# Unsure how the following summation simplifies down to this known result?

How does this:

$$\frac{1}{c+1} + (c-1)\cdot\left(\frac{1}{c(c+2)}+\dotsb+\frac{1}{(n-2)n}\right) + \left(\frac{c-1}{n-1}\right)\cdot\frac{1}{2}$$

Become this:

$$= \frac{2cn-c^2+c-n}{cn}$$

PS. $$1 \leq c \leq n$$. Not sure if that's needed or not.

Edit:

Am I right with:

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

When I sub numbers in for $$c$$ and $$n$$, this does not seem to give me the above final line?

$$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$$ and the entire bracket is a telescoping series
UPDATE Your case is $$\frac{1}{k(k+2)} = \frac{1}{2} \left[\frac{1}{k} - \frac{1}{k+2}\right]$$
After you apply this, you get $$\begin{split} S &= \sum_{k=c}^{n-2}\frac{1}{k(k+2)} = \frac12 \sum_{k=c}^{n-2}\left[\frac{1}{k} - \frac{1}{k+2}\right]\\ &= \frac12 \left[ \sum_{k=c}^{n-2} \frac{1}{k} - \sum_{k=c+2}^n \frac{1}{k} \right] \\ &= \frac12 \left[ \frac1c + \frac{1}{c+1} -\frac1{n-1} - \frac{1}n\right] \end{split}$$ Hence $$\begin{split} V &= \frac{1}{c+1} + (c-1)\sum_{k=c}^{n-2}\frac{1}{k(k+2)} + \left(\frac{c-1}{n-1}\right)\cdot\frac{1}{2} \\ &= \frac{1}{c+1} + \frac{c-1}{2} \left[ \frac1c + \frac{1}{c+1} -\frac1{n-1} - \frac{1}n\right] + \left(\frac{c-1}{n-1}\right)\cdot\frac{1}{2} \\ &= \frac{1}{c+1} + \frac{c-1}{2} \left[ \frac1c + \frac{1}{c+1} - \frac{1}n\right]\\ \end{split}$$ Could you now complete this?