$ \mathop {\lim }\limits_{n \to + \infty } \frac{{v_{n + 1} }}{{v_n }} = 2$ help me please
true or fulse 
(1)$$\mathop {\lim }\limits_{n \to  + \infty } \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}} = 1?
$$
\begin{array}{l}
 u_n  = \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}}\quad ;u_n  > 0 \\ 
  \Leftrightarrow \ln \left( {u_n } \right) = \ln \left( {n\left( {n + 1} \right) \cdots \left( {n + n} \right)} \right)^{\frac{1}{n}}  \\ 
  \Leftrightarrow \ln \left( {u_n } \right) = \frac{1}{n}\ln \left( {n\left( {n + 1} \right) \cdots \left( {n + n} \right)} \right) \\ 
  \Leftrightarrow \ln \left( {u_n } \right) = \frac{{\ln \left( n \right)}}{n} + \frac{{\ln \left( {n + 1} \right)}}{n} +  \cdots  + \frac{{\ln \left( {n + n} \right)}}{n} \\ 
  \Leftrightarrow \mathop {\lim }\limits_{n \to  + \infty } \left( {\ln \left( {u_n } \right)} \right) = \mathop {\lim }\limits_{n \to  + \infty } \left[ {\frac{{\ln \left( n \right)}}{n} + \frac{{\ln \left( {n + 1} \right)}}{n} +  \cdots  + \frac{{\ln \left( {n + n} \right)}}{n}} \right] = 0 \\ 
 \mathop {\lim }\limits_{n \to  + \infty } \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}} = 1 \\ 
 \end{array}
(2)
$$
\begin{array}{l}
 t_n  = \frac{{1 \times 3 \times  \cdots (2n - 1)}}{{n^n }} \\ 
  \Rightarrow \mathop {\lim }\limits_{n \to  + \infty } \frac{{t_{n + 1} }}{{t_n }} = \mathop {\lim }\limits_{n \to  + \infty } \frac{{1 \times 3 \times  \cdots (2n - 1)(2n + 1)}}{{\left( {n + 1} \right)^{n + 1} }} \times \frac{{n^n }}{{1 \times 3 \times  \cdots (2n - 1)}} \\ 
  \Rightarrow \mathop {\lim }\limits_{n \to  + \infty } \frac{{t_{n + 1} }}{{t_n }} = \mathop {\lim }\limits_{n \to  + \infty } \frac{{(2n + 1)}}{{\left( {n + 1} \right)^{n + 1} }} \times \frac{{n^n }}{1} \\ 
  = \mathop {\lim }\limits_{n \to  + \infty } \left( {2\left( {\frac{n}{{n + 1}}} \right)^{n + 1}  + \frac{{n^n }}{{\left( {n + 1} \right)^{n + 1} }}} \right) = 2e^{ - 1} ;\quad \left( {\mathop {\lim }\limits_{n \to  + \infty } \left( {1 + \frac{x}{n}} \right)^n  = e^x } \right) \\ 
  \Rightarrow \mathop {\lim }\limits_{n \to  + \infty } \sqrt[n]{{t_n }} = \mathop {\lim }\limits_{n \to  + \infty } \sqrt[n]{{\frac{{1 \times 3 \times  \cdots (2n - 1)}}{{n^n }}}} = 2e^{ - 1}  \\ 
 \end{array}
$$
(3) $$\mathop {\lim }\limits_{n \to  + \infty } \frac{1}{{n^2 }}\sqrt[n]{{\frac{{3n!}}{{n!}}}}$$
$$\left( 4 \right)\mathop {\lim }\limits_{n \to  + \infty } \left( {n\sqrt[n]{{\frac{{\left( {2n} \right)!}}{{\left( {n!} \right)^3 }}}}} \right) = !?
$$
 A: You can rewrite the definition of $v_n$ as
$$
v_1=2\\[2ex]
v_{n+1}=\frac{(2n+1)(2n+2)}{n}v_n
$$
so
$$
\frac{v_{n+1}}{v_n}=\frac{(2n+1)(2n+2)}{n}
$$
Another way to see this is noting that
$$
v_n=\frac{(2n)!}{(n-1)!}
$$
For $t_n$ you have
\begin{align}
t_n
&=\frac{1\cdot3\cdot\ldots\cdot(2n-1)}{n^n}\\
&=\frac{1}{n^n}
  \frac{(1\cdot3\cdot\ldots\cdot(2n-1))(2\cdot 4\cdot\ldots\cdot 2n)}
       {2\cdot 4\cdot\ldots\cdot 2n}\\
&=\frac{1}{n^n}\frac{(2n)!}{2^n\cdot n!}\\
\end{align}
so
$$
\frac{t_{n+1}}{t_n}=
\frac{1}{(n+1)^{n+1}}\frac{(2n+2)!}{2^{n+1}\cdot (n+1)!}
n^n\frac{2^n\cdot n!}{(2n)!}
=\frac{1}{2}\frac{n^n}{(n+1)^{n+1}}\frac{(2n+2)(2n+1)}{n+1}
$$
Thus we need to look at
$$
\lim_{n\to\infty}\frac{n^n}{(n+1)^{n+1}}(2n+1)
$$
and it's easier doing the limit of the inverse:
$$
\lim_{n\to\infty}\frac{(n+1)^{n+1}}{n^n}\frac{1}{2n+1}=
\lim_{n\to\infty}\frac{(n+1)^n}{n^n}\frac{n+1}{2n+1}=
\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\frac{n+1}{2n+1}=\frac{e}{2}
$$
so your given limit is $2/e$.
A: $$\underset{n\rightarrow\infty}{\lim}\frac{v_{n+1}}{v_{n}}=\underset{n\rightarrow\infty}{\lim}\frac{\left(n+1\right)\cdots\left(n+1+n-1\right)\left(n+1+n\right)\left(n+1+n+1\right)}{n\cdots\left(n+n\right)}=$$
 $$=\underset{n\rightarrow\infty}{\lim}\frac{\left(n+1+n\right)\left(n+1+n+1\right)}{n}=\infty$$ because you have essentialy $n^{2}$ at numerator.$$\underset{n\rightarrow\infty}{\lim}\frac{t_{n+1}}{t_{n}}=\underset{n\rightarrow\infty}{\lim}\frac{1\cdot3\cdots\left(2n-1\right)\left(2n+1\right)}{\left(n+1\right)^{n+1}}\cdot\frac{n^{n}}{1\cdot3\cdots\left(2n-1\right)}=$$
 $$=\underset{n\rightarrow\infty}{\lim}\left(2n+1\right)\frac{n^{n}}{\left(n+1\right)^{n+1}}=\underset{n\rightarrow\infty}{\lim}\left(2\left(\frac{n}{n+1}\right)^{n+1}+\frac{n^{n}}{\left(n+1\right)^{n+1}}\right)=2e^{-1}$$
 because $\underset{n\rightarrow\infty}{\lim}\left(1+\frac{x}{n}\right)^{n}=e^{x}.$
A: $$
\begin{array}{l}
 n! = \sqrt {n\pi e} \left( {\frac{n}{e}} \right)^n  \\ 
  \Rightarrow \mathop {\lim }\limits_{n \to  + \infty } \left( {n\sqrt[n]{{\frac{{\left( {2n} \right)!}}{{\left( {n!} \right)^3 }}}}} \right) = \mathop {\lim }\limits_{n \to  + \infty } \left( {n\sqrt[n]{{\frac{{\sqrt {\left( {2n} \right)\pi e} \left( {\frac{{2n}}{e}} \right)^{2n} }}{{\sqrt {\left( {3n} \right)\pi e} \left( {\frac{{3n}}{e}} \right)^{3n} }}}}} \right) \\ 
  \Rightarrow \mathop {\lim }\limits_{n \to  + \infty } \left( {n\sqrt[n]{{\frac{{\left( {2n} \right)!}}{{\left( {n!} \right)^3 }}}}} \right) = \mathop {\lim }\limits_{n \to  + \infty } \left( {\frac{4}{{27}}e\sqrt[n]{{\sqrt {\frac{2}{3}} }}} \right) = \frac{4}{{27}}e\quad  \\ 
 ;\mathop {\lim }\limits_{n \to  + \infty } \left( {\sqrt[n]{{\sqrt {\frac{2}{3}} }}} \right) = \mathop {\lim }\limits_{n \to  + \infty } \left( {\frac{2}{3}} \right)^{\frac{1}{{2n}}}  = 1 \\ 
 \end{array}
$$
