Solve the following limit of the sequence Another sequence limit I'm stuck.
$$\lim_{n\rightarrow\infty}{\frac{\prod_{k=1}^n(2k-1)}{(2n)^n}}$$
Any idea ?
 A: The inside of the limit can be written as $$\frac{1}{2n} \times \frac{3}{2n} \times \frac{5}{2n} \times \ldots \times \frac{2n-1}{2n}$$
What can you say about each of those fractions? Hence what happens when I start to have infinitely many of them multiplying?
A: Well you can take a look at this as
$$
\lim_{n \to \infty} \frac{(2(1) - 1)(2(2) - 1)(2(3) - 1)\cdots(2(n-1)-1)(2(n) - 1)}{(2n)\;(2n) \; (2n) \; \cdots \; (2n) \; (2n)}
$$
and so you can notice that for each term in the numerator there's a corresponding $(2n)$ in the denominator.
If this isn't clear enough you can notice that
$$
0 < \frac{\prod_{k=1}^n (2k-1)}{(2n)^n} < \frac{\prod_{k=1}^{n}2k}{(2n)^n} = \frac{2^n \prod_{k=1}^n k}{2^n n^n} = \frac{\prod_{k=1}^n k}{n^n}
$$
and that for each $k$ in the numerator there is an $n$ in the denominator and note that $n \ge k$ for each $k$. 
EDIT: Just saw that other answers were posted so I'll add some more info to make my answer seem better ;)
We can rewrite the rhs of the top inequality as
$$
\prod_{k=1}^n \frac{k}{n}
$$
Now you can easily observe that each term of this product is $\le 1$ so
$$
\prod_{k=1}^n \frac{k}{n} \le \frac{1}{n}
$$
Now you can use the squeeze/sandwich theorem to do what you need
A: Let $a_n$ be the $n$-th term of the sequence. We want to find out about the behaviour of $\dfrac{a_{n+1}}{a_n}$. There is immediate cancellation, and we get
$$\frac{a_{n+1}}{a_n}=\frac{(2n+1)(2n)^n}{(2n+2)^{n+1}}.$$
A little manipulation changes this to 
$$\frac{2n+1}{2n+2} \left(\frac{n}{n+1}\right)^n.$$
Now think a little about the behaviour of $\left(\dfrac{n}{n+1}\right)^n$ for large $n$.
A: Hint: try to rewrite it as 
$$
\frac{(2n)!}{(2n)^n}\frac{1}{2^n n!}
$$
and at that point, you can for instance use Stirling's (or any method of your choice) to deal with the factorials.
