Find $\lim \limits_{n \to \infty}{1*4*7*\dots(3n+1) \over 2*5*8* \dots (3n+2)}$ I am trying to find
$$\lim \limits_{n \to \infty}{1*4*7*\dots(3n+1) \over 2*5*8* \dots (3n+2)}$$
My first guess is to look at the reciprocal and isolate factors:
$${2 \over 1}{5 \over 4}{8 \over 7} \dots {3n+2 \over 3n+1}= {\left(1+1\right)}\left(1+{1 \over 4}\right)\left(1+{1 \over 7}\right) \dots \left(1+{1 \over 3n+1}\right)$$
Now we take the natural log:
$$\ln{ + \ln\left(1+1\right)} + \ln\left(1+{1 \over 4}\right) +\ln\left(1+{1 \over 7}\right) \dots +\ln\left(1+{1 \over 3n+1}\right)$$
Using $\ln(1+x) \le x$, we get:
$$\ln{ + \ln\left(1+1\right)} + \ln\left(1+{1 \over 4}\right) +\ln\left(1+{1 \over 7}\right) \dots +\ln\left(1+{1 \over 3n+1}\right) \le 1 + {1 \over 4} + {1 \over 7} + \dots + {1 \over 3n+1}$$
Now, I'm stuck. I suppose I might use the fact that the RHS is similar to the harmonic series and show that it converges to some log, but I'm not sure how to do that. Do you have any clues?

Taking the logarithm and using $\ln(1+x) \le x$ is a good idea. It becomes simpler if you  apply it to the original product and not to the reciprocal:
$$
 \log a_n = \sum_{k=0}^n\ln\Bigl(1-\frac1{3k+2}\Bigr)
\le - \sum_{k=0}^n\frac{1}{3k+2} < -\frac{1}{3}\sum_{k=0}^n\frac{1}{k+1}
$$
and the right-hand side diverges to $-\infty$, so that $a_n \to 0$.
