5
$\begingroup$

$\lim n^n/(e^nn!)=0$ using Stirling approximation it is obvious. But can we do it without using Stirling approximation.

Now series with terms $x^n n^n/n!$ has ROC $1/e$. What we can say about behaviour of power series at endpoints?

$\endgroup$
2
  • 1
    $\begingroup$ please use Latex $\endgroup$
    – Alex
    Mar 19, 2015 at 8:47
  • $\begingroup$ A previous answer of mine proves this in a completely elementary fashion: math.stackexchange.com/a/131084/1102. Falls right out of Proposition E. $\endgroup$
    – Aryabhata
    Mar 26, 2015 at 7:15

6 Answers 6

6
+50
$\begingroup$

If we set $\left(1+\frac1n\right)^n=1$ for $n=0$, we have $$ \begin{align} \left.\frac{(n+1)^{n+1}}{e^{n+1}\,(n+1)!}\middle/\frac{n^n}{e^n\,n!}\right. &=\frac{\left(1+\frac1n\right)^n}{e} \end{align} $$ In this answer, it is shown that $$ e-\left(1+\frac1n\right)^n\ge\frac e{2n+3} $$ which implies that $$ \frac{\left(1+\frac1n\right)^n}{e}\le\frac{2n+2}{2n+3} $$ Therefore, $$ \begin{align} \frac{n^n}{e^n\,n!} &=\prod_{k=0}^{n-1}\frac{\left(1+\frac1k\right)^k}{e}\\ &\le\prod_{k=0}^{n-1}\frac{2k+2}{2k+3}\\ &\le\prod_{k=0}^{n-1}\left(\frac{2k+2}{2k+3}\frac{2k+3}{2k+4}\right)^{1/2}\\ &=\left(\prod_{k=0}^{n-1}\frac{k+1}{k+2}\right)^{1/2}\\ &=\frac1{\sqrt{n+1}} \end{align} $$ Thus, $$ \begin{align} \lim_{n\to\infty}\frac{n^n}{e^n\,n!} &\le\lim_{n\to\infty}\frac1{\sqrt{n+1}}\\ &=0 \end{align} $$

$\endgroup$
0
3
$\begingroup$

We can do it by proving Stirling's theorem without the constant. We need the following Lemma which has appeared here before, and is easily proven by clever partial integration:

Lemma. If $a<b$ and $f(a)=f(b)=0$ then $$\int_a^b f(t)\>dt=-{(b-a)^3\over12}f''(\xi)$$ for some $\xi\in\ ]a,b[\ $.

We now the compute the integral $$\int_1^n \log t\>dt=n\log n -n+1\tag{1}$$ using the trapezoidal rule: $$\int_1^n \log t\>dt=0+\sum_{k=2}^{n-1}\log k +{1\over2}\log n+R_n=\log\bigl(n!\bigr)-{1\over2}\log n+R_n\ .\tag{2}$$ The error $R_n$ is positive, since $\log$ is concave. The difference $f$ between $\log$ and the interpolating piecewise linear function is $=0$ at all integer points, and its second derivative in between these points is given by $$f''(t)=\log''(t)=-{1\over t^2}\ .$$ It then follows from the above Lemma that $$0<R_n=\sum_{k=1}^{n-1}\int_k^{k+1}f(t)\>dt\leq{1\over12}\sum_{k=1}^{n-1}{1\over k^2}<{\pi^2\over72}<1\ .$$ Using $(1)$ and $(2)$ we now obtain $$n(\log n-1)-\log\bigl(n!\bigr)=-{1\over2}\log n +R_n-1\to-\infty\qquad(n\to\infty)\ ,$$ which is what you wanted.

$\endgroup$
2
$\begingroup$

The ratio of terms $n+1$ and $n$ is $$\frac1e\left(1+\frac1n\right)^n.$$

By the binomial theorem,

$$\left(1+\frac1n\right)^n=1+\frac nn\frac{n(n-1)}{2n^2}+\frac{n(n-1)(n-2)}{3!n^3}\cdots<2+\frac{n-1}n\left(\frac12+\frac1{3!}\cdots\right)\\=2+\frac{n-1}n(e-2),$$ so that $$\frac1e\left(1+\frac1n\right)^n<1-\frac{e-2}{en}<1-\frac1{4n}.$$

Then by telescoping the product of the ratios tends to $0$.


UPDATE: simpler derivation.

The function $(1+x)^{1/x}$ is convex between $0$ and $1$, so that

$$\frac1e(1+x)^{1/x}\le 1-(1-\frac2e)x<1-\frac x4,$$ and $$\frac{n^n}{e^nn!}=\prod_{k=1}^{n-1}\frac{(k+1)^{k+1}}{e^{k+1}(k+1)!}\frac{e^kk!}{k^k}=\prod_{k=1}^{n-1}\frac1e\left(1+\frac1k\right)^k<\prod_{k=1}^{n-1}\left(1-\frac1{4k}\right),$$which converges to $0$.

$\endgroup$
1
$\begingroup$

Since $$ n=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right) \tag{1}$$ it follows that: $$ \frac{n^n}{n!e^n}=e^{-n}\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k\tag{2}$$ hence to prove that the limit is zero it is enough to prove that: $$ \left(1+\frac{1}{k}\right)^{k+\alpha} \leq e \tag{3}$$ for some positive $\alpha$ and any $k\in\mathbb{N}^+$.

By taking logarithms, we need to prove that for some $\alpha > 0$ $$ (k+\alpha) \int_{k}^{k+1}\frac{dx}{x} \leq 1 \tag{4}$$ holds. By the Hermite-Hadamard inequality we have: $$ \int_{k}^{k+1}\frac{dx}{x}\leq \frac{1}{2}\left(\frac{1}{k}+\frac{1}{k+1}\right) \tag{5} $$ hence the choice $\alpha=\frac{1}{3}$ is perfectly fine.

$\endgroup$
0
$\begingroup$

Maybe is not the most directly method, but expanding $$\frac{x^{x}e^{-x}}{\Gamma\left(x+1\right)}$$ in Puiseux series, we have $$\frac{x^{x}e^{-x}}{\Gamma\left(x+1\right)}=O\left(\sqrt{\frac{1}{x}}\right)$$ at $x\rightarrow\infty.$ Another way is to note that holds, by continuity, $$\lim_{n\rightarrow\infty}\frac{n^{n}e^{-n}}{n!}=\exp\left(\lim_{n\rightarrow\infty}-n+n\log\left(n\right)-\log\left(n!\right)\right) .$$ Now here you can find a proof, without Stirling's approximation, that $$\lim_{n\rightarrow\infty}\frac{\log\left(n!\right)}{n\log\left(n\right)}=1$$hence $$\lim_{n\rightarrow\infty}\frac{n^{n}e^{-n}}{n!}=\exp\left(\lim_{n\rightarrow\infty}-n\right)=0.$$

$\endgroup$
0
$\begingroup$

As an alternative, let consider

$$a_n=\dfrac{n! e^n}{n^n} \implies \log a_n = \sum_{k=1}^n\log k+n-n\log n \\\implies \frac{\log a_n}{\log n}=\frac{\sum_{k=1}^n\log k+n-n\log n}{\log n}\to \frac12$$

indeed by Cesaro-Stoltz

$$\frac{\sum_{k=1}^{n+1}\log k+(n+1)-(n+1)\log (n+1)-\sum_{k=1}^n\log k-n+n\log n}{\log (n+1)-\log n}=$$

$$=\frac{1-n\log\left(1+\frac1n\right)}{\log\left(1+\frac1n\right)}\to \frac12$$

which can be easily shown by Taylor's series or by l'Hopital.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .