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

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.

• partial fractions Aug 13, 2015 at 11:49
• This is my problem. I have done the partial fractions. But what happens then? Even with partial fractions we have "i" terms in the denominators. So it becomes like a harmonic series. How am i supposed to evaluate the sum of these? What am i missing? Aug 13, 2015 at 11:51
• try telescoping them Aug 13, 2015 at 11:54
• Hmm i have to look some things up. Aug 13, 2015 at 11:56

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

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

\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}

• I solved the problem the long "classic" way. I saw the terms that were cancelling out etc. and got this result. A shorthand way such as your would be great, but i don't understand what you did at first. I'll have to see it more later. Aug 13, 2015 at 15:15
• In general if the summand is of the form $P/[i(i+1)(i+2)...(i+m)]$ it can be expressed as the difference between two consecutive terms where the denominator is one degree less than that of the summand i.e. $Q/[(i+1)(i+2)...(i+m)]−Q/[i(i+1)(i+2)...(i+m−1)]$. As there is an $i$ in the numerator, we add that accordingly and assign an arbitrary coefficient to be determined. Aug 13, 2015 at 16:04