$\lim\limits_{n\to\infty} \frac {n}{\sqrt[n]{n!}} $ and $\lim\limits_{n\to\infty} C_{2 n}^n$ Compute $$\lim\limits_{n\to\infty} \frac {n}{\sqrt[n]{n!}} $$ and $$\lim\limits_{n\to\infty} {2n \choose n}$$
I suppose that the first limit is equal to $e $.
 A: First Limit
Using Riemann-Sums, we get
$$
\begin{align}
\lim_{n\to\infty}\frac1n\log\left(\frac{n!}{n^n}\right)
&=\lim_{n\to\infty}\sum_{k=1}^n\log\left(\frac kn\right)\frac1n\\
&=\int_0^1\log(x)\,\mathrm{d}x\\[9pt]
&=-1
\end{align}
$$
Thus,
$$
\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}=\frac1e
$$
and therefore,
$$
\lim_{n\to\infty}\frac{n}{(n!)^{1/n}}=e
$$

Second Limit
Hint:
$$
\binom{2n}{n}=\binom{2n-2}{n-1}\left(4-\frac2n\right)
$$
A: Hint for the first limit:
$$
\ln \left( \lim\limits_{n\to\infty} \frac {n}{\sqrt[n]{n!}} \right)
= \lim\limits_{n\to\infty} \left( \ln \frac {n}{\sqrt[n]{n!}} \right)
= \lim\limits_{n\to\infty} \left( \ln n- \frac{1}{n}\sum_{j=1}^n \ln j \right).
$$
And I suppose there is something wrong with your second limit since it is obviously divergent.
A: Let $a_n=\frac{n^n}{n!}.$
Thus, $$\lim_{n\rightarrow+\infty}\frac{n}{\sqrt[n]{n!}}=\lim_{n\rightarrow+\infty}\sqrt[n]{a_n}=\lim_{n\rightarrow+\infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow+\infty}\left(1+\frac{1}{n}\right)^n=e.$$
The second:
$$\binom{2n}{n}=\frac{2n\cdot(2n-1)\cdot...\cdot(n+1)}{n\cdot(n-1)\cdot...\cdot1}\geq2\cdot2\cdot...\cdot2=2^n\rightarrow+\infty.$$
A: From Taylor expansion we have $\frac{x^n}{n!} \leq e^x$ for any $x ≥ 0$. Spe- cialising this to $x = n$ we obtain a crude lower bound
$n! ≥ n^ne^{−n}$.
In the other direction, we trivially have
$n! ≤ n^n$
so we know already that n! is within an exponential factor of $n^n$.
One can do better by starting with the identity
$log n! = \sum_{m=1}^n  log m$
and viewing the right-hand side as a Riemann integral approximation to  $\int_1^nlog x\  dx$. Indeed a simple area comparison yields the inequalities
$\int_1^nlog x dx ≤\sum_{m=1}^n log m ≤ log n +\int_1^n log x dx$
which leads to the inequalities $en^ne^{−n} ≤ n! ≤ en × n^ne^{−n}$
