Interesting proof of sequence limit

In summary, the author has attempted to prove that the sequence x_n=\frac{n}{\sqrt[n]{n!}} is increasing and \lim_{n\rightarrow\infty}{x_n}=e, but has been unsuccessful.
  • #1
paniurelis
12
0
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...
 
Last edited:
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  • #2
And another related problem with one above: Prove sequence [tex](y_n), y_n=\frac{n}{\left(\sqrt[n]{n!}\right)^2}[/tex] is decreasing and [tex]\lim y_n = 0 [/tex].
I see [tex]y_n=\frac{x_n^2}{n}[/tex] and [tex]\lim y_n = \frac{\lim x_n^2}{\lim_{n\rightarrow\infty}n}=\frac{e^2}{\lim_{\rightarrow\infty}n}=0.[/tex]

But how to show sequence is decreasing ?
 
  • #3
paniurelis said:
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] ?

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

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

This is just a geometric mean. Maybe that helps you?
 
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  • #4
might be easier to take logs, so
[tex]
\log x_n = \sum_{k=1}^{n} (-\log(k/n)) (1/n)
[/tex]
which will approach the limit [itex]\int_0^1(-\log(x))dx[/itex] from below. You still need to show that the approximations converge monotonically to the limit as n increases.
 
  • #5
Thank you, gel, I have corrected x_n with ! in my first post. Also, I have found error in my original post:
[tex]\lim_{n\rightarrow\infty}z_n=e \Longrightarrow \lim_{n\rightarrow\infty}x_n=e[/tex]

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

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

Related to Interesting proof of sequence limit

What is a sequence limit?

A sequence limit is the value that a sequence approaches as the number of terms in the sequence increases. It is also known as the limit of a sequence.

Why is the proof of a sequence limit interesting?

The proof of a sequence limit is interesting because it helps us understand the behavior of a sequence and provides a mathematical basis for many real-world applications. It also allows us to determine if a sequence converges or diverges.

What are the different methods used to prove a sequence limit?

There are several methods used to prove a sequence limit, including the epsilon-delta method, the monotone convergence theorem, and the squeeze theorem. Each method has its own set of requirements and is used depending on the nature of the sequence.

Can a sequence have more than one limit?

No, a sequence can only have one limit. If a sequence has more than one limit, it is considered to be divergent.

What is the importance of understanding sequence limits in scientific research?

Understanding sequence limits is crucial in scientific research as it allows us to make predictions and draw conclusions based on the behavior of a sequence. It also helps in understanding the convergence or divergence of data, which is essential in many fields of research such as physics, engineering, and economics.

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