How to use telescoping series to find: $\sum_{r=1}^{n}\frac{1}{r+2}$

Question

I am a bit confused in this one, how do i do the required modification in this case?

Answer 1

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 $1 - \frac{1}{n + 1} = \frac{n}{n + 1}$.

Answer 2

As many people have mentioned, this sum does not telescope. A summation series telescopes only if there is addition and subtraction involved. The addition and subtraction causes terms to cancel out. For example, the summation series $\sum_{n=0}^\infty \frac1n - \frac1{n+1}$ telescopes because the subtraction allows parts of the series to cancel with parts of the series that come after. For a product series, it should have multiplication and division so that the operations cancel each other out. For example, the product $\prod_{n=0}^\infty \frac1n*({n+1})$ uses multiplication in the previous term to cancel out with division in the current term.