<s>[INST] How to use telescoping series to find: $\sum_{r=1}^{n}\frac{1}{r+2}$ I am a bit confused in this one, how do i do the required modification in this case?
 [/INST] As stated in the comments, $\sum_{r = 1}^{n} \frac{1}{r + 2}$ does not telescope. As for an example of a sum that does telescope, consider $\sum_{r = 1}^{n} \frac{1}{r(r + 1)}$. We can rewrite this summation as
$$\sum_{r = 1}^{n} \frac{1}{r(r + 1)} = \sum_{r = 1}^{n} \left(\frac{1}{r} - \frac{1}{r + 1}\right) = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} -  \frac{1}{n + 1}\right),$$
and we can pair off terms to see that the entire sum ends up being $1 - \frac{1}{n + 1} = \frac{n}{n + 1}$.
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