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?

You are on the right track. Use the inequality $\ln(1+x)\ge x-x^2/2$ for $x\ge0$:
$$
\sum_{k=0}^n\ln\Bigl(1+\frac1{3\,k+1}\Bigr)\ge\sum_{k=1}^n\Bigl(\frac{1}{3\,k+1}-\frac12\,\frac{1}{(3\,k+1)^2}\Bigr)=\sum_{k=1}^n\frac{1}{3\,k+1}-\frac12\sum_{k=1}^n\frac{1}{(3\,k+1)^2}.
$$
The first sum goes to $+\infty$ because the series $\sum_{k=0}^\infty1/(3\,k+1)$ is divergent, and the second sum is bounded because the series $\sum_{k=0}^\infty1/(3\,k+1)^2$ is convergent. This means that the original limit is equal to $0$.
