Limit of a sequence that tends to $1/e$ 
Possible Duplicate:
Showing that $\frac{\sqrt[n]{n!}}{n}$ $\rightarrow \frac{1}{e}$ 

Any suggestions to find the following limit:
$$\displaystyle\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}}$$
with basic tools of the calculus!
 A: Below is a quick derivation of part of Stirling's formula, which will enable you to conclude that the limit is $1/e$.
First note that from Abel summation, we have that
$$\sum_{n=1}^N a(n) b(n) = \sum_{n=1}^N b(n)(A(n)-A(n-1)) = \sum_{n=1}^{N} b(n) A(n) - \sum_{n=1}^N b(n)A(n-1)\\
= \sum_{n=1}^{N} b(n) A(n) - \sum_{n=0}^{N-1}^N b(n+1)A(n) = b(N) A(N) - b(1)A(0) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$
where $A(n) = \displaystyle \sum_{k=1}^n a(k)$.
In our case, take $a(n) = 1$, $b(n) = \log(n)$. This gives us $A(n) = n$.
\begin{align}
\log(n!) = \sum_{k=1}^n \log(k) & = \sum_{k=1}^n a(k) b(k) = n \log(n) + \sum_{k=1}^{n-1}k (\log(k) - \log(k+1))\\
& = n \log(n) + \sum_{k=1}^{n-1}k \log \left(1 - \dfrac1{k+1} \right)\\
& = n \log(n) + \sum_{k=1}^{n-1}k \left(-\dfrac1{k+1} + \mathcal{O}\left(\dfrac1{(k+1)^2} \right)\right)\\
& = n \log(n) + \sum_{k=1}^{n-1} \left(-\dfrac{k}{k+1} + \mathcal{O}\left(\dfrac{k}{(k+1)^2} \right)\right)\\
& = n \log(n) + \sum_{k=1}^{n-1} \left(-1 + \mathcal{O}\left(\dfrac1{(k+1)} \right)\right)\\
& = n \log(n) - n + \mathcal{O}\left(\log(n) \right)\\
\end{align}
Hence, we get that
$$\log \left(\dfrac{n!}{n^n}\right) = - n + \mathcal{O}(\log (n))$$
Hence,
$$\dfrac1n \log \left(\dfrac{n!}{n^n}\right) = - 1 + \mathcal{O}\left(\dfrac{\log (n)}n \right)$$
Hence, we get that
$$\lim_{n \to \infty} \dfrac1n \log \left(\dfrac{n!}{n^n}\right) = -1$$
And since $\log$ is continuous, we get that
$$\lim_{n \to \infty}\left(\dfrac{n!}{n^n}\right)^{1/n} = \dfrac1e$$
A: I could not find the following proof in the multiple duplicates, so here it is.
By integral comparison, $\int_{k-1}^k\ln t dt\leq \ln k \leq \int_k^n \ln t dt$ for all $k=2,\ldots,n$.
Summing these $n-1$ inequations yields:
$$
\int_1^n\ln t dt \leq \ln (n!) \leq \int_2^{n+1}\ln t dt.
$$
Using that $t\ln t-t$ is an antiderivative of $\ln t$, we get:
$$
n\ln n-n+1\leq \ln (n!) \leq (n+1)\ln(n+1)-n+C
$$
where $C=1-2\ln 2$.
Next 
$$
-n+1\leq \ln (n!) - n\ln n\leq n\ln(1+1/n) + \ln(n+1) -n+C
$$
whence
$$
-1+\frac{1}{n} \leq \frac{ \ln (n!) - n\ln n}{n} \leq -1 + \ln(1+1/n) +\frac{\ln(n+1)+C}{n}.
$$
By the Squeeze Theorem, it follows that
$$
\lim_{n\rightarrow +\infty}\frac{ \ln (n!) - n\ln n}{n}=-1
$$
and finally
$$
\lim_{n\rightarrow +\infty} \left( \frac{n!}{n^n}\right)^{1/n}=\lim_{n\rightarrow +\infty} \exp\left( \frac{ \ln (n!) - n\ln n}{n} \right)=\exp(-1).
$$
