Sum of given infinite series

Notice that

$$\frac k{\prod_{m=1}^k(3m+1)}=\frac1{3\prod_{m = 1}^{k-1} (3m+1)}-\frac{1}{3\prod_{m = 1}^k (3m+1)}$$

Which gives us a telescoping series:$$S_N=\frac{1}{3} - \frac{1}{3\prod_{m = 1}^N (3m+1)}$$

which tends to $1/3$ as suspected.