$\lim_{n\to\infty}\left( \frac1{4\cdot 7}+\frac1{7\cdot 10}+\ldots+\frac1{(3n+1)(3n+4)} \right) $ I am having trouble finding the infinite sum
$$
\lim_{n\to\infty}\left( \frac1{4\cdot 7}+\frac1{7\cdot 10}+\ldots+\frac1{(3n+1)(3n+4)} \right).
$$
I know that
$$
\lim_{n\to\infty}\frac1{(3n+1)(3n+4)} =0,
$$
but I have no ideas for a further solution.

$$S_n=\sum_{k=1}^{n} \frac{1}{(3k+1)(3k+4)}=\frac{1}{3}\sum_{k=1}^{n} \left(\frac{1}{3k+1}- \frac{1}{3k+3}\right)= \frac{1}{3} \sum_{k=0}^{n} \int_{1}^{1}[t^{3k}- t^{3k+3}] dt.$$
Use IGP: $$\sum_{k=1}^n r^k=\frac{r}{1-r}, |r|<1$$
$$S_{\infty}= \frac{1}{3} \int_{0}^{1} t^3\frac{1-t^3}{1-t^3} dt=\frac{1}{3}\int_{0}^{1} t^3 dt=\frac{1}{12}.$$
