Calculus by Spivak, Chapter 2, Problem 6, Part 3

[Nafi Khandaker]

In this problem, Spivak shows how to derive formulas to summations. They start by showing the method for
1^2 + 2^2 + ... + n^2 as follows:

(k + 1)^3 - k^3 = 3k^2 + 3k + 1

Writing this formula for k = 1, 2, ..., n and adding, we obtain

2^3 - 1^3 = 3*1^2 + 3*1 + 1
3^3 - 2^3 = 3*2^2 + 3*2 + 1
.
.
.
(n + 1)^3 - n^3 = 3*n^2 + 3*n + 1
(n + 1)^3 - 1 = 3[1^2 + ... + n^2] + 3[1 + ... + n] + n.
​

I understood the method and used it for 1^3 + ... + n^3 and 1^4 + ... + n^4, but I am completely lost as to how to apply this to 1/(1*2) + 1/(2*3) + ... 1/(n (n + 1)). The book's answer doesn't help much:
1/k - 1/(k + 1) = 1/(k (k + 1).

Any ideas as to how to derive this result using the method previously given?

 

[tommyxu3]

The best way I came up with to solve your problem is also the way you states:
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...+\frac{1}{n(n+1)}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{1}{1}-\frac{1}{n+1}.$$
The way you mentioned to solve the previous problems is great, but seems not feasible on this case.

 

[Nafi Khandaker]

The best way I came up with to solve your problem is also the way you states:
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...+\frac{1}{n(n+1)}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{1}{1}-\frac{1}{n+1}.$$
The way you mentioned to solve the previous problems is great, but seems not feasible on this case.

Thanks for replying. I asked someone else who has more math experience than me and he said something similar.