Find Sum of n terms of the Series: $ \frac{1}{1\times2\times3} + \frac{3}{2\times3\times4} + \frac{5}{3\times4\times5} + \cdots $

Question

I want to find the sum to n terms of the following series. $$ \frac{1}{1\times2\times3} + \frac{3}{2\times3\times4} + \frac{5}{3\times4\times5} + \frac{7}{4\times5\times6} + \cdots $$ I have found that the general term of the sum is \begin{equation} \sum\limits_{i=1}^n \frac{2i - 1}{i(i+1)(i+2)} \end{equation} so the nth term must be \begin{equation} \frac{2n - 1}{n(n+1)(n+2)} \end{equation} Now.. how should i evaluate the sum of the n terms? What is the method i must use in order to do so? Thank you in advance.

Answer 1

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

Now $$\displaystyle A = \frac{2}{(n+1)(n+2)} = 2\left[\frac{(n+2)-(n+1)}{(n+1)(n+2)}\right] = 2\left[\frac{1}{n+1}-\frac{1}{n+2}\right]$$

And $$\displaystyle B = \frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left[\frac{(n+2)-(n)}{n(n+1)(n+2)}\right] = \frac{1}{2}\left[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right]$$

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

Now $$\displaystyle \sum_{r=1}^{n}S_{n} = 2\sum_{r=1}^{n}\left[\frac{1}{r+1}-\frac{1}{r+2}\right]-\frac{1}{2}\sum_{r=1}^{n}\left[\frac{1}{r}-\frac{1}{r+1}\right]+\frac{1}{2}\sum_{r=1}^{n}\left[\frac{1}{r+1}-\frac{1}{r+2}\right]$$

Now Use Telescopic Sum

Answer 2

$$\begin{align} \sum\limits_{i=1}^n \frac{2i - 1}{i(i+1)(i+2)} &=\sum_{i=1}^n\frac {Ai+B}{i(i+1)}-\frac{A(i+1)+B}{(i+1)(i+2)}\\ &=\sum_{i=1}^n\frac{2i-\frac12}{i(i+1)}-\frac{2(i+1)-\frac12}{(i+1)(i+2)}\\ &=\frac34-\frac{4n+3}{2(n+1)(n+2)}\qquad\text{by telescoping}\\ &=\frac{n(3n+1)}{4(n+1)(n+2)}\qquad\blacksquare \end{align}$$