Question

Sum the following series to n terms and to infinity $\frac{1}{\mathrm{1.2.3}}+\frac{3}{\mathrm{2.3.4}}+\frac{5}{\mathrm{3.4.5}}+\frac{7}{\mathrm{4.5.6}}+....$.

Answer 1

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

We can write

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

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

$=2A_n-B_n$

where $A_n=\frac{1}{(n+1)(n+2)}$ and $B_n=\frac{1}{n(n+1)(n+2)}$

By partial fraction

$A_n=[\frac{1}{(n+1)}-\frac{1}{(n+2)}]$

Then, $\sum _{n=1}^n{A}_n=\sum _{n=1}^n[\frac{1}{n+1}-\frac{1}{n+2}]$

$=[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}.....\frac{1}{n+1}-\frac{1}{n+2}]$

$=[\frac{1}{2}-\frac{1}{n+2}]$

Similarly for $B_n$,

$B_n=\frac{1}{2}[\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2}]$

$=\frac{1}{2}\{[\frac{1}{n}-\frac{1}{n+1}]-[\frac{1}{n+1}-\frac{1}{n+2}]\}$

So,

$\sum _{n+1}^n{B}_n=\frac{1}{2}\{\sum _{n=1}^n[\frac{1}{n}-\frac{1}{n+1}]-\sum _{n=1}^n[\frac{1}{n+1}-\frac{1}{n+2}]\}$

$=\frac{1}{2}[\frac{1}{2}-\frac{1}{n+1}+\frac{1}{n+2}]$

Combining $\sum _{n=1}^n(2A_n-B_n)$, we get

$S_n=2[\frac{1}{2}-\frac{1}{n+2}]-\frac{1}{2}[\frac{1}{2}-\frac{1}{n+1}+\frac{1}{n+2}]$

Let $n\to \infty$

$S_{\infty }=2[\frac{1}{2}-0]-\frac{1}{2}[\frac{1}{2}-0-0]$

$=1-\frac{1}{4}$

$=\frac{3}{4}$