Interesting proof of sequence limit

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 &gt; 0, x_2 &gt; 0, ..., x_n&gt;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!}&lt;\frac{(1+n)n}{2n}=\frac{n+1}{2}<br /> \Longrightarrow x_n=\frac{n}{\sqrt[n]{n!}}&gt;\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}&gt;1

Can I say, x_n is increasing, because y_n is increasing and x_n &gt; y_n, n \in N ?

Next, we have n!&gt;\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!&gt;\left({\frac{n}{e}}\right)^n \Longrightarrow e&gt;\frac{n}{\sqrt[n]{n!}}

Next, \forall n \in N: \sqrt[n]{n!} &lt; \frac{(n+1)n}{2n} = \left(1 + \frac{1}{n}\right) \frac{n}{2}&lt;\left(1 + \frac{1}{n}\right)n=\frac{1}{1-\frac{1}{n+1}}n \Longrightarrow x_n=\frac{n}{\sqrt[n]{n!}}&gt;1-\frac{1}{n+1}&gt;1-\frac{1}{n}&gt;\left(1-\frac{1}{n}\right)^{n}=z_n

Finally, we have \forall n \in N: e&gt; x_n=\frac{n}{\sqrt[n]{n!}}&gt;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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And another related problem with one above: Prove sequence (y_n), y_n=\frac{n}{\left(\sqrt[n]{n!}\right)^2} is decreasing and \lim y_n = 0.
I see y_n=\frac{x_n^2}{n} and \lim y_n = \frac{\lim x_n^2}{\lim_{n\rightarrow\infty}n}=\frac{e^2}{\lim_{\rightarrow\infty}n}=0.

But how to show sequence is decreasing ?
 
paniurelis said:
Can I say, x_n is increasing, because y_n is increasing and x_n &gt; y_n, n \in N ?

No you can't.
Also, you missed the ! in the original equation for x_n.

The first thing I see with expression for x_n (once I realized that it should be n! not n in the denominator) is that if you expand out the n! you get

<br /> x_n^{-1}=\sqrt[n]{(1/n)(2/n)(3/n)\cdots(n/n)}<br />

This is just a geometric mean. Maybe that helps you?
 
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might be easier to take logs, so
<br /> \log x_n = \sum_{k=1}^{n} (-\log(k/n)) (1/n)<br />
which will approach the limit \int_0^1(-\log(x))dx from below. You still need to show that the approximations converge monotonically to the limit as n increases.
 
Thank you, gel, I have corrected x_n with ! in my first post. Also, I have found error in my original post:
\lim_{n\rightarrow\infty}z_n=e \Longrightarrow \lim_{n\rightarrow\infty}x_n=e

Should be: \lim_{n\rightarrow\infty}z_n=\frac{1}{e} \Longrightarrow e\geq \lim_{n\rightarrow\infty}x_n\geq \frac{1}{e}

So, now I have proven x_n is bounded, but not the convergence to e :-(
I'll think about geometric mean and logarithms, maybe I will come up with something :-)
 

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