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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?

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HINT

$$ \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} $$ 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?

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  • $\begingroup$ Thanks. I've used this to simplify it, but I seem to have made a mistake? It doesn't seem to match up with the final line. $\endgroup$ – Penguin47 Apr 10 at 16:19
  • $\begingroup$ @Penguin47 please see the update $\endgroup$ – gt6989b Apr 10 at 16:36
  • $\begingroup$ Thank you. I see my mistake now. $\endgroup$ – Penguin47 Apr 10 at 16:37

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