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.

  • 3
    $\begingroup$ 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. $\endgroup$ Jun 26, 2019 at 1:38
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    $\begingroup$ 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! $\endgroup$
    – user600016
    Jun 26, 2019 at 1:39
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    $\begingroup$ It looks OK to me as you are using the same method. $\endgroup$
    – ArsenBerk
    Jun 26, 2019 at 1:40

1 Answer 1


It is a perfectly fine answer. The above method which is being illustrated is known as Telescopic method of summation. You can read more about it here.




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