The sum of infinite terms of the series  is 

  $$\begin{gathered} {\mathrm{n}}^{\text{th }}\text{ term }{\mathrm{T}}_{\mathrm{n}}=\frac{1}{(3\mathrm{n}-2)(3\mathrm{n}+1)(3\mathrm{n}+4)} \\ \begin{array}{r}{\mathrm{T}}_{\mathrm{n}}=\frac{1}{6}\left[\frac{1}{(3\mathrm{n}-2)(3\mathrm{n}+1)}-\frac{1}{(3\mathrm{n}+1)(3\mathrm{n}+4)}\right]\\ {\mathrm{T}}_1=\frac{1}{6}\left[\frac{1}{1.4}-\frac{1}{4.7}\right]\\ {\mathrm{T}}_2=\frac{1}{6}\left[\frac{1}{4.7}-\frac{1}{7.10}\right]\\ {\mathrm{T}}_{\mathrm{n}}=\frac{1}{6}\left[\frac{1}{(3\mathrm{n}-2)(3\mathrm{n}+1)}-\frac{1}{(3\mathrm{n}+1)(3\mathrm{n}+4)}\right]\\ \text{ Then, }{\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{n}=1}^{\mathrm{n}} {\mathrm{T}}_{\mathrm{n}}=\frac{1}{6}\left[\frac{1}{1.4}-\frac{1}{(3n+1)(3\mathrm{n}+4)}\right]\\ =\frac{1}{24}-\frac{1}{6(3\mathrm{n}+1)(3\mathrm{n}+4)}\end{array} \\ \underset{\mathrm{n}\to \infty }{\lim }{\mathrm{S}}_{\mathrm{n}}=\frac{1}{6}\left[\frac{1}{4}-0\right]=\frac{1}{24} \end{gathered}$$