Why is $\frac {1\cdot2\cdot3\cdot...\cdot n}{(n+1)(n+2)...(2n)}\le \frac 1 {n+1}$ I'm trying to find the limit: $\displaystyle\lim_{n\to\infty}\frac {1\cdot2\cdot3\cdot...\cdot n}{(n+1)(n+2)...(2n)}$
I thought of taking a pretty obvious binding from above expression: $\frac {n^n} {(n+1)^n}$ which is $n$ times the largest numerator and $n$ times the smallest denominator, but this limit isn't zero and it looks like the given does go to zero.
So maybe 'multiplying' $\frac 1 {n+1}$ by $1$, $n$ times is a good bound but it isn't very obvious that:  $\frac {1\cdot2\cdot3\cdot...\cdot n}{(n+1)(n+2)...(2n)}\le \frac 1 {n+1}$ so why is it true?
 A: The numerator has a term 1, so upper bound $n^{n-1}$ is acceptable.
Since you found denominator is at least $(n+1)^n$, you get
$$ \frac{1\cdots n}{(n+1)\cdots 2n} \leq \frac{n^{n-1}}{(n+1)^n}\leq \frac1{n+1}.$$
A: Write your left-hand side as
$${1 \over (n+1)} \times {2 \cdot 3 \cdot \cdots \cdot n \over (n+2)(n+3) \cdot \cdots \cdot 2n}$$
Now the first factor is $1/(n+1)$, and the second factor is obviously less than 1 (since the numerator is smaller than the denominator).  So the product is less than $1/(n+1)$.
This throws out a lot of information, but your bound is very weak - as has been observed, your function is approximately $\sqrt{\pi n}/4^n$, which is much smaller than $1/(n+1)$.
Edited to add: we can rewrite the product as $(n!)^2/(2n)!$.  Using Stirling's approximation $n! \approx \sqrt{2\pi n} (n/e)^n$ you get
$$(n!)^2/(2n)! \approx { (2\pi n) (n/e)^{2n} \over \sqrt{4 \pi n} (2n/e)^{2n}}$$
and you can simplify that to get the approximation $\sqrt{\pi n}/4^n$.
A: Try to express your term with binomial coefficients!
A: Hint: $\binom{2n}{k}$ is unimodal by moving k with maximum at $\binom{2n}{n}$. See also that $n+1\leq 2n=\binom{2n}{1}$.
