Question

Sum the following series to n terms and to infinity $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...$.

Answer 1

$S_n=\frac{1}{1\cdot 4}+\frac{1}{4\cdot 7}+\frac{1}{7\cdot 10}+\dots \dots +\frac{1}{(3n-2)(3n+1)}$

$S_n=\frac{1}{3}[\frac{4-1}{1\cdot 4}+\frac{7-4}{4\cdot 7}+\frac{10-7}{7\cdot 10}+\dots \dots +\frac{(3n+1)-(3n-2)}{(3n-2)(3n+1)}]$

$S_n=\frac{1}{3}[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\dots \dots +\frac{1}{3n-2}-\frac{1}{3n+1}]$

$S_n=\frac{1}{3}-\frac{1}{3(3n+1)}=\frac{n}{3n+1}$ and $S_{\infty }=\frac{1}{3}$