sequences and series - 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

Answer

$\frac{1}{n}\left[\prod_{i=1}^n(n+i)\right]^{1/n}=\left[\prod_{i=1}^n \frac{1}{n}(n+i)\right]^{1/n}=\left[\prod_{i=1}^n (1+\frac{i}{n})\right]^{1/n}$

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 $e^{\ln 4 - 1}$.