Question

Sum the following series to n terms and to infinity $\frac{1}{\mathrm{1.4.7}}+\frac{1}{\mathrm{4.7.10}}+\frac{1}{\mathrm{7.10.13}}+...$.

Answer 1

$n^{th}$ term $T_n=\frac{1}{(3n-2)(3n+1)(3n+4)}$

By partial fraction we get

$T_n=\frac{1}{6}[\frac{1}{(3n-2)(3n+1)}-\frac{1}{(3n+1)(3n+4)}]$

$T_1=\frac{1}{6}[\frac{1}{1.4}-\frac{1}{4.7}]$

$T_2=\frac{1}{6}[\frac{1}{4.7}-\frac{1}{7.10}]$

$T_n=\frac{1}{6}[\frac{1}{(3n-2)(3n+1)}-\frac{1}{(3n+1)(3n+4)}]$

Then, $S_n=\sum _{n=1}^n{T}_n=\frac{1}{6}[\frac{1}{1.4}-\frac{1}{(3n+1)(3n+4)}]$

$=\frac{1}{24}-\frac{1}{6(3n+1)(3n+4)}$

As $n\to \infty$

$S_n=\frac{1}{24}-0$

$=\frac{1}{24}$