Find the value of $a \in R$ such that $\langle x_n \rangle$ converges to a positive real number when $x_n=\frac{1}{3}\frac{4}{6}...\frac{3n-2}{3n}n^a$ Find the value of $a \in \mathbb{R}$ such that $\langle x_n \rangle$ converges to a positive real number when $x_n=\dfrac{1}{3}\dfrac{4}{6}\cdots\dfrac{3n-2}{3n}n^a$
Here is my approach. First of all, let $a_n=\dfrac{x_n}{n^a}=\dfrac{1}{3}\dfrac{4}{6}\cdots\dfrac{3n-2}{3n}$
Then, let $b_n=\dfrac{2}{4}\dfrac{5}{7}\cdots\dfrac{3n-1}{3n+1}$ and $c_n=\dfrac{3}{5}\dfrac{6}{8}\cdots\dfrac{3n}{3n+2}$
Since $0<a_n<b_n$ and $0<a_n<c_n$
$0<a_n^3<a_nb_nc_n=\dfrac{1}{3}\dfrac{2}{4}\dfrac{3}{5}\cdots\dfrac{3n-2}{3n}\dfrac{3n-1}{3n+1}\dfrac{3n}{3n+2}=\dfrac{2}{(3n+1)(3n+2)}$
Therefore, 
$0<x_n^3<\dfrac{2n^{3a}}{(3n+1)(3n+2)}$
So, since the limit of $x_n^3$ should be greater than 0 and less than some positive real number,
the limit of $\dfrac{2n^{3a}}{(3n+1)(3n+2)}$ should be neither 0 nor infinity. Therefore, 3a=2, a=2/3.
Anything wrong? Or better idea?

Expand $$\log(3n-2) - \log(3n) = \log(1 - \frac23 \frac1n) = - \frac23 \frac1n + \frac49 \epsilon_n \frac1{n^2}$$
where $\epsilon_n$ is a bounded sequence.
Then :
$$\log x_n = -\frac23 \sum_{k = 1}^n \frac1k + a \log n + \frac49 \sum_{k = 1}^n \epsilon_k \frac1{k^2}$$
Le last sum converges : $$\big|\epsilon_k \frac1{k^2} \big| \leq (\max_m |\epsilon_m|) \times \frac1{k^2}$$
The harmonic serie satisfies :
$$\sum_{k = 1}^n \frac1k  = \log n - \gamma + \epsilon^1_n$$
where $\epsilon^1_n$ converges to $0$.
(to prove that write $\log n = \sum_{k = 1}^n \log(k+1) - \log k = \sum_{k = 1}^n \log(1+\frac1k)$ and expand)
Thus, $\log x_n$ converges to a finite limite iff :
$$-\frac23 \sum_{k = 1}^n \frac1k + a \log n = (a - \frac23) \log n + \frac23 \gamma + \frac23 \epsilon^1_n$$
converges to a finite limit. Finally $a = 2/3$, as you expected.
