# find the sum to n term of $\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ...$

$$\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ...$$

$$=\sum \limits_{k=1}^{n} \frac{2k-1}{k\cdot(k+1)\cdot(k+2)}$$ $$= \sum \limits_{k=1}^{n} - \frac{1}{2}\cdot k + \sum \limits_{k=1}^{n} \frac{3}{k+1} - \sum \limits_{k=1}^{n}\frac{5}{2\cdot(k+2)}$$

I do not know how to get a telescoping series from here to cancel terms.

• You can use $$\frac{2k-1}{k(k+1)(k+2)}=\frac{2k}{k(k+1)(k+2)}-\frac1{k(k+1)(k+2)}.$$ Commented Oct 26, 2018 at 18:11
• You do mean an infinite sum, not a finite sum, right? Commented Oct 26, 2018 at 18:15
• @ConnorHarris I think this means writing the partial sums explicitly. Commented Oct 26, 2018 at 18:16
• @ConnorHarris, a sum that can be expressed in n. Commented Oct 26, 2018 at 18:17
• Note that the $\frac{3}{k+1}$ term for some value of $k$ cancels the $\frac{-1}{2k}$ and $\frac{-5}{2(k+2)}$ terms for adjacent values of $k$. Commented Oct 26, 2018 at 18:18

HINT:

Note that we have

\begin{align} \frac{2k-1}{k(k+1)(k+2)}&=\color{blue}{\frac{3}{k+1}}-\frac{5/2}{k+2}-\frac{1/2}{k}\\\\ &=\color{blue}{\frac12}\left(\color{blue}{\frac{1}{k+1}}-\frac1k\right)+\color{blue}{\frac52}\left(\color{blue}{\frac{1}{k+1}}-\frac{1}{k+2}\right) \end{align}

• There's a typo after the first $=$ sign, it should be $\frac{5/2}{k+2}$. Can't edit though, since edits must be 6 characters at minimum. Commented Oct 27, 2018 at 12:50
• @a_guest Thank you. I've edited accordingly. Commented Oct 27, 2018 at 16:29

Let the fractions be $$\frac{a}{k}$$, $$\frac{b}{k+1}$$, and $$\frac{c}{k+2}$$.

$$\frac{a}{k}+\frac{b}{k+1}+\frac{c}{k+2}=\frac{a(k+1)(k+2)+bk(k+2)+ck(k+1)}{k(k+1)(k+2)}=\frac{2k-1}{k(k+1)(k+2)}$$

We want the following

$$a+b+c=0$$

$$3a+2b+c=2$$

$$2a=-1$$

Solve, $$a=-\frac{1}{2}$$, $$b=3$$, and $$c=-\frac{5}{2}$$.

The rest is standard.

You are almost there. You can merge the parts of the series for which the denominator is similar and you will see they cancel each other. Then you are left with the terms for which the denominator is either smaller than $$3$$ or greater than $$n$$.

\begin{aligned} & \sum_{k=1}^n\frac{-1}{2k} + \sum_{k=1}^n\frac{3}{k+1} - \sum_{k=1}^n\frac{5}{2}\frac{1}{k+2} \\ & = \left[-\frac{1}{2} - \frac{1}{4} + \frac{1}{2}\sum_{k=3}^n\frac{-1}{k}\right] + \left[\frac{3}{2} + \frac{1}{2}\sum_{k=2}^n\frac{6}{k+1}\right] - \left[\frac{1}{2}\sum_{k=1}^n\frac{5}{k+2}\right] \\ & = \frac{3}{4} + \left[\frac{1}{2}\sum_{k=3}^n\frac{-1}{k}\right] + \left[\frac{1}{2}\sum_{k=3}^{n+1}\frac{6}{k}\right] - \left[\frac{1}{2}\sum_{k=3}^{n+2}\frac{5}{k}\right] \\ & = \frac{3}{4} + \left[\frac{1}{2}\sum_{k=3}^n\frac{-1}{k}\right] + \left[\frac{1}{2}\sum_{k=3}^{n}\frac{6}{k} + \frac{6}{2}\frac{1}{n+1}\right] - \left[\frac{1}{2}\sum_{k=3}^{n}\frac{5}{k} + \frac{5}{2}\frac{1}{n+1} + \frac{5}{2}\frac{1}{n+2}\right] \\ & = \frac{3}{4} + \frac{1}{2}\sum_{k=3}^{n}\left[\frac{-1 + 6 - 5}{k}\right] + \frac{6}{2}\frac{1}{n+1} - \frac{5}{2}\frac{1}{n+1} - \frac{5}{2}\frac{1}{n+2} \\ & = \frac{3}{4} + \frac{1}{2(n+1)} - \frac{5}{2(n+2)} \end{aligned}

When terms are in A.P in denominator, we use difference of last and first terms in product to distribute the terms of denominator over numerator, to change into telescoping series.