paniurelis
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Prove that sequence x_n=\frac{n}{\sqrt[n]{n!}} is increasing and \lim_{n\rightarrow\infty}{x_n} = e
My attempt:
First, I try to prove \forall n\in N:\frac{x_{n+1}}{x_n}>1.
\forall n\in :\frac{x_{n+1}}{x_n}=\frac{n+1}{n}\frac{\sqrt[n]{n!}}{\sqrt[n+1]{(n+1)!}}<br /> =\sqrt[n(n+1)]{\frac{(n+1)^{n(n+1)}(n!)^{n+1}}{n^{n(n+1)}((n+1)!)^n}}<br /> =\sqrt[n(n+1)]{\frac{(n+1)^{n^2}n!}{n^{n^2+n}}} and this is dead end.
Another attempt, we have 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}, where equality is then x_1=x_2=...=x_n
So, \forall n\in N: \sqrt[n]{n!}<\frac{(1+n)n}{2n}=\frac{n+1}{2}<br /> \Longrightarrow x_n=\frac{n}{\sqrt[n]{n!}}>\frac{2n}{1+n}=y_n
Sequence (y_n) is increasing, because \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
Can I say, x_n is increasing, because y_n is increasing and x_n > y_n, n \in N ?
Next, we have n!>\left({\frac{n}{e}}\right)^n, n \in N. I will omit the proof of this inequality, it is proven very simply by using math. induction.
So, we have \forall n\in N:n!>\left({\frac{n}{e}}\right)^n \Longrightarrow e>\frac{n}{\sqrt[n]{n!}}
Next, \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
Finally, we have \forall n \in N: e> x_n=\frac{n}{\sqrt[n]{n!}}>z_n
and \lim_{n\rightarrow\infty}z_n=e \Longrightarrow \lim_{n\rightarrow\infty}x_n=e
Q.E.D.
Is it correct ? Any suggestions ? I am missing something here...
My attempt:
First, I try to prove \forall n\in N:\frac{x_{n+1}}{x_n}>1.
\forall n\in :\frac{x_{n+1}}{x_n}=\frac{n+1}{n}\frac{\sqrt[n]{n!}}{\sqrt[n+1]{(n+1)!}}<br /> =\sqrt[n(n+1)]{\frac{(n+1)^{n(n+1)}(n!)^{n+1}}{n^{n(n+1)}((n+1)!)^n}}<br /> =\sqrt[n(n+1)]{\frac{(n+1)^{n^2}n!}{n^{n^2+n}}} and this is dead end.
Another attempt, we have 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}, where equality is then x_1=x_2=...=x_n
So, \forall n\in N: \sqrt[n]{n!}<\frac{(1+n)n}{2n}=\frac{n+1}{2}<br /> \Longrightarrow x_n=\frac{n}{\sqrt[n]{n!}}>\frac{2n}{1+n}=y_n
Sequence (y_n) is increasing, because \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
Can I say, x_n is increasing, because y_n is increasing and x_n > y_n, n \in N ?
Next, we have n!>\left({\frac{n}{e}}\right)^n, n \in N. I will omit the proof of this inequality, it is proven very simply by using math. induction.
So, we have \forall n\in N:n!>\left({\frac{n}{e}}\right)^n \Longrightarrow e>\frac{n}{\sqrt[n]{n!}}
Next, \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
Finally, we have \forall n \in N: e> x_n=\frac{n}{\sqrt[n]{n!}}>z_n
and \lim_{n\rightarrow\infty}z_n=e \Longrightarrow \lim_{n\rightarrow\infty}x_n=e
Q.E.D.
Is it correct ? Any suggestions ? I am missing something here...
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