Finding limit of a product. Prove:$$\lim_{n \to\infty  }\frac{1}{n}\left[\prod_{i=1}^{n}(n+i)   \right ]^{\frac{1}{n}}=\frac{4}{e}$$
I tried using Squeeze Theorem but can't go beyond $1<L<2$. $$\lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^n=e$$might be useful too.
 A: Wo consider the log. of $\frac{1}{n}\left[\prod_{i=1}^{n}(n+i)   \right ]^{\frac{1}{n}}$. If you take logarithm then you get
$$-\log n+\frac{1}{n}\sum_{k=1}^\infty \log(n+k)=-\log n +\frac{1}{n}\log \frac{(2n)!}{n!}$$
By Stirling's formula,
$$
\begin{aligned}
-\log n +\frac{1}{n}\log \frac{(2n)!}{n!}&\sim -\log n +\frac{1}{n} \log \frac{\sqrt{4\pi n}2^{2n} n^{2n} e^{-2n}}{\sqrt{2\pi n}n^n e^{-n}} \\
&= -\log n +\frac{1}{n} \log \left( \sqrt{2}\cdot 4^n n^ne^{-n}\right)\\
&= -\log n +\frac{1}{n}\left( \log\sqrt{2} +n\log 4+n\log n-n\right)\\
&= \frac{1}{n}\log \sqrt{2} +\log 4 -1
\end{aligned}$$
So $\lim_{n\to\infty}-\log n+\frac{1}{n}\sum_{k=1}^\infty \log(n+k)=\log 4 -1$. It follows desired formula.
A: $\frac{1}{n}\left[\prod_{i=1}^n(n+i)\right]^{1/n}=\left[\prod_{i=1}^n \frac{1}{n}(n+i)\right]^{1/n}=\left[\prod_{i=1}^n (1+\frac{i}{n})\right]^{1/n}$
Taking log we get
$\frac{1}{n}\sum_{i=1}^n\ln (1+\frac{i}{n}) \to \int_0^1 \ln(1+x)dx,  n \to \infty$
Integrating by parts gives $\int_0^1 \ln(1+x)dx=\ln 4 -1.$
Now the limit of the product is $e^{\ln 4 - 1}$.
A: Note that  $$\ln\left\{\frac 1n\left[\prod_{i=1}^{n}(n+i)   \right ]^{\frac{1}{n}}\right\}=\ln(1/n)+(1/n)S$$
where 
$$S=\sum_{k=n+1}^{2n+1}\ln k.$$
Here, you can use 
$$\int_{n}^{2n}\ln xdx\lt S\lt \int_{n+1}^{2n+1}\ln xdx$$
$$\iff f(2n)-f(n)\lt S\lt f(2n+1)-f(n+1)$$
where $$f(n)=n(\ln(n)-1).$$
Now you'll see
$$\lim_{n\to\infty}\left(\ln\frac 1n+\frac 1n(f(2n)-f(n))\right)=-1+\ln 4,$$
$$\lim_{n\to\infty}\left(\ln\frac 1n+\frac 1n(f(2n+1)-f(n+1))\right)=-1+\ln 4.$$
Hence, by Squeeze Theorem, you'll get the answer you wrote.
