nth root of $S_n=\frac{n^n}{(n+1)(n+2)...(n+n)}$ Let $S_n=\frac{n^n}{(n+1)(n+2)...(n+n)}, n\geq 1$, then $S_n^{\frac{1}{n}}$ converges to


*

*$e/2$

*$e/4$

*$e/8$

*$0$
Clearly $S_n^{\frac{1}{n}}=\frac{n}{(n+1)^{\frac{1}{n}}(n+2)^{\frac{1}{n}}...(n+n)^{\frac{1}{n}}}$, then how can we proceed
 A: We have $\frac{S_{n+1}}{S_n}=(1+\frac{1}{n})^n \frac{(n+1)^2}{(2n+1)(2n+2)} \to e/4.$
Hence $\lim S_n^{1/n}= \lim \frac{S_{n+1}}{S_n}=e/4.$
A: You can rewrite $S_n$ as $S_n = \frac{n^n \cdot n!}{(2n)!}$ and then use Stirling's approximation: 
$$n!\sim\sqrt{2\pi n} (\frac{n}{e})^n$$
$$(2n)! \sim \sqrt{4\pi n} (\frac{2n}{e})^{2n}$$
Substituting these you get that: 
$$S_n \sim \frac{n^n \cdot \sqrt{2\pi n} (\frac{n}{e})^n}{\sqrt{4\pi n} (\frac{2n}{e})^{2n}} = \frac{e^n}{\sqrt2 \cdot 2^{2n}}$$ 
Taking the $n^{th}$ root, you get that $$S_n^{\frac{1}{n}} \rightarrow \frac{e}{4}$$
A: We have $$(n+1)^n \lt (n+1)(n+2)...(n+n) \lt(2n)^n$$
and therefore
$$\frac12=\frac{n}{2n}\le S_n^{\frac{1}{n}}\le \frac{n}{n+1}\le 1$$
and further
$$\frac12 \le \lim\limits_{n\to 1}S_n^{\frac{1}{n}}\le 1$$
if that limit exists.
Only one of the proposed solution satisfies these inequalities. From $e=2.7\ldots$ follows $2.7<e<2.8$  and therefore


*

*$\frac{2.7}2 < \frac e 2 < \frac{2.8}{2}\implies 1.35 <\frac e2<1.4$ 

*$\frac{2.7}4 < \frac e 4 < \frac{2.8}{4}\implies 0.675 <\frac e4<0.7$ 

*$\frac{2.7}8 < \frac e 8 < \frac{2.8}{8}\implies 0.3375 <\frac e8<0.35$ 

*$0$
So if this limit exists and if it is one of these four numbers then it must be $\frac e 4$.
A: Another approach:
$$\ln(S_n^{1/n})=-\frac1n\sum_{k=1}^n\ln\left(1+\frac kn\right)$$
which is a Riemann sum for the integral
$$-\int_0^1\ln(1+x)\,dx.$$
So
$$\lim_{n\to\infty}S_n^{1/n}=\exp\left(-\int_0^1\ln(1+x)\,dx\right)$$
so you just have to evaluate this integral.
