Product of infinite terms $\lim_{n\to\infty}\left(\frac{(n+1)(n+2)...(3n)}{n^{2n}}\right)$ Let there be a series
$$\lim_{n\rightarrow \infty}\left(\frac{(n+1)(n+2)...(3n)}{n^{2n}}\right)$$ 
is equal to?
For this type of problem I am unable to approach.The numerator starts from $(n+1)$ to $3n$ where as the denominator has ${n^{2n}}$ terms
 A: There are $2n$ factors in the numerator.  Of them, the first $n$ are each greater than $n$, and the next $n$ are each greater than $2n$.  Thus, the numerator is greater than $n^n (2n)^n$.  The denominator is $n^{2n}$ and so the ratio is greater than $2^n$, which of course tends to $\infty$ as $n\to\infty$.
A: Hint:) With Stirling's approximation
$$n!\sim\sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n$$
we have
$$\lim_{n\rightarrow \infty}\left(\frac{(n+1)(n+2)...3n}{n^{2n}}.\dfrac{n!}{n!} \right)=\lim_{n\rightarrow \infty}\dfrac{(3n)!}{n!n^{2n}}=\lim_{n\rightarrow \infty}\sqrt{3}\left(\dfrac{27}{e^2}\right)^n=\color{blue}{\infty}$$
A: One possibility that I usually do for products is to take logarithms. For any finite $n$, we have 
$$
\ln\left(\frac{(n+1)\dots(3n)}{n^{2n}}\right)=\sum_{k=1}^{2n}\ln\left(\frac{n+k}{n}\right)=\sum_{n=1}^{2n}\ln\left(1+\frac{k}{n}\right)=n\sum_{k=1}^{2n}\frac{\ln(1+\tfrac{k}{n})}{n}
$$
In the limit, the sum 
$$
\sum_{k=1}^{2n}\frac{\ln(1+\tfrac{k}{n})}{n}
$$
approaches a positive, nonzero definite integral, whereas $n$ approaches infinity. Thus the logarithm approaches infinity, so the original product does as well.
A: Let us denote
$$a_n=\frac{(n+1)(n+2)...(3n)}{n^{2n}}.$$
Then you have
$$\frac{a_{n+1}}{a_n} = 
\frac{(3n+1)(3n+2)(3n+3)}{(n+1)^3}  \left(\frac{n}{n+1}\right)^{2n}.$$
I should be relatively easy to check that
$$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} > 1.$$
(In fact, the limit is equal to $27/e^2$ unless I missed something.)
Can you tell something about $a_n$ combining the facts that it is positive and starting from some $n_0$ you have $a_{n+1}/a_n>1+\varepsilon$ (for some $\varepsilon>0$)? 
In the other words, can you show this: If $a_n>0$ for each and $\lim\limits_{n\to\infty} a_{n+1}/a_n>1$, then $\lim\limits_{n\to\infty} a_n=\infty$?
