Question

Sum the following series to $n$ terms and to infinity $\frac{1}{\mathrm{3.4.5}}+\frac{2}{\mathrm{4.5.6}}+\frac{3}{\mathrm{5.6.7}}+...$.

Answer 1

$S_n=\frac{1}{3\cdot 4\cdot 5}+\frac{2}{4\cdot 5\cdot 6}+\frac{3}{5\cdot 6\cdot 7}+\dots \dots +\frac{n}{(n+2)(n+3)(n+4)}$

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

$S_1=\sum _{n=1}^n\frac{1}{(n+3)(n+4)},S_2=\sum _{n=1}^n\frac{2}{(n+2)(n+3)(n+4)}$

$S_1=\frac{1}{4\cdot 5}+\frac{1}{5\cdot 6}+\frac{1}{6\cdot 7}+\dots \dots +\frac{1}{(n+3)(n+4)}$

$S_1=\frac{5-4}{4\cdot 5}+\frac{6-5}{5\cdot 6}+\frac{7-6}{6\cdot 7}+\dots \dots +\frac{(n+4)-(n+3)}{(n+3)(n+4)}$

$S_1=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\dots \dots +\frac{1}{n+3}-\frac{1}{n+4}$

$S_1=\frac{1}{4}-\frac{1}{n+4}=\frac{n}{4(n+4)}$

$S_2=\frac{2}{3\cdot 4\cdot 5}+\frac{2}{4\cdot 5\cdot 6}+\frac{2}{5\cdot 6\cdot 7}+\dots \dots +\frac{2}{(n+2)(n+3)(n+4)}$

$S_2=\frac{5-3}{3\cdot 4\cdot 5}+\frac{6-4}{4\cdot 5\cdot 6}+\frac{7-5}{5\cdot 6\cdot 7}+\dots \dots +\frac{(n+4)-(n+2)}{(n+2)(n+3)(n+4)}$

$S_2=\frac{1}{3\cdot 4}-\frac{1}{4\cdot 5}+\frac{1}{4\cdot 5}-\frac{1}{5\cdot 6}+\frac{1}{5\cdot 6}-\frac{1}{6\cdot 7}+\dots \dots +\frac{1}{(n+2)(n+3)}-\frac{1}{(n+3)(n+4)}$

$S_2=\frac{1}{3\cdot 4}-\frac{1}{(n+3)(n+4)}=\frac{n}{4(n+4)}$

$S_n=S_1-S_2=\frac{1}{4}-\frac{1}{n+4}-\frac{1}{3\cdot 4}+\frac{1}{(n+3)(n+4)}=\frac{1}{6}-\frac{1}{n+3}+\frac{2}{(n+3)(n+4)}$

$S_{\infty }=\underset{n\to \infty }{lim}{S}_n=\frac{1}{6}$