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 $$\frac{2k-1}{k(k+1)(k+2)}=\frac{2k}{k(k+1)(k+2)}-\frac1{k(k+1)(k+2)}.$$ – Sungjin Kim Oct 26 '18 at 18:11
• You do mean an infinite sum, not a finite sum, right? – Connor Harris Oct 26 '18 at 18:15
• @ConnorHarris I think this means writing the partial sums explicitly. – Sungjin Kim Oct 26 '18 at 18:16
• @ConnorHarris, a sum that can be expressed in n. – LetzerWille Oct 26 '18 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 $k$. – Connor Harris Oct 26 '18 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. – a_guest Oct 27 '18 at 12:50
• @a_guest Thank you. I've edited accordingly. – Mark Viola Oct 27 '18 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.