Exist or Not ? $\displaystyle \lim_{x\to\infty} \dfrac{1}{x} - \dfrac{1}{x-k}$, $1 \leq k \leq x-1$

Question

I was trying to evaluate $$ \lim_{n\to \infty} \left(\dfrac{n!}{n^n}\right)^{\dfrac{1}{n}} $$ without resorting to a certain theorem that states : $$ \limsup_{n\to \infty}\; a_n\,^{\frac{\Large 1}{\Large n}} = \lim_{n\to \infty} \left(\dfrac{a_{n+1}}{a_n}\right). $$

With my own idea:

• By using the discreet definition of factorial using the product notation (also converting $n^n$ into discreet product) :

$$ n! = \prod_{0\leq k \leq n-1} (n-k)= n\cdot\left(\prod_{1\leq k \leq n-1} (n-k)\right) $$

$$ n^n = \prod_{1\leq k \leq n} n = n\cdot\left(\prod_{1\leq k \leq n-1} n \right) $$

And by dividing $n-k$ by $n$ and duplicating the limit (one for evaluating the limit inside product notation and another for reminding myself that $x$ goes to $\infty$ ) into two I get this limit :

$\displaystyle \lim_{n\to\infty} \prod_{1\leq k \leq n-1} \lim_{n\to\infty}\left(1-\dfrac{k}{n}\right)^{\dfrac{1}{n}}$ (1)

Evaluating the inner limit by logarithm and L'Hopital get me into this point :

$a = \displaystyle \lim_{n\to\infty} \dfrac{1}{n} - \dfrac{1}{n-k}$, $1 < k < n-1$

where $y = \displaystyle \lim_{n\to\infty} \left(1-\dfrac{k}{n}\right)^{\dfrac{1}{n}}$, and $a = \ln(y)$

My Question :

$1$) Does $a$ exist as $k$ varies and goes to $n-1$ ? (Continue to Q$2$)

$2$) How to evaluate it ? (Continuation of Q$1$)

$3$) Is my way of doing the limit at equation (1) valid (Duplicating the limit) ??

$4$) Any other way to do $\displaystyle \lim_{n\to \infty} \left(\dfrac{n!}{n^n}\right)^{\dfrac{1}{n}}$ without that sequence theorem I mentioned earlier

My opinion about Q$1$ : because the main limit (at the very beginning) exist and has a value, I think somehow $a$ exist, but what confuse me is that $a$ varies between $0$ and $-1$ ($a = -1$ when $k$ goes to $n-1$).

Q$2$ : I don't have any idea as $k$ value is varying (at least I know that when $k$ is small then $a$ goes to $0$, and when $k$ value is approaching $n-1$ then $a$ goes to $-1$)

Answer 1

Another way: $$ {n!\over n^n} = \prod_{k=1}^n {k\over n}$$ $$ \ln \left({n!\over n^n}\right)^{1/n} = \frac1n \sum_{k=1}^n \ln{k\over n}$$ It can be noticed that it is a Riemann integral sum: $$\lim_{n\to\infty} \frac1n \sum_{k=1}^n \ln{k\over n} = \int_0^1\ln x\,dx = (x\ln x-x)|_0^1 = -1 $$ We have $$ \lim_{n\to\infty} \ln \left({n!\over n^n}\right)^{1/n} = -1$$ so $$ \lim_{n\to\infty} \left({n!\over n^n}\right)^{1/n} = e^{-1}$$

Answer 2

Quick approach - Sterling's formula:

$(\frac{n!}{n^n})^\frac{1}{n}\approx \frac{1}{e}(\sqrt{2\pi n})^\frac{1}{n}\to \frac{1}{e}$.

Answer 3

We have $\ln \left( \frac {n!}{n^n}\right)^{1/n}=\frac {1}{n}\ln (n!)-\ln n.$

For $n\ge 2$ we can obtain crude but useful lower & upper estimates for $\ln (n!)$ as follows: $$(n\ln n)-n+1=\int_1^n\ln x\,dx=$$ $$=\sum_{j=1}^{n-1}\int_j^{j+1}\ln x\,dx<$$ $$<\sum_{j=1}^{n-1}\int_j^{j+1}\ln (j+1)\,dx=$$ $$=\sum_{j=1}^{n-1}\ln (j+1)=\ln (n!).$$ ......and............ $$\ln (n!)=\sum_{j=1}^n\ln j=$$ $$=\sum_{j=1}^n\int_j^{j+1} \ln j\,dx<$$ $$<\sum_{j=1}^n\int_j^{j+1} \ln (x+1)\,dx=$$ $$=\int_1^n\ln (x+1)\,dx=$$ $$=(n+1)\ln (n+1)-n+1-2\ln 2.$$

From this, and from $\lim_{n\to\infty}\frac {\ln (n+1)}{n}=0$ we immediately obtain $\lim_{n\to \infty}\ln \left( \frac {n!}{n^n}\right)^{1/n}=-1.$

Stirling's Formula $n!(1-\frac {1}{6n})<S(n)<n!(1+\frac {1}{6n})$ for $n>0,$ where $S(n)=(\frac {n}{e})^n \sqrt {2\pi n}$, is more difficult to prove and provides much more precision than is needed for your Q.