# Interesting proof of sequence limit

1. Feb 9, 2008

### paniurelis

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 $$\forall n\in N:\frac{x_{n+1}}{x_n}>1$$.
$$\forall n\in :\frac{x_{n+1}}{x_n}=\frac{n+1}{n}\frac{\sqrt[n]{n!}}{\sqrt[n+1]{(n+1)!}} =\sqrt[n(n+1)]{\frac{(n+1)^{n(n+1)}(n!)^{n+1}}{n^{n(n+1)}((n+1)!)^n}} =\sqrt[n(n+1)]{\frac{(n+1)^{n^2}n!}{n^{n^2+n}}}$$ 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, $$\forall n\in N: \sqrt[n]{n!}<\frac{(1+n)n}{2n}=\frac{n+1}{2} \Longrightarrow x_n=\frac{n}{\sqrt[n]{n!}}>\frac{2n}{1+n}=y_n$$
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...

Last edited: Feb 9, 2008
2. Feb 9, 2008

### paniurelis

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 ?

3. Feb 9, 2008

### gel

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 realised that it should be n! not n in the denominator) is that if you expand out the n! you get

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

This is just a geometric mean. Maybe that helps you?

Last edited: Feb 9, 2008
4. Feb 9, 2008

### gel

might be easier to take logs, so
$$\log x_n = \sum_{k=1}^{n} (-\log(k/n)) (1/n)$$
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.

5. Feb 9, 2008

### paniurelis

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 :-)