

    \[\lim_{n \rightarrow +\infty}c_n=l.\]

↩

 

 


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

(3)   \begin{equation*} \lim_{n \to +\infty} \dfrac{2\cdot n^{3n}+7n\cdot(2n)!+1000^n}{10^{3n}-(2n+1)!+(n+1)^{3n}}. \end{equation*}

 

Svolgimento esercizio 2. Riscriviamo (3) come segue

    \[\begin{aligned} \lim_{n \to +\infty} \dfrac{2\cdot n^{3n}+7n\cdot(2n)!+1000^n}{10^{3n}-(2n+1)!+(n+1)^{3n}} & =\lim_{n \rightarrow +\infty }\dfrac{n^{3n}\left( 1+\frac{7n(2n)!}{n^{3n}}+\frac{1000^n}{n^{3n}}\right) }{10^{3n}-(2n+1)(2n)!+n^{3n}(1+\frac{1}{n})^{3n}}=\\\\ &=\lim_{n \rightarrow +\infty }\dfrac{\left( 2+\frac{7n(2n)!}{n^{3n}}+\frac{1000^n}{n^{3n}}\right) }{(1+\frac{1}{n})^{3n}+\frac{10^{3n}}{n^{3n}}-\frac{(2n+1)(2n)!}{n^{3n}}}. \end{aligned}\]

Osserviamo che

    \[10^{3n}<n^{3n}\quad \text{per }\,\, n \rightarrow +\infty\]

e quindi

    \[\lim_{n \rightarrow +\infty}\dfrac{10^{3n}}{n^{3n}}=0.\]

Ora notiamo che[2]

    \[\begin{aligned} &\lim_{n \rightarrow +\infty}\dfrac{n(2n)!}{n^{3n}}=\lim_{n \rightarrow +\infty}\dfrac{n\sqrt{2\pi(2n)}(2n)^{2n}}{e^{2n}n^{3n}}\left(1+o\left(1\right)\right)=\lim_{n\rightarrow +\infty }\dfrac{n2^{2n}n^{2n}\sqrt{4\pi n}}{e^{2n}n^{3n}}\left( 1+o(1)\right)=\\\\ &=2\sqrt{4 \pi }\lim_{n \rightarrow +\infty }\dfrac{2^{2n}}{e^{2n}}\cdot \dfrac{n^{\frac{3}{2}}}{n^{2n}}\cdot \left(1+o(1)\right)=0 \end{aligned}\]

e quindi tornando a (3) e applicando (1), tenendo conto dei risultati ottenuti, abbiamo

    \[\lim_{n \rightarrow +\infty }\dfrac{\left( 2+\frac{7n(2n)!}{n^{3n}}+\frac{1000^n}{n^{3n}}\right) }{(1+\frac{1}{n})^{3n}+\frac{10^{3n}}{n^{3n}}-\frac{(2n+1)(2n)!}{n^{3n}}}=\dfrac{2}{e^3}.\]

Si conclude che

    \[\boxcolorato{analisi}{ \lim_{n \to +\infty} \dfrac{2\cdot n^{3n}+7n\cdot(2n)!+1000^n}{10^{3n}-(2n+1)!+(n+1)^{3n}}=\dfrac{2}{e^3}}\]

 

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

 

 

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

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

 

 

Svolgimento esercizio 3. Risolviamo (4) applicando (1):

    \[\begin{aligned} \lim_{n \to +\infty} \dfrac{(n+1)^{n+\frac{1}{n}}+n!}{(n+2)^n-n^n} & = \lim_{n \to +\infty} \dfrac{(n+1)^n\cdot (n+1)^{\frac{1}{n}}+n!}{n^n\cdot \left( 1+\frac{2}{n}\right) ^n-n^n}=\\\\ & \lim_{n \to +\infty}\dfrac{n^n\cdot\left( 1+\frac{1}{n}\right) ^n\cdot e^{\frac{1}{n}\ln(n+1)}+n!}{n^n\left( \left( 1+\frac{2}{n}\right) ^n-1\right) }=\\\\ & =\lim_{n \rightarrow +\infty}\dfrac{\left( 1+\frac{1}{n}\right)^n\cdot e^{\frac{1}{n}\ln \left( n+1 \right) }+\frac{n!}{n^n} }{\left(1+\frac{2}{n}\right)^n-1}=\dfrac{e}{e^2-1}. \end{aligned}\]

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

 

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

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

 

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

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

Si conclude che:

    \[\boxcolorato{analisi}{ \lim_{n \to +\infty} \dfrac{\left( \sqrt{n^n}+n^{\sqrt{n}}+\left(\sqrt{n}\right) ^n \right) ^2}{(n+1)^n}=\dfrac{4}{e}.}\]

 

 

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

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

 

Svolgimento esercizio 5. iscriviamo (6) come segue

    \[\begin{aligned} \lim_{n \to +\infty} \dfrac{\left( 1+\dfrac{1}{n}\right) ^{n!}+2^{n!}}{\left( 2-\dfrac{1}{n^2}\right) ^{n!}+\left(2+\dfrac{1}{n!} \right) ^{n!}} & =\lim_{n \rightarrow +\infty} \dfrac{1+\left( \frac{1}{2}+\frac{1}{2n}\right)^{n!}}{\left(1-\frac{1}{2n^2}\right)^{n!}+\left( 1 +\frac{1}{2n!}\right)^{n!}}=\\ &=\lim_{n \to +\infty}\dfrac{1}{\left( \left(1+\frac{1}{2n!}\right)^{2n!}\right)^{\frac{1}{2}}}\cdot \dfrac{1+\frac{1}{2^{n!}}\left(1+\frac{1}{n} \right)^{n!}}{1+\dfrac{\left(\left( 1-\frac{1}{2n^2}\right)^{-2n^2} \right)^{-\frac{n!}{2n^2}}}{\left(1+\frac{1}{2n!} \right)^{n!}}}. \end{aligned}\]

Ora osserviamo che \ln \left(1+\dfrac{1}{n} \right)\leq \ln 2 per n\geq 1, quindi \displaystyle \lim_{n \rightarrow +\infty}\dfrac{\left(1+\frac{1}{n}\right)^{n!}}{2^{n!}}=0.
Tornando al limite e applicando (1) si ha che

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

Si conclude che:

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

 

 

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