Show that $\lim_{n\to\infty}\left(\prod_{k=1}^n\frac{2k+1}{3k-2}\right)^{\frac{1}{n}}=\frac{2}{3}$ How can I show 
$$\lim_{n\to\infty}\left(\prod\limits_{k=1}^n\frac{2k+1}{3k-2}\right)^{\frac{1}{n}}=\frac{2}{3}$$
with the aid of Gamma function？

We have 
$$ \left( \prod_{k=1}^{n} \frac{2k+1}{3k-2} \right)^{\frac{1}{n}}
= \frac{2}{3} \left( \prod_{k=1}^{n} \left( 1 + \frac{7}{6k-4} \right) \right)^{\frac{1}{n}}. $$
Therefore the claim follows from
$$ 1
\leq \left( \prod_{k=1}^{n} \left( 1 + \frac{7}{6k-4} \right) \right)^{\frac{1}{n}}
\leq \exp\left(\frac{1}{n}\sum_{k=1}^{n} \frac{7}{6k-4} \right)
\xrightarrow[n\to\infty]{}1. $$
Here we utilized the inequality $1+x \leq e^{x}$.
