1
$\begingroup$

I want to prove the following and wanted to ask, if my proof is correct.

$$\lim \limits_{n \rightarrow \infty} \frac{2^n n!}{n^{n+1}} = 0$$

Remark (i): For $n \in \mathbb{N}$ and $n > 1$ holds: $2 < (1+\frac{1}{n})^n$

Define $b_n := \left\{\frac{2^n n!}{n^{n}}\right\}_{n \in \mathbb{N}}$.

Then $\left| \frac{b_{n+1}}{b_n} \right| = \left| \frac{\frac{2^{n+1} (n+1)!}{(n+1)^{n+1}}}{\frac{2^n n!}{n^{n}}} \right| = 2 \left(\frac{n}{n+1}\right)^n = 2 \left(1-\frac{1}{n+1}\right)^n = \frac{2 \left(1-\frac{1}{n+1}\right)^n \cdot (1+\frac{1}{n})^n}{(1+\frac{1}{n})^n} = \frac{2}{(1+\frac{1}{n})^n} \overbrace{<}^\text{(i)} 1$ holds for $n \in \mathbb{N}$ and $n > 1$

Hence: $\lim \limits_{n \rightarrow \infty} \left|\frac{b_{n+1}}{b_n}\right| < 1$, which yields: $\lim \limits_{n \rightarrow \infty} b_n = 0$

Define $a_n := \left\{\frac{2^n n!}{n^{n+1}}\right\}_{n \in \mathbb{N}}$.

Obviously $0 \leq a_n \leq b_n$ holds for all $n \in \mathbb{N}$.

Hence with the sandwich-theorem: $\lim \limits_{n \rightarrow \infty} 0 = \lim \limits_{n \rightarrow \infty} b_n = \lim \limits_{n \rightarrow \infty} a_n = \lim \limits_{n \rightarrow \infty} \frac{2^n n!}{n^{n+1}} = 0$

$\endgroup$
  • $\begingroup$ Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. $\endgroup$ – AlexR Dec 2 '14 at 18:59
1
$\begingroup$

At $n\to\infty$, $n!$ can be written as $\sqrt{2\pi n}\left(\dfrac ne\right)^n$. Now limit can be written as $$\lim_{n\to\infty}{\dfrac{2^n\sqrt{2\pi n}\left(\dfrac ne\right)^n}{n^{n+1}}}$$ Now we need to simplify it. $$e^{\lim_{n\to\infty}{\left((\ln2-1)n-\dfrac12\ln n+\dfrac12\ln2\pi\right)}}$$ So, we need to prove that $\lim_{n\to\infty}{\left((\ln2-1)n-\dfrac12\ln n\right)=-\infty}$
Function $(\ln2-1)n$ at $n\to\infty$ is $-\infty$ and $-\dfrac12\ln n$ at $n\to\infty$ is $-\infty$, so sum of this functions must be $-\infty$.

$\endgroup$
0
$\begingroup$

If $b_n\rightarrow 0$ then naturally $b_n/n\rightarrow 0$ by distributing the limits. You might want to also mention why $b_{n+1}/b_n$ has a limit in the first place (even though it seems implicit in your assertion that the ratio is <1). Basically, the proof is fine.

$\endgroup$
0
$\begingroup$

It is correct, but I find this a bit faster:

$$\frac{a_{n+1}}{a_n}=\frac{2n}{n+1}\left(\frac n{n+1}\right)^{n+1}\to\frac 2e<1$$

Nevertheless, yours is more elementary since it does not use the limit $\lim\left(1+\frac1n\right)^n=e$

$\endgroup$
0
$\begingroup$

I present an elementary solution.

Write $$\ell = \lim_{n \to \infty} \frac {2^n n!} {n^{n + 1}} = \lim_{n \to \infty} \frac {1} {n} \prod_{k = 1}^{n} \frac {2k} {n}.$$ For $k < n/2$, write $2k/n = 1 - a_k$, where $0 < a_k < 1$. Then, obviously, for $k > n/2$ we'll have $2k/n = 1 + a_k$. Hence, $$\ell = \lim_{n \to \infty} \frac {1} {n} \prod_{k = 1}^{n/2} (1 - a_k^2) = \lim_{n \to \infty} \frac {1} {n} \cdot \lim_{n \to \infty} \prod_{k = 1}^{n/2} (1 - a_k^2) = 0 \cdot 0 = 0.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.