- #1

paniurelis

- 12

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Prove that sequence [tex]x_n=\frac{n}{\sqrt[n]{n!}}[/tex] is increasing and [tex]\lim_{n\rightarrow\infty}{x_n} = e[/tex]

My attempt:

First, I try to prove [tex]\forall n\in N:\frac{x_{n+1}}{x_n}>1[/tex].

[tex]\forall n\in :\frac{x_{n+1}}{x_n}=\frac{n+1}{n}\frac{\sqrt[n]{n!}}{\sqrt[n+1]{(n+1)!}}

=\sqrt[n(n+1)]{\frac{(n+1)^{n(n+1)}(n!)^{n+1}}{n^{n(n+1)}((n+1)!)^n}}

=\sqrt[n(n+1)]{\frac{(n+1)^{n^2}n!}{n^{n^2+n}}}[/tex] and this is dead end.

Another attempt, we have [tex]x_1 > 0, x_2 > 0, ..., x_n>0:\sqrt[n]{x_1x_2..x_n}\leq\frac{x_1+x_2+...+x_n}{n}[/tex], where equality is then [tex]x_1=x_2=...=x_n[/tex]

So, [tex]\forall n\in N: \sqrt[n]{n!}<\frac{(1+n)n}{2n}=\frac{n+1}{2}

\Longrightarrow x_n=\frac{n}{\sqrt[n]{n!}}>\frac{2n}{1+n}=y_n[/tex]

Sequence [tex](y_n)[/tex] is increasing, because [tex]\forall n\in N:\frac{y_{n+1}}{y_n}=\frac{2(n+1)}{n+2}\frac{1+n}{n+2}=\frac{n^2+2n+1}{n+2}=n+\frac{1}{n+2}>1[/tex]

Can I say, [tex]x_n[/tex] is increasing, because [tex]y_n[/tex] is increasing and [tex]x_n > y_n, n \in N[/tex] ?

Next, we have [tex]n!>\left({\frac{n}{e}}\right)^n, n \in N[/tex]. I will omit the proof of this inequality, it is proven very simply by using math. induction.

So, we have [tex]\forall n\in N:n!>\left({\frac{n}{e}}\right)^n \Longrightarrow e>\frac{n}{\sqrt[n]{n!}}[/tex]

Next, [tex]\forall n \in N: \sqrt[n]{n!} < \frac{(n+1)n}{2n} = \left(1 + \frac{1}{n}\right) \frac{n}{2}<\left(1 + \frac{1}{n}\right)n=\frac{1}{1-\frac{1}{n+1}}n \Longrightarrow x_n=\frac{n}{\sqrt[n]{n!}}>1-\frac{1}{n+1}>1-\frac{1}{n}>\left(1-\frac{1}{n}\right)^{n}=z_n[/tex]

Finally, we have [tex]\forall n \in N: e> x_n=\frac{n}{\sqrt[n]{n!}}>z_n[/tex]

and [tex]\lim_{n\rightarrow\infty}z_n=e \Longrightarrow \lim_{n\rightarrow\infty}x_n=e[/tex]

Q.E.D.

Is it correct ? Any suggestions ? I am missing something here...

My attempt:

First, I try to prove [tex]\forall n\in N:\frac{x_{n+1}}{x_n}>1[/tex].

[tex]\forall n\in :\frac{x_{n+1}}{x_n}=\frac{n+1}{n}\frac{\sqrt[n]{n!}}{\sqrt[n+1]{(n+1)!}}

=\sqrt[n(n+1)]{\frac{(n+1)^{n(n+1)}(n!)^{n+1}}{n^{n(n+1)}((n+1)!)^n}}

=\sqrt[n(n+1)]{\frac{(n+1)^{n^2}n!}{n^{n^2+n}}}[/tex] and this is dead end.

Another attempt, we have [tex]x_1 > 0, x_2 > 0, ..., x_n>0:\sqrt[n]{x_1x_2..x_n}\leq\frac{x_1+x_2+...+x_n}{n}[/tex], where equality is then [tex]x_1=x_2=...=x_n[/tex]

So, [tex]\forall n\in N: \sqrt[n]{n!}<\frac{(1+n)n}{2n}=\frac{n+1}{2}

\Longrightarrow x_n=\frac{n}{\sqrt[n]{n!}}>\frac{2n}{1+n}=y_n[/tex]

Sequence [tex](y_n)[/tex] is increasing, because [tex]\forall n\in N:\frac{y_{n+1}}{y_n}=\frac{2(n+1)}{n+2}\frac{1+n}{n+2}=\frac{n^2+2n+1}{n+2}=n+\frac{1}{n+2}>1[/tex]

Can I say, [tex]x_n[/tex] is increasing, because [tex]y_n[/tex] is increasing and [tex]x_n > y_n, n \in N[/tex] ?

Next, we have [tex]n!>\left({\frac{n}{e}}\right)^n, n \in N[/tex]. I will omit the proof of this inequality, it is proven very simply by using math. induction.

So, we have [tex]\forall n\in N:n!>\left({\frac{n}{e}}\right)^n \Longrightarrow e>\frac{n}{\sqrt[n]{n!}}[/tex]

Next, [tex]\forall n \in N: \sqrt[n]{n!} < \frac{(n+1)n}{2n} = \left(1 + \frac{1}{n}\right) \frac{n}{2}<\left(1 + \frac{1}{n}\right)n=\frac{1}{1-\frac{1}{n+1}}n \Longrightarrow x_n=\frac{n}{\sqrt[n]{n!}}>1-\frac{1}{n+1}>1-\frac{1}{n}>\left(1-\frac{1}{n}\right)^{n}=z_n[/tex]

Finally, we have [tex]\forall n \in N: e> x_n=\frac{n}{\sqrt[n]{n!}}>z_n[/tex]

and [tex]\lim_{n\rightarrow\infty}z_n=e \Longrightarrow \lim_{n\rightarrow\infty}x_n=e[/tex]

Q.E.D.

Is it correct ? Any suggestions ? I am missing something here...

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