# How to I find a formula for $\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{n(n+1)}$ using the given method?

The following is problem $$6$$ $$(iii)$$ from chapter $$2$$ of Spivak's Calculus:

The formula for $$1^{2} + \cdots + n^{2}$$ may be derived as follows. We begin with the formula $$(k+1)^{3} - k^{3} = 3k^{2}+3k +1$$.

Writing this formula for $$k=1, \cdots , n$$ and adding

\begin{align*} 2^{3}-1^{3}&=3\cdot1^{2} + 3\cdot1 + 1 \\ 3^{3}-2^{3}&=3\cdot2^{2} + 3\cdot2 + 1 \\ &\vdots\\ (n+1)^{3}-n^{3}&=3\cdot n^{2} + 3\cdot n +1 \\ \end{align*}

we obtain $$(n+1)^{3}-1 = 3[1^{2} + \cdots + n^{2}] + 3[1 + \cdots + n] + n.$$

Thus we can find $$\sum_{k=1}^n k^{2}$$ if we already know $$\sum_{k=1}^n k .$$ Use this method to find

$$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{n(n+1)}$$.

So originally, I attempted $$\sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n \frac{1}{k+1}.$$ This means that we have

$$(1 + \frac{1}{2} + \cdots + \frac{1}{n}) - (\frac{1}{2} + \frac{1}{3} + \cdots \frac{1}{n}+\frac{1}{n+1})$$

which gives us $$1-\dfrac{1}{n+1}$$. Although, I'm not sure if the method I've used suffices as the method required for the problem; that is, it's supposed to look similar to the example using $$(k+1)^{3}$$ above.

• I know you need to use this method, but I much prefer your version to Spivak's. Partial fraction decomposition is nice. Another nice way to do this is by induction on n. Commented Jun 26, 2019 at 1:38
• I think it should be ok as the method that they are trying to illustrate is just the summation of difference of two terms, which is nothing but telescoping series! Commented Jun 26, 2019 at 1:39
• It looks OK to me as you are using the same method. Commented Jun 26, 2019 at 1:40