Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I find the formula for the following equation?

$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\ldots +\frac{1}{n(n+1)}$$

More importantly, how would you approach finding the formula? I have found that every time, the denominator number seems to go up by n+2, but that's about as far as I have been able to get:

$1/2 + 1/6 + 1/12 + 1/20 + 1/30...$ the denominator increases by $4,6,8,10,12...$ etc.

So how should I approach finding the formula? Thanks!

share|cite|improve this question
up vote 8 down vote accepted

If you simplify your partial sums, you get $\frac12,\frac23,\frac34,\frac45,....$ Does this give you any ideas?

share|cite|improve this answer
Thanks! This was very helpful :) Now I just gotta prove it by induction! – Charles Jan 24 '13 at 18:31

Hint: Use the fact that $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and find $S_n=\sum_1^n\left(\frac{1}{k}-\frac{1}{k+1}\right)$.

share|cite|improve this answer
Simple and nice (+1) – user 1618033 Jan 24 '13 at 18:59
@Chris'ssister: Thanks a lot for your consideration. – Babak S. Jan 24 '13 at 19:03
Nicely said, Babak(+1)! – amWhy Jan 24 '13 at 19:21

While exploitation of the resulting telescoping series after partial fraction expansion is a very simple way forward, I thought it might be instructive to present another way forward. Here, we write

$$\begin{align} \sum_{k=1}^N \frac{1}{k(k+1)}&=\sum_{k=1}^\infty \int_0^1y^{k-1}\,dy \int_0^1 x^k\,dx\\\\ &=\int_0^1\int_0^1x\sum_{k=1}^N (xy)^{k-1}\,dx\\\\ &=\int_0^1\int_0^1 x\frac{1-(xy)^N}{1-xy}\,dx\,dy\\\\ &=\int_0^1\int_0^x \frac{1-y^N}{1-y}\,dy\,dx\\\\ &=\int_0^1\int_y^1\frac{1-y^N}{1-y}\,dx\,dy\\\\ &=\int_0^1(1-y^N)\,dy\\\\ &=1-\frac1{N+1} \end{align}$$

as expected!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.